Show that finite simple group $G$ with $n$ involutions satisfies $n < |G| / 3$











up vote
4
down vote

favorite
3












I want to solve the following problem. Let $G$ be a finite simple group of even order greater than $2$. Let $n$ be the number of involutions in $G$. Show that $n < |G| / 3$.



Edit: Adding the information I do know about $G$, although not sure whether it is useful for this problem. I can see that $G$ is non-abelian since all abelian simple groups have prime order.



Also before in the text book before this exercise there is a proposition that states that if $t in G$ is an involution with $m = |C_{G} (t) | $, then $C_{G} (t) $ is a proper subgroup of $G$ and $ |G | leq ( frac{1}{2} m(m+1) ) ! $.










share|cite|improve this question
























  • Can you provide a screenshot of the statement of that theorem? Or just write down exactly what it says.
    – the_fox
    Nov 22 at 17:54












  • You should accept an answer if it addresses your question to your satisfaction.
    – the_fox
    Nov 25 at 10:54










  • Sorry for not replying earlier, I have no had time to return to this until now. The theorem is: Let $G$ be a simple group of even order greater than 2, let $t$ be any involution in $G$, and let $m = |C_{G} (t) | $. Then $C_{G} (t) < G$ and $|G| leq ( frac{1}{2} m (m+1) )! $
    – John Doe
    Nov 28 at 11:31















up vote
4
down vote

favorite
3












I want to solve the following problem. Let $G$ be a finite simple group of even order greater than $2$. Let $n$ be the number of involutions in $G$. Show that $n < |G| / 3$.



Edit: Adding the information I do know about $G$, although not sure whether it is useful for this problem. I can see that $G$ is non-abelian since all abelian simple groups have prime order.



Also before in the text book before this exercise there is a proposition that states that if $t in G$ is an involution with $m = |C_{G} (t) | $, then $C_{G} (t) $ is a proper subgroup of $G$ and $ |G | leq ( frac{1}{2} m(m+1) ) ! $.










share|cite|improve this question
























  • Can you provide a screenshot of the statement of that theorem? Or just write down exactly what it says.
    – the_fox
    Nov 22 at 17:54












  • You should accept an answer if it addresses your question to your satisfaction.
    – the_fox
    Nov 25 at 10:54










  • Sorry for not replying earlier, I have no had time to return to this until now. The theorem is: Let $G$ be a simple group of even order greater than 2, let $t$ be any involution in $G$, and let $m = |C_{G} (t) | $. Then $C_{G} (t) < G$ and $|G| leq ( frac{1}{2} m (m+1) )! $
    – John Doe
    Nov 28 at 11:31













up vote
4
down vote

favorite
3









up vote
4
down vote

favorite
3






3





I want to solve the following problem. Let $G$ be a finite simple group of even order greater than $2$. Let $n$ be the number of involutions in $G$. Show that $n < |G| / 3$.



Edit: Adding the information I do know about $G$, although not sure whether it is useful for this problem. I can see that $G$ is non-abelian since all abelian simple groups have prime order.



Also before in the text book before this exercise there is a proposition that states that if $t in G$ is an involution with $m = |C_{G} (t) | $, then $C_{G} (t) $ is a proper subgroup of $G$ and $ |G | leq ( frac{1}{2} m(m+1) ) ! $.










share|cite|improve this question















I want to solve the following problem. Let $G$ be a finite simple group of even order greater than $2$. Let $n$ be the number of involutions in $G$. Show that $n < |G| / 3$.



Edit: Adding the information I do know about $G$, although not sure whether it is useful for this problem. I can see that $G$ is non-abelian since all abelian simple groups have prime order.



Also before in the text book before this exercise there is a proposition that states that if $t in G$ is an involution with $m = |C_{G} (t) | $, then $C_{G} (t) $ is a proper subgroup of $G$ and $ |G | leq ( frac{1}{2} m(m+1) ) ! $.







group-theory finite-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 at 13:48









the_fox

2,3191430




2,3191430










asked Nov 22 at 13:03









John Doe

574




574












  • Can you provide a screenshot of the statement of that theorem? Or just write down exactly what it says.
    – the_fox
    Nov 22 at 17:54












  • You should accept an answer if it addresses your question to your satisfaction.
    – the_fox
    Nov 25 at 10:54










  • Sorry for not replying earlier, I have no had time to return to this until now. The theorem is: Let $G$ be a simple group of even order greater than 2, let $t$ be any involution in $G$, and let $m = |C_{G} (t) | $. Then $C_{G} (t) < G$ and $|G| leq ( frac{1}{2} m (m+1) )! $
    – John Doe
    Nov 28 at 11:31


















  • Can you provide a screenshot of the statement of that theorem? Or just write down exactly what it says.
    – the_fox
    Nov 22 at 17:54












  • You should accept an answer if it addresses your question to your satisfaction.
    – the_fox
    Nov 25 at 10:54










  • Sorry for not replying earlier, I have no had time to return to this until now. The theorem is: Let $G$ be a simple group of even order greater than 2, let $t$ be any involution in $G$, and let $m = |C_{G} (t) | $. Then $C_{G} (t) < G$ and $|G| leq ( frac{1}{2} m (m+1) )! $
    – John Doe
    Nov 28 at 11:31
















Can you provide a screenshot of the statement of that theorem? Or just write down exactly what it says.
– the_fox
Nov 22 at 17:54






Can you provide a screenshot of the statement of that theorem? Or just write down exactly what it says.
– the_fox
Nov 22 at 17:54














You should accept an answer if it addresses your question to your satisfaction.
– the_fox
Nov 25 at 10:54




You should accept an answer if it addresses your question to your satisfaction.
– the_fox
Nov 25 at 10:54












Sorry for not replying earlier, I have no had time to return to this until now. The theorem is: Let $G$ be a simple group of even order greater than 2, let $t$ be any involution in $G$, and let $m = |C_{G} (t) | $. Then $C_{G} (t) < G$ and $|G| leq ( frac{1}{2} m (m+1) )! $
– John Doe
Nov 28 at 11:31




Sorry for not replying earlier, I have no had time to return to this until now. The theorem is: Let $G$ be a simple group of even order greater than 2, let $t$ be any involution in $G$, and let $m = |C_{G} (t) | $. Then $C_{G} (t) < G$ and $|G| leq ( frac{1}{2} m (m+1) )! $
– John Doe
Nov 28 at 11:31










2 Answers
2






active

oldest

votes

















up vote
3
down vote



accepted










Corollary $(2I)$ in the paper "On groups of even order" by Brauer and Fowler says that if $G$ is a simple group which contains $n$ involutions and $t= frac{|G|}{n}$ then $|G| < lceil t(t+1)/2 rceil !$.



This is a simple consequence of Theorem $(2F)$ or Theorem $(2H)$ in that paper so check if something along these lines is in your book. (Which book is it by the way?)



Arguing by contradiction, suppose that $n geq frac{|G|}{3}$. Then $t leq 3$ so $|G|<720$. There are just $5$ non-abelian simple groups of order less than $720$, which are $A_5$, $A_6$, $operatorname{PSL}_2(7)$, $operatorname{PSL}_2(8)$ and $operatorname{PSL}_2(11)$, and these have $15$, $45$, $21$, $63$, and $55$ involutions respectively. In no case does $n geq frac{|G|}{3}$ hold, which is a contradiction.





Added. You can use directly Theorem $6.7$ in Rose's "A course on Group Theory" which says: Let $G$ be a group of even order with precisely $n$ involutions, and suppose that $|Z(G)|$ is odd. Let $a = |G|/n$. Then $G$ has a proper subgroup $H$ such that either $|G:H|=2$ or $|G:H|<frac{1}{2}a(a+1)$.



Now suppose that $G$ is a finite simple group with precisely $n$ involutions. Since $|Z(G)|=1$ the preceding theorem applies. Note that $|G:H| neq 2$ since otherwise $H$ is normal in $G$. In fact, the stronger claim is true (which Derek mentioned in his answer), that $|G:H| geq 5$.



Assume for a contradiction that $n geq |G|/3$. Then $a := |G|/n leq 3$, so $G$ has a proper subgroup $H$ such that $|G:H|<6$ by Thm. $6.7$, thus $|G:H|=5$ by the preceding observation. But $|G:H|=5$ is only possible if $G cong A_5$ (do you see why?) and you are given that $A_5$ has less than $60/3=20$ involutions. That is a contradiction, however, and the proof is complete.






share|cite|improve this answer























  • Hi, the book is A Course on Group Theory by John Rose. I have not looked at the specific simple groups you have mentioned, so ideally I am looking for a proof that does not rely on these particular groups (although I am sure what you have said will be correct)
    – John Doe
    Nov 28 at 11:28










  • @JohnDoe Thanks for the reference. I have updated my answer.
    – the_fox
    Nov 28 at 13:41






  • 1




    Only part I'm not completely sure about is that $|G:H| = 5$ Implies $G$ is isomorphic to $A_{5} $, although I think that can be shown as follows. If $G$ is a finite non-abelian simple group and $H < G$ with $|G : H | = n $, then $G / H_{G} $ can be embedded in $S_{n} $, where $H_{G} $ is the core of $H$. Since $H_{G} $ is a normal subgroup of $G$, we have $H_{G} = { 1 } $. So $G$ can be embedded in $S_{n} $.
    – John Doe
    Nov 29 at 13:42








  • 1




    Coninuted... The only simple group embedded in $S_{5} $ is $A_{5} $ since this is the only possible finite non-abelian simple group of order less than $100$ (and there is no such simple group of order $120$, so $S_{5} $ is not simple). So $G cong A_{5} $.
    – John Doe
    Nov 29 at 13:46




















up vote
2
down vote













I think the following is essentially the same argument as in the proof of the Brauer-Fowler result cited by the_fox.



Let $I$ be the set of involutions in $G$ and $m=|I|$, $n=|G|$.



For $x in G$, define $C^*_G(x) = { g in G mid g^{-1}xg = x^{pm 1} }$. Then either $C^*_G(x) = C_G(x)$, or $|C^*_G(x):C_G(x)|=2$.
Notice that, if $u,v in I$ and $x=uv notin I$, then $u,v in C^*_G(x) setminus C_G(x)$.



Now, for $x in G$, let $beta(x)$ be the number of ordered pairs $(u,v) in I times I$ with $uv=x$. Then $m^2 = sum_{x in G}beta(x)$.



To estimate $beta(x)$, we consider three cases.



If $x = 1$, then clearly $beta(x)=m$.



If $x in I$ and $uv = x$ with $u,v in I$, then $u,v in C_G(x)$, so $beta(x)$ is the number of involutions in $C_G(x) setminus {x}$, which is at most $|C_G(x)|-2$.



Otherwise, if $x ne 1$ and $x notin I$, then either $beta(x)=0$, or $beta(x) le |C^*_G(x)| - |C_G(x)| = |C_G(x)|$.



We will use the well-known result that the only finite nonabelian simple group with a proper subgroup of index at most 5 is $A_5$, and that the result is true in $A_5$.
(In fact $|I|=|G|/4$ in $A_5$.)



So, for $m in I$, we have $|C_G(x)| le n/6$, and for $m in G setminus (I cup {1})$, $|C^*_G(x)| le n/6$ and hence $|C_G(x)| le n/12$. So we get



$$m^2 le m + m(n/6-2) + (n-m-1)n/12,$$
and then, putting $k=n/m$, we have
$$n^2/k^2 le n/k + n(n/6-2)/k + (n-n/k-1)n/12= n^2/(12k) - n/k +n^2/12- n/12,$$
so $$nleft(frac{1}{k^2} -frac{1}{12k}-frac{1}{12}right) le -frac{1}{k} - frac{1}{12},$$
but the left hand side is positive for $k le 3$, so $|I| < |G|/3$, QED.






share|cite|improve this answer























  • Thanks for the response. I am not sure about the conclusion though. Toward the end you say that the inequality is positive for $K geq 3$, where it is in fact negative (I think this may have just been a typo though). The only possible values of $k$ where it is positive is for $0 < k leq - frac{1}{2} + frac{1}{2} sqrt{41} $. Since we want the left hand side to be negative, the inequality is only valid when $k > - frac{1}{2} + frac{1}{2} sqrt{41} = 2.701 dots $. It seems that this only gives the inequality $2.701 dots < |G| / m $, which doesn't quite get to the full answer.
    – John Doe
    Nov 28 at 11:38












  • Yes you are right, I miscalculated. But I think it is OK now - I just needed to assume that proper subgroups have index at least $6$ rather than $5$.
    – Derek Holt
    Nov 28 at 16:46











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009105%2fshow-that-finite-simple-group-g-with-n-involutions-satisfies-n-g-3%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










Corollary $(2I)$ in the paper "On groups of even order" by Brauer and Fowler says that if $G$ is a simple group which contains $n$ involutions and $t= frac{|G|}{n}$ then $|G| < lceil t(t+1)/2 rceil !$.



This is a simple consequence of Theorem $(2F)$ or Theorem $(2H)$ in that paper so check if something along these lines is in your book. (Which book is it by the way?)



Arguing by contradiction, suppose that $n geq frac{|G|}{3}$. Then $t leq 3$ so $|G|<720$. There are just $5$ non-abelian simple groups of order less than $720$, which are $A_5$, $A_6$, $operatorname{PSL}_2(7)$, $operatorname{PSL}_2(8)$ and $operatorname{PSL}_2(11)$, and these have $15$, $45$, $21$, $63$, and $55$ involutions respectively. In no case does $n geq frac{|G|}{3}$ hold, which is a contradiction.





Added. You can use directly Theorem $6.7$ in Rose's "A course on Group Theory" which says: Let $G$ be a group of even order with precisely $n$ involutions, and suppose that $|Z(G)|$ is odd. Let $a = |G|/n$. Then $G$ has a proper subgroup $H$ such that either $|G:H|=2$ or $|G:H|<frac{1}{2}a(a+1)$.



Now suppose that $G$ is a finite simple group with precisely $n$ involutions. Since $|Z(G)|=1$ the preceding theorem applies. Note that $|G:H| neq 2$ since otherwise $H$ is normal in $G$. In fact, the stronger claim is true (which Derek mentioned in his answer), that $|G:H| geq 5$.



Assume for a contradiction that $n geq |G|/3$. Then $a := |G|/n leq 3$, so $G$ has a proper subgroup $H$ such that $|G:H|<6$ by Thm. $6.7$, thus $|G:H|=5$ by the preceding observation. But $|G:H|=5$ is only possible if $G cong A_5$ (do you see why?) and you are given that $A_5$ has less than $60/3=20$ involutions. That is a contradiction, however, and the proof is complete.






share|cite|improve this answer























  • Hi, the book is A Course on Group Theory by John Rose. I have not looked at the specific simple groups you have mentioned, so ideally I am looking for a proof that does not rely on these particular groups (although I am sure what you have said will be correct)
    – John Doe
    Nov 28 at 11:28










  • @JohnDoe Thanks for the reference. I have updated my answer.
    – the_fox
    Nov 28 at 13:41






  • 1




    Only part I'm not completely sure about is that $|G:H| = 5$ Implies $G$ is isomorphic to $A_{5} $, although I think that can be shown as follows. If $G$ is a finite non-abelian simple group and $H < G$ with $|G : H | = n $, then $G / H_{G} $ can be embedded in $S_{n} $, where $H_{G} $ is the core of $H$. Since $H_{G} $ is a normal subgroup of $G$, we have $H_{G} = { 1 } $. So $G$ can be embedded in $S_{n} $.
    – John Doe
    Nov 29 at 13:42








  • 1




    Coninuted... The only simple group embedded in $S_{5} $ is $A_{5} $ since this is the only possible finite non-abelian simple group of order less than $100$ (and there is no such simple group of order $120$, so $S_{5} $ is not simple). So $G cong A_{5} $.
    – John Doe
    Nov 29 at 13:46

















up vote
3
down vote



accepted










Corollary $(2I)$ in the paper "On groups of even order" by Brauer and Fowler says that if $G$ is a simple group which contains $n$ involutions and $t= frac{|G|}{n}$ then $|G| < lceil t(t+1)/2 rceil !$.



This is a simple consequence of Theorem $(2F)$ or Theorem $(2H)$ in that paper so check if something along these lines is in your book. (Which book is it by the way?)



Arguing by contradiction, suppose that $n geq frac{|G|}{3}$. Then $t leq 3$ so $|G|<720$. There are just $5$ non-abelian simple groups of order less than $720$, which are $A_5$, $A_6$, $operatorname{PSL}_2(7)$, $operatorname{PSL}_2(8)$ and $operatorname{PSL}_2(11)$, and these have $15$, $45$, $21$, $63$, and $55$ involutions respectively. In no case does $n geq frac{|G|}{3}$ hold, which is a contradiction.





Added. You can use directly Theorem $6.7$ in Rose's "A course on Group Theory" which says: Let $G$ be a group of even order with precisely $n$ involutions, and suppose that $|Z(G)|$ is odd. Let $a = |G|/n$. Then $G$ has a proper subgroup $H$ such that either $|G:H|=2$ or $|G:H|<frac{1}{2}a(a+1)$.



Now suppose that $G$ is a finite simple group with precisely $n$ involutions. Since $|Z(G)|=1$ the preceding theorem applies. Note that $|G:H| neq 2$ since otherwise $H$ is normal in $G$. In fact, the stronger claim is true (which Derek mentioned in his answer), that $|G:H| geq 5$.



Assume for a contradiction that $n geq |G|/3$. Then $a := |G|/n leq 3$, so $G$ has a proper subgroup $H$ such that $|G:H|<6$ by Thm. $6.7$, thus $|G:H|=5$ by the preceding observation. But $|G:H|=5$ is only possible if $G cong A_5$ (do you see why?) and you are given that $A_5$ has less than $60/3=20$ involutions. That is a contradiction, however, and the proof is complete.






share|cite|improve this answer























  • Hi, the book is A Course on Group Theory by John Rose. I have not looked at the specific simple groups you have mentioned, so ideally I am looking for a proof that does not rely on these particular groups (although I am sure what you have said will be correct)
    – John Doe
    Nov 28 at 11:28










  • @JohnDoe Thanks for the reference. I have updated my answer.
    – the_fox
    Nov 28 at 13:41






  • 1




    Only part I'm not completely sure about is that $|G:H| = 5$ Implies $G$ is isomorphic to $A_{5} $, although I think that can be shown as follows. If $G$ is a finite non-abelian simple group and $H < G$ with $|G : H | = n $, then $G / H_{G} $ can be embedded in $S_{n} $, where $H_{G} $ is the core of $H$. Since $H_{G} $ is a normal subgroup of $G$, we have $H_{G} = { 1 } $. So $G$ can be embedded in $S_{n} $.
    – John Doe
    Nov 29 at 13:42








  • 1




    Coninuted... The only simple group embedded in $S_{5} $ is $A_{5} $ since this is the only possible finite non-abelian simple group of order less than $100$ (and there is no such simple group of order $120$, so $S_{5} $ is not simple). So $G cong A_{5} $.
    – John Doe
    Nov 29 at 13:46















up vote
3
down vote



accepted







up vote
3
down vote



accepted






Corollary $(2I)$ in the paper "On groups of even order" by Brauer and Fowler says that if $G$ is a simple group which contains $n$ involutions and $t= frac{|G|}{n}$ then $|G| < lceil t(t+1)/2 rceil !$.



This is a simple consequence of Theorem $(2F)$ or Theorem $(2H)$ in that paper so check if something along these lines is in your book. (Which book is it by the way?)



Arguing by contradiction, suppose that $n geq frac{|G|}{3}$. Then $t leq 3$ so $|G|<720$. There are just $5$ non-abelian simple groups of order less than $720$, which are $A_5$, $A_6$, $operatorname{PSL}_2(7)$, $operatorname{PSL}_2(8)$ and $operatorname{PSL}_2(11)$, and these have $15$, $45$, $21$, $63$, and $55$ involutions respectively. In no case does $n geq frac{|G|}{3}$ hold, which is a contradiction.





Added. You can use directly Theorem $6.7$ in Rose's "A course on Group Theory" which says: Let $G$ be a group of even order with precisely $n$ involutions, and suppose that $|Z(G)|$ is odd. Let $a = |G|/n$. Then $G$ has a proper subgroup $H$ such that either $|G:H|=2$ or $|G:H|<frac{1}{2}a(a+1)$.



Now suppose that $G$ is a finite simple group with precisely $n$ involutions. Since $|Z(G)|=1$ the preceding theorem applies. Note that $|G:H| neq 2$ since otherwise $H$ is normal in $G$. In fact, the stronger claim is true (which Derek mentioned in his answer), that $|G:H| geq 5$.



Assume for a contradiction that $n geq |G|/3$. Then $a := |G|/n leq 3$, so $G$ has a proper subgroup $H$ such that $|G:H|<6$ by Thm. $6.7$, thus $|G:H|=5$ by the preceding observation. But $|G:H|=5$ is only possible if $G cong A_5$ (do you see why?) and you are given that $A_5$ has less than $60/3=20$ involutions. That is a contradiction, however, and the proof is complete.






share|cite|improve this answer














Corollary $(2I)$ in the paper "On groups of even order" by Brauer and Fowler says that if $G$ is a simple group which contains $n$ involutions and $t= frac{|G|}{n}$ then $|G| < lceil t(t+1)/2 rceil !$.



This is a simple consequence of Theorem $(2F)$ or Theorem $(2H)$ in that paper so check if something along these lines is in your book. (Which book is it by the way?)



Arguing by contradiction, suppose that $n geq frac{|G|}{3}$. Then $t leq 3$ so $|G|<720$. There are just $5$ non-abelian simple groups of order less than $720$, which are $A_5$, $A_6$, $operatorname{PSL}_2(7)$, $operatorname{PSL}_2(8)$ and $operatorname{PSL}_2(11)$, and these have $15$, $45$, $21$, $63$, and $55$ involutions respectively. In no case does $n geq frac{|G|}{3}$ hold, which is a contradiction.





Added. You can use directly Theorem $6.7$ in Rose's "A course on Group Theory" which says: Let $G$ be a group of even order with precisely $n$ involutions, and suppose that $|Z(G)|$ is odd. Let $a = |G|/n$. Then $G$ has a proper subgroup $H$ such that either $|G:H|=2$ or $|G:H|<frac{1}{2}a(a+1)$.



Now suppose that $G$ is a finite simple group with precisely $n$ involutions. Since $|Z(G)|=1$ the preceding theorem applies. Note that $|G:H| neq 2$ since otherwise $H$ is normal in $G$. In fact, the stronger claim is true (which Derek mentioned in his answer), that $|G:H| geq 5$.



Assume for a contradiction that $n geq |G|/3$. Then $a := |G|/n leq 3$, so $G$ has a proper subgroup $H$ such that $|G:H|<6$ by Thm. $6.7$, thus $|G:H|=5$ by the preceding observation. But $|G:H|=5$ is only possible if $G cong A_5$ (do you see why?) and you are given that $A_5$ has less than $60/3=20$ involutions. That is a contradiction, however, and the proof is complete.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 at 23:34

























answered Nov 22 at 18:44









the_fox

2,3191430




2,3191430












  • Hi, the book is A Course on Group Theory by John Rose. I have not looked at the specific simple groups you have mentioned, so ideally I am looking for a proof that does not rely on these particular groups (although I am sure what you have said will be correct)
    – John Doe
    Nov 28 at 11:28










  • @JohnDoe Thanks for the reference. I have updated my answer.
    – the_fox
    Nov 28 at 13:41






  • 1




    Only part I'm not completely sure about is that $|G:H| = 5$ Implies $G$ is isomorphic to $A_{5} $, although I think that can be shown as follows. If $G$ is a finite non-abelian simple group and $H < G$ with $|G : H | = n $, then $G / H_{G} $ can be embedded in $S_{n} $, where $H_{G} $ is the core of $H$. Since $H_{G} $ is a normal subgroup of $G$, we have $H_{G} = { 1 } $. So $G$ can be embedded in $S_{n} $.
    – John Doe
    Nov 29 at 13:42








  • 1




    Coninuted... The only simple group embedded in $S_{5} $ is $A_{5} $ since this is the only possible finite non-abelian simple group of order less than $100$ (and there is no such simple group of order $120$, so $S_{5} $ is not simple). So $G cong A_{5} $.
    – John Doe
    Nov 29 at 13:46




















  • Hi, the book is A Course on Group Theory by John Rose. I have not looked at the specific simple groups you have mentioned, so ideally I am looking for a proof that does not rely on these particular groups (although I am sure what you have said will be correct)
    – John Doe
    Nov 28 at 11:28










  • @JohnDoe Thanks for the reference. I have updated my answer.
    – the_fox
    Nov 28 at 13:41






  • 1




    Only part I'm not completely sure about is that $|G:H| = 5$ Implies $G$ is isomorphic to $A_{5} $, although I think that can be shown as follows. If $G$ is a finite non-abelian simple group and $H < G$ with $|G : H | = n $, then $G / H_{G} $ can be embedded in $S_{n} $, where $H_{G} $ is the core of $H$. Since $H_{G} $ is a normal subgroup of $G$, we have $H_{G} = { 1 } $. So $G$ can be embedded in $S_{n} $.
    – John Doe
    Nov 29 at 13:42








  • 1




    Coninuted... The only simple group embedded in $S_{5} $ is $A_{5} $ since this is the only possible finite non-abelian simple group of order less than $100$ (and there is no such simple group of order $120$, so $S_{5} $ is not simple). So $G cong A_{5} $.
    – John Doe
    Nov 29 at 13:46


















Hi, the book is A Course on Group Theory by John Rose. I have not looked at the specific simple groups you have mentioned, so ideally I am looking for a proof that does not rely on these particular groups (although I am sure what you have said will be correct)
– John Doe
Nov 28 at 11:28




Hi, the book is A Course on Group Theory by John Rose. I have not looked at the specific simple groups you have mentioned, so ideally I am looking for a proof that does not rely on these particular groups (although I am sure what you have said will be correct)
– John Doe
Nov 28 at 11:28












@JohnDoe Thanks for the reference. I have updated my answer.
– the_fox
Nov 28 at 13:41




@JohnDoe Thanks for the reference. I have updated my answer.
– the_fox
Nov 28 at 13:41




1




1




Only part I'm not completely sure about is that $|G:H| = 5$ Implies $G$ is isomorphic to $A_{5} $, although I think that can be shown as follows. If $G$ is a finite non-abelian simple group and $H < G$ with $|G : H | = n $, then $G / H_{G} $ can be embedded in $S_{n} $, where $H_{G} $ is the core of $H$. Since $H_{G} $ is a normal subgroup of $G$, we have $H_{G} = { 1 } $. So $G$ can be embedded in $S_{n} $.
– John Doe
Nov 29 at 13:42






Only part I'm not completely sure about is that $|G:H| = 5$ Implies $G$ is isomorphic to $A_{5} $, although I think that can be shown as follows. If $G$ is a finite non-abelian simple group and $H < G$ with $|G : H | = n $, then $G / H_{G} $ can be embedded in $S_{n} $, where $H_{G} $ is the core of $H$. Since $H_{G} $ is a normal subgroup of $G$, we have $H_{G} = { 1 } $. So $G$ can be embedded in $S_{n} $.
– John Doe
Nov 29 at 13:42






1




1




Coninuted... The only simple group embedded in $S_{5} $ is $A_{5} $ since this is the only possible finite non-abelian simple group of order less than $100$ (and there is no such simple group of order $120$, so $S_{5} $ is not simple). So $G cong A_{5} $.
– John Doe
Nov 29 at 13:46






Coninuted... The only simple group embedded in $S_{5} $ is $A_{5} $ since this is the only possible finite non-abelian simple group of order less than $100$ (and there is no such simple group of order $120$, so $S_{5} $ is not simple). So $G cong A_{5} $.
– John Doe
Nov 29 at 13:46












up vote
2
down vote













I think the following is essentially the same argument as in the proof of the Brauer-Fowler result cited by the_fox.



Let $I$ be the set of involutions in $G$ and $m=|I|$, $n=|G|$.



For $x in G$, define $C^*_G(x) = { g in G mid g^{-1}xg = x^{pm 1} }$. Then either $C^*_G(x) = C_G(x)$, or $|C^*_G(x):C_G(x)|=2$.
Notice that, if $u,v in I$ and $x=uv notin I$, then $u,v in C^*_G(x) setminus C_G(x)$.



Now, for $x in G$, let $beta(x)$ be the number of ordered pairs $(u,v) in I times I$ with $uv=x$. Then $m^2 = sum_{x in G}beta(x)$.



To estimate $beta(x)$, we consider three cases.



If $x = 1$, then clearly $beta(x)=m$.



If $x in I$ and $uv = x$ with $u,v in I$, then $u,v in C_G(x)$, so $beta(x)$ is the number of involutions in $C_G(x) setminus {x}$, which is at most $|C_G(x)|-2$.



Otherwise, if $x ne 1$ and $x notin I$, then either $beta(x)=0$, or $beta(x) le |C^*_G(x)| - |C_G(x)| = |C_G(x)|$.



We will use the well-known result that the only finite nonabelian simple group with a proper subgroup of index at most 5 is $A_5$, and that the result is true in $A_5$.
(In fact $|I|=|G|/4$ in $A_5$.)



So, for $m in I$, we have $|C_G(x)| le n/6$, and for $m in G setminus (I cup {1})$, $|C^*_G(x)| le n/6$ and hence $|C_G(x)| le n/12$. So we get



$$m^2 le m + m(n/6-2) + (n-m-1)n/12,$$
and then, putting $k=n/m$, we have
$$n^2/k^2 le n/k + n(n/6-2)/k + (n-n/k-1)n/12= n^2/(12k) - n/k +n^2/12- n/12,$$
so $$nleft(frac{1}{k^2} -frac{1}{12k}-frac{1}{12}right) le -frac{1}{k} - frac{1}{12},$$
but the left hand side is positive for $k le 3$, so $|I| < |G|/3$, QED.






share|cite|improve this answer























  • Thanks for the response. I am not sure about the conclusion though. Toward the end you say that the inequality is positive for $K geq 3$, where it is in fact negative (I think this may have just been a typo though). The only possible values of $k$ where it is positive is for $0 < k leq - frac{1}{2} + frac{1}{2} sqrt{41} $. Since we want the left hand side to be negative, the inequality is only valid when $k > - frac{1}{2} + frac{1}{2} sqrt{41} = 2.701 dots $. It seems that this only gives the inequality $2.701 dots < |G| / m $, which doesn't quite get to the full answer.
    – John Doe
    Nov 28 at 11:38












  • Yes you are right, I miscalculated. But I think it is OK now - I just needed to assume that proper subgroups have index at least $6$ rather than $5$.
    – Derek Holt
    Nov 28 at 16:46















up vote
2
down vote













I think the following is essentially the same argument as in the proof of the Brauer-Fowler result cited by the_fox.



Let $I$ be the set of involutions in $G$ and $m=|I|$, $n=|G|$.



For $x in G$, define $C^*_G(x) = { g in G mid g^{-1}xg = x^{pm 1} }$. Then either $C^*_G(x) = C_G(x)$, or $|C^*_G(x):C_G(x)|=2$.
Notice that, if $u,v in I$ and $x=uv notin I$, then $u,v in C^*_G(x) setminus C_G(x)$.



Now, for $x in G$, let $beta(x)$ be the number of ordered pairs $(u,v) in I times I$ with $uv=x$. Then $m^2 = sum_{x in G}beta(x)$.



To estimate $beta(x)$, we consider three cases.



If $x = 1$, then clearly $beta(x)=m$.



If $x in I$ and $uv = x$ with $u,v in I$, then $u,v in C_G(x)$, so $beta(x)$ is the number of involutions in $C_G(x) setminus {x}$, which is at most $|C_G(x)|-2$.



Otherwise, if $x ne 1$ and $x notin I$, then either $beta(x)=0$, or $beta(x) le |C^*_G(x)| - |C_G(x)| = |C_G(x)|$.



We will use the well-known result that the only finite nonabelian simple group with a proper subgroup of index at most 5 is $A_5$, and that the result is true in $A_5$.
(In fact $|I|=|G|/4$ in $A_5$.)



So, for $m in I$, we have $|C_G(x)| le n/6$, and for $m in G setminus (I cup {1})$, $|C^*_G(x)| le n/6$ and hence $|C_G(x)| le n/12$. So we get



$$m^2 le m + m(n/6-2) + (n-m-1)n/12,$$
and then, putting $k=n/m$, we have
$$n^2/k^2 le n/k + n(n/6-2)/k + (n-n/k-1)n/12= n^2/(12k) - n/k +n^2/12- n/12,$$
so $$nleft(frac{1}{k^2} -frac{1}{12k}-frac{1}{12}right) le -frac{1}{k} - frac{1}{12},$$
but the left hand side is positive for $k le 3$, so $|I| < |G|/3$, QED.






share|cite|improve this answer























  • Thanks for the response. I am not sure about the conclusion though. Toward the end you say that the inequality is positive for $K geq 3$, where it is in fact negative (I think this may have just been a typo though). The only possible values of $k$ where it is positive is for $0 < k leq - frac{1}{2} + frac{1}{2} sqrt{41} $. Since we want the left hand side to be negative, the inequality is only valid when $k > - frac{1}{2} + frac{1}{2} sqrt{41} = 2.701 dots $. It seems that this only gives the inequality $2.701 dots < |G| / m $, which doesn't quite get to the full answer.
    – John Doe
    Nov 28 at 11:38












  • Yes you are right, I miscalculated. But I think it is OK now - I just needed to assume that proper subgroups have index at least $6$ rather than $5$.
    – Derek Holt
    Nov 28 at 16:46













up vote
2
down vote










up vote
2
down vote









I think the following is essentially the same argument as in the proof of the Brauer-Fowler result cited by the_fox.



Let $I$ be the set of involutions in $G$ and $m=|I|$, $n=|G|$.



For $x in G$, define $C^*_G(x) = { g in G mid g^{-1}xg = x^{pm 1} }$. Then either $C^*_G(x) = C_G(x)$, or $|C^*_G(x):C_G(x)|=2$.
Notice that, if $u,v in I$ and $x=uv notin I$, then $u,v in C^*_G(x) setminus C_G(x)$.



Now, for $x in G$, let $beta(x)$ be the number of ordered pairs $(u,v) in I times I$ with $uv=x$. Then $m^2 = sum_{x in G}beta(x)$.



To estimate $beta(x)$, we consider three cases.



If $x = 1$, then clearly $beta(x)=m$.



If $x in I$ and $uv = x$ with $u,v in I$, then $u,v in C_G(x)$, so $beta(x)$ is the number of involutions in $C_G(x) setminus {x}$, which is at most $|C_G(x)|-2$.



Otherwise, if $x ne 1$ and $x notin I$, then either $beta(x)=0$, or $beta(x) le |C^*_G(x)| - |C_G(x)| = |C_G(x)|$.



We will use the well-known result that the only finite nonabelian simple group with a proper subgroup of index at most 5 is $A_5$, and that the result is true in $A_5$.
(In fact $|I|=|G|/4$ in $A_5$.)



So, for $m in I$, we have $|C_G(x)| le n/6$, and for $m in G setminus (I cup {1})$, $|C^*_G(x)| le n/6$ and hence $|C_G(x)| le n/12$. So we get



$$m^2 le m + m(n/6-2) + (n-m-1)n/12,$$
and then, putting $k=n/m$, we have
$$n^2/k^2 le n/k + n(n/6-2)/k + (n-n/k-1)n/12= n^2/(12k) - n/k +n^2/12- n/12,$$
so $$nleft(frac{1}{k^2} -frac{1}{12k}-frac{1}{12}right) le -frac{1}{k} - frac{1}{12},$$
but the left hand side is positive for $k le 3$, so $|I| < |G|/3$, QED.






share|cite|improve this answer














I think the following is essentially the same argument as in the proof of the Brauer-Fowler result cited by the_fox.



Let $I$ be the set of involutions in $G$ and $m=|I|$, $n=|G|$.



For $x in G$, define $C^*_G(x) = { g in G mid g^{-1}xg = x^{pm 1} }$. Then either $C^*_G(x) = C_G(x)$, or $|C^*_G(x):C_G(x)|=2$.
Notice that, if $u,v in I$ and $x=uv notin I$, then $u,v in C^*_G(x) setminus C_G(x)$.



Now, for $x in G$, let $beta(x)$ be the number of ordered pairs $(u,v) in I times I$ with $uv=x$. Then $m^2 = sum_{x in G}beta(x)$.



To estimate $beta(x)$, we consider three cases.



If $x = 1$, then clearly $beta(x)=m$.



If $x in I$ and $uv = x$ with $u,v in I$, then $u,v in C_G(x)$, so $beta(x)$ is the number of involutions in $C_G(x) setminus {x}$, which is at most $|C_G(x)|-2$.



Otherwise, if $x ne 1$ and $x notin I$, then either $beta(x)=0$, or $beta(x) le |C^*_G(x)| - |C_G(x)| = |C_G(x)|$.



We will use the well-known result that the only finite nonabelian simple group with a proper subgroup of index at most 5 is $A_5$, and that the result is true in $A_5$.
(In fact $|I|=|G|/4$ in $A_5$.)



So, for $m in I$, we have $|C_G(x)| le n/6$, and for $m in G setminus (I cup {1})$, $|C^*_G(x)| le n/6$ and hence $|C_G(x)| le n/12$. So we get



$$m^2 le m + m(n/6-2) + (n-m-1)n/12,$$
and then, putting $k=n/m$, we have
$$n^2/k^2 le n/k + n(n/6-2)/k + (n-n/k-1)n/12= n^2/(12k) - n/k +n^2/12- n/12,$$
so $$nleft(frac{1}{k^2} -frac{1}{12k}-frac{1}{12}right) le -frac{1}{k} - frac{1}{12},$$
but the left hand side is positive for $k le 3$, so $|I| < |G|/3$, QED.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 28 at 16:47

























answered Nov 22 at 20:19









Derek Holt

52.3k53570




52.3k53570












  • Thanks for the response. I am not sure about the conclusion though. Toward the end you say that the inequality is positive for $K geq 3$, where it is in fact negative (I think this may have just been a typo though). The only possible values of $k$ where it is positive is for $0 < k leq - frac{1}{2} + frac{1}{2} sqrt{41} $. Since we want the left hand side to be negative, the inequality is only valid when $k > - frac{1}{2} + frac{1}{2} sqrt{41} = 2.701 dots $. It seems that this only gives the inequality $2.701 dots < |G| / m $, which doesn't quite get to the full answer.
    – John Doe
    Nov 28 at 11:38












  • Yes you are right, I miscalculated. But I think it is OK now - I just needed to assume that proper subgroups have index at least $6$ rather than $5$.
    – Derek Holt
    Nov 28 at 16:46


















  • Thanks for the response. I am not sure about the conclusion though. Toward the end you say that the inequality is positive for $K geq 3$, where it is in fact negative (I think this may have just been a typo though). The only possible values of $k$ where it is positive is for $0 < k leq - frac{1}{2} + frac{1}{2} sqrt{41} $. Since we want the left hand side to be negative, the inequality is only valid when $k > - frac{1}{2} + frac{1}{2} sqrt{41} = 2.701 dots $. It seems that this only gives the inequality $2.701 dots < |G| / m $, which doesn't quite get to the full answer.
    – John Doe
    Nov 28 at 11:38












  • Yes you are right, I miscalculated. But I think it is OK now - I just needed to assume that proper subgroups have index at least $6$ rather than $5$.
    – Derek Holt
    Nov 28 at 16:46
















Thanks for the response. I am not sure about the conclusion though. Toward the end you say that the inequality is positive for $K geq 3$, where it is in fact negative (I think this may have just been a typo though). The only possible values of $k$ where it is positive is for $0 < k leq - frac{1}{2} + frac{1}{2} sqrt{41} $. Since we want the left hand side to be negative, the inequality is only valid when $k > - frac{1}{2} + frac{1}{2} sqrt{41} = 2.701 dots $. It seems that this only gives the inequality $2.701 dots < |G| / m $, which doesn't quite get to the full answer.
– John Doe
Nov 28 at 11:38






Thanks for the response. I am not sure about the conclusion though. Toward the end you say that the inequality is positive for $K geq 3$, where it is in fact negative (I think this may have just been a typo though). The only possible values of $k$ where it is positive is for $0 < k leq - frac{1}{2} + frac{1}{2} sqrt{41} $. Since we want the left hand side to be negative, the inequality is only valid when $k > - frac{1}{2} + frac{1}{2} sqrt{41} = 2.701 dots $. It seems that this only gives the inequality $2.701 dots < |G| / m $, which doesn't quite get to the full answer.
– John Doe
Nov 28 at 11:38














Yes you are right, I miscalculated. But I think it is OK now - I just needed to assume that proper subgroups have index at least $6$ rather than $5$.
– Derek Holt
Nov 28 at 16:46




Yes you are right, I miscalculated. But I think it is OK now - I just needed to assume that proper subgroups have index at least $6$ rather than $5$.
– Derek Holt
Nov 28 at 16:46


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009105%2fshow-that-finite-simple-group-g-with-n-involutions-satisfies-n-g-3%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Ellipse (mathématiques)

Quarter-circle Tiles

Mont Emei