How many groups with order $(17^2)(5^2)$ exist?
$begingroup$
Its the first time I'm seeing such question, I tried something but I can't really confirm it since its pretty new to me:
Its easy to prove with sylow that there is only one subgroup $H$ with order $17^2$, hence, $H$ is normal. Since $17$ is prime, and $|H|=17^2$ then there are $2$ groups up to isomorphism with that order. $G/H$ is a group of order $5^2$ (since $H$ is normal), and again, we have $2$ groups with that order up to isomorphism.
So the answer is $2times 2+1=5$? (Added one for the case when $G$ is cyclic). is there anything else I missed here?
abstract-algebra group-theory finite-groups
edited Dec 13 '18 at 13:40
Tianlalu
3,08621038
asked Dec 13 '18 at 13:13
S.CoS.Co
687
$endgroup$
add a comment |
$begingroup$
Its the first time I'm seeing such question, I tried something but I can't really confirm it since its pretty new to me:
Its easy to prove with sylow that there is only one subgroup $H$ with order $17^2$, hence, $H$ is normal. Since $17$ is prime, and $|H|=17^2$ then there are $2$ groups up to isomorphism with that order. $G/H$ is a group of order $5^2$ (since $H$ is normal), and again, we have $2$ groups with that order up to isomorphism.
So the answer is $2times 2+1=5$? (Added one for the case when $G$ is cyclic). is there anything else I missed here?
abstract-algebra group-theory finite-groups
edited Dec 13 '18 at 13:40
Tianlalu
3,08621038
asked Dec 13 '18 at 13:13
S.CoS.Co
687
$endgroup$
1
$begingroup$
The wording here is imprecise, and in my reading of the intermediate steps are not fully justified. Anyway, one can just as well prove that there is only one Sylow $5$-group $K$, of order $25$, and looking at element orders shows that $H cap K = {0}$, so $G = H times K$; in particular $G$ is abelian. Since $H$ has order $17^2$, it is either $C_{17}^2$ or $C_{17^2}$ and similarly for $K$, leaving $2 cdot 2 = 4$ groups of this order.
$endgroup$
– Travis
Dec 13 '18 at 13:33
$begingroup$
Adding one for the case when $G$ is cyclic makes no sense at all. Your argument is really counting the groups that are direct products of their Sylow subgroups i.e. the nilpotent groups, and as it happens all groups of this order are nilpotent, but that is not always the case.
$endgroup$
– Derek Holt
Dec 13 '18 at 13:46
add a comment |
$begingroup$
Its the first time I'm seeing such question, I tried something but I can't really confirm it since its pretty new to me:
Its easy to prove with sylow that there is only one subgroup $H$ with order $17^2$, hence, $H$ is normal. Since $17$ is prime, and $|H|=17^2$ then there are $2$ groups up to isomorphism with that order. $G/H$ is a group of order $5^2$ (since $H$ is normal), and again, we have $2$ groups with that order up to isomorphism.
So the answer is $2times 2+1=5$? (Added one for the case when $G$ is cyclic). is there anything else I missed here?
abstract-algebra group-theory finite-groups
edited Dec 13 '18 at 13:40
Tianlalu
3,08621038
asked Dec 13 '18 at 13:13
S.CoS.Co
687
$endgroup$
Its the first time I'm seeing such question, I tried something but I can't really confirm it since its pretty new to me:
Its easy to prove with sylow that there is only one subgroup $H$ with order $17^2$, hence, $H$ is normal. Since $17$ is prime, and $|H|=17^2$ then there are $2$ groups up to isomorphism with that order. $G/H$ is a group of order $5^2$ (since $H$ is normal), and again, we have $2$ groups with that order up to isomorphism.
So the answer is $2times 2+1=5$? (Added one for the case when $G$ is cyclic). is there anything else I missed here?
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
edited Dec 13 '18 at 13:40
Tianlalu
3,08621038
asked Dec 13 '18 at 13:13
S.CoS.Co
687
edited Dec 13 '18 at 13:40
Tianlalu
3,08621038
asked Dec 13 '18 at 13:13
S.CoS.Co
687
edited Dec 13 '18 at 13:40
Tianlalu
3,08621038
edited Dec 13 '18 at 13:40
Tianlalu
3,08621038
edited Dec 13 '18 at 13:40
Tianlalu
3,08621038
3,08621038
asked Dec 13 '18 at 13:13
S.CoS.Co
687
asked Dec 13 '18 at 13:13
S.CoS.Co
687
asked Dec 13 '18 at 13:13
S.CoS.Co
687
687
1
$begingroup$
The wording here is imprecise, and in my reading of the intermediate steps are not fully justified. Anyway, one can just as well prove that there is only one Sylow $5$-group $K$, of order $25$, and looking at element orders shows that $H cap K = {0}$, so $G = H times K$; in particular $G$ is abelian. Since $H$ has order $17^2$, it is either $C_{17}^2$ or $C_{17^2}$ and similarly for $K$, leaving $2 cdot 2 = 4$ groups of this order.
$endgroup$
– Travis
Dec 13 '18 at 13:33
$begingroup$
Adding one for the case when $G$ is cyclic makes no sense at all. Your argument is really counting the groups that are direct products of their Sylow subgroups i.e. the nilpotent groups, and as it happens all groups of this order are nilpotent, but that is not always the case.
$endgroup$
– Derek Holt
Dec 13 '18 at 13:46
add a comment |
1
$begingroup$
The wording here is imprecise, and in my reading of the intermediate steps are not fully justified. Anyway, one can just as well prove that there is only one Sylow $5$-group $K$, of order $25$, and looking at element orders shows that $H cap K = {0}$, so $G = H times K$; in particular $G$ is abelian. Since $H$ has order $17^2$, it is either $C_{17}^2$ or $C_{17^2}$ and similarly for $K$, leaving $2 cdot 2 = 4$ groups of this order.
$endgroup$
– Travis
Dec 13 '18 at 13:33
$begingroup$
Adding one for the case when $G$ is cyclic makes no sense at all. Your argument is really counting the groups that are direct products of their Sylow subgroups i.e. the nilpotent groups, and as it happens all groups of this order are nilpotent, but that is not always the case.
$endgroup$
– Derek Holt
Dec 13 '18 at 13:46
1
1
$begingroup$
The wording here is imprecise, and in my reading of the intermediate steps are not fully justified. Anyway, one can just as well prove that there is only one Sylow $5$-group $K$, of order $25$, and looking at element orders shows that $H cap K = {0}$, so $G = H times K$; in particular $G$ is abelian. Since $H$ has order $17^2$, it is either $C_{17}^2$ or $C_{17^2}$ and similarly for $K$, leaving $2 cdot 2 = 4$ groups of this order.
$endgroup$
– Travis
Dec 13 '18 at 13:33
$begingroup$
The wording here is imprecise, and in my reading of the intermediate steps are not fully justified. Anyway, one can just as well prove that there is only one Sylow $5$-group $K$, of order $25$, and looking at element orders shows that $H cap K = {0}$, so $G = H times K$; in particular $G$ is abelian. Since $H$ has order $17^2$, it is either $C_{17}^2$ or $C_{17^2}$ and similarly for $K$, leaving $2 cdot 2 = 4$ groups of this order.
$endgroup$
– Travis
Dec 13 '18 at 13:33
$begingroup$
Adding one for the case when $G$ is cyclic makes no sense at all. Your argument is really counting the groups that are direct products of their Sylow subgroups i.e. the nilpotent groups, and as it happens all groups of this order are nilpotent, but that is not always the case.
$endgroup$
– Derek Holt
Dec 13 '18 at 13:46
$begingroup$
Adding one for the case when $G$ is cyclic makes no sense at all. Your argument is really counting the groups that are direct products of their Sylow subgroups i.e. the nilpotent groups, and as it happens all groups of this order are nilpotent, but that is not always the case.
$endgroup$
– Derek Holt
Dec 13 '18 at 13:46
add a comment |
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$begingroup$
The wording here is imprecise, and in my reading of the intermediate steps are not fully justified. Anyway, one can just as well prove that there is only one Sylow $5$-group $K$, of order $25$, and looking at element orders shows that $H cap K = {0}$, so $G = H times K$; in particular $G$ is abelian. Since $H$ has order $17^2$, it is either $C_{17}^2$ or $C_{17^2}$ and similarly for $K$, leaving $2 cdot 2 = 4$ groups of this order.
$endgroup$
– Travis
Dec 13 '18 at 13:33
$begingroup$
Adding one for the case when $G$ is cyclic makes no sense at all. Your argument is really counting the groups that are direct products of their Sylow subgroups i.e. the nilpotent groups, and as it happens all groups of this order are nilpotent, but that is not always the case.
$endgroup$
– Derek Holt
Dec 13 '18 at 13:46