Show that finite simple group $G$ with $n$ involutions satisfies $n < |G| / 3$
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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
add a comment |
up vote
4
down vote
favorite
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
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
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
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
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
group-theory finite-groups
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
add a comment |
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
add a comment |
2 Answers
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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.
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
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up vote
2
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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.
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
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2 Answers
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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.
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
add a comment |
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.
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
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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.
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
add a comment |
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.
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
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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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