Solvable group $G$ isn't a cyclic $p$-group
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Let $G$ be a nontrivial finite solvable group. We want to show that $G$ has proper subgroups like $A$ and $B$ such that $G=AB$ iff $G$ isn't a cyclic $p$-group.
I use the below fact to solve the problem.
" Let $G$ be a nontrivial finite group. $G$ has one and only one maximal subgroup iff $G$ be a cyclic group of order $p^{m}$ such that $p$ is a prime number. "
I can solve one side but I have problem on the other side.
I suppose that $G$ isn't a cyclic $p$-group. Therefore $G$ has more than one maximal subgroup. Let $M$ and $N$ be maximal subgroups of $G$ such that $M neq N$. I want to show $G = MN $ but I don't know how.
group-theory finite-groups p-groups solvable-groups
asked Nov 15 at 16:19
Yasmin
10310
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show 2 more comments
up vote
-1
down vote
favorite
Let $G$ be a nontrivial finite solvable group. We want to show that $G$ has proper subgroups like $A$ and $B$ such that $G=AB$ iff $G$ isn't a cyclic $p$-group.
I use the below fact to solve the problem.
" Let $G$ be a nontrivial finite group. $G$ has one and only one maximal subgroup iff $G$ be a cyclic group of order $p^{m}$ such that $p$ is a prime number. "
I can solve one side but I have problem on the other side.
I suppose that $G$ isn't a cyclic $p$-group. Therefore $G$ has more than one maximal subgroup. Let $M$ and $N$ be maximal subgroups of $G$ such that $M neq N$. I want to show $G = MN $ but I don't know how.
group-theory finite-groups p-groups solvable-groups
asked Nov 15 at 16:19
Yasmin
10310
1
$MN$ is a subgroup of $G$ and contains both $M$ and $N$. Does that help?
– user3482749
Nov 15 at 16:21
@user3482749Why $MN$ is a subgroup?If so, yes, we have $ M lneqq MN leq G $ and as $M$ is a maximal subgroup of $G$, we have $G=MN$.
– Yasmin
Nov 15 at 16:43
The result is false when $G = C_{p^infty}$, which is an infinite abelian group in which all proper subgroups are finite. And your fact is wrong when $G = C_{p^infty} times C_p$, which has the unique maximal subgroup $C_{p^inf ty}$.
– Derek Holt
Nov 15 at 21:13
1
Your original version said nothing about finiteness, so why would you expect anyone to think that you were talking about finite groups?
– Derek Holt
Nov 16 at 8:05
1
Anyway, you can solve your problem by taking $M$ to be a normal subgroup of prime index, and then $MN$ is a subgroup.
– Derek Holt
Nov 16 at 8:05
|
show 2 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $G$ be a nontrivial finite solvable group. We want to show that $G$ has proper subgroups like $A$ and $B$ such that $G=AB$ iff $G$ isn't a cyclic $p$-group.
I use the below fact to solve the problem.
" Let $G$ be a nontrivial finite group. $G$ has one and only one maximal subgroup iff $G$ be a cyclic group of order $p^{m}$ such that $p$ is a prime number. "
I can solve one side but I have problem on the other side.
I suppose that $G$ isn't a cyclic $p$-group. Therefore $G$ has more than one maximal subgroup. Let $M$ and $N$ be maximal subgroups of $G$ such that $M neq N$. I want to show $G = MN $ but I don't know how.
group-theory finite-groups p-groups solvable-groups
asked Nov 15 at 16:19
Yasmin
10310
Let $G$ be a nontrivial finite solvable group. We want to show that $G$ has proper subgroups like $A$ and $B$ such that $G=AB$ iff $G$ isn't a cyclic $p$-group.
I use the below fact to solve the problem.
" Let $G$ be a nontrivial finite group. $G$ has one and only one maximal subgroup iff $G$ be a cyclic group of order $p^{m}$ such that $p$ is a prime number. "
I can solve one side but I have problem on the other side.
I suppose that $G$ isn't a cyclic $p$-group. Therefore $G$ has more than one maximal subgroup. Let $M$ and $N$ be maximal subgroups of $G$ such that $M neq N$. I want to show $G = MN $ but I don't know how.
group-theory finite-groups p-groups solvable-groups
group-theory finite-groups p-groups solvable-groups
asked Nov 15 at 16:19
Yasmin
10310
asked Nov 15 at 16:19
Yasmin
10310
edited Nov 16 at 6:12
asked Nov 15 at 16:19
Yasmin
10310
asked Nov 15 at 16:19
Yasmin
10310
asked Nov 15 at 16:19
Yasmin
10310
10310
1
$MN$ is a subgroup of $G$ and contains both $M$ and $N$. Does that help?
– user3482749
Nov 15 at 16:21
@user3482749Why $MN$ is a subgroup?If so, yes, we have $ M lneqq MN leq G $ and as $M$ is a maximal subgroup of $G$, we have $G=MN$.
– Yasmin
Nov 15 at 16:43
The result is false when $G = C_{p^infty}$, which is an infinite abelian group in which all proper subgroups are finite. And your fact is wrong when $G = C_{p^infty} times C_p$, which has the unique maximal subgroup $C_{p^inf ty}$.
– Derek Holt
Nov 15 at 21:13
1
Your original version said nothing about finiteness, so why would you expect anyone to think that you were talking about finite groups?
– Derek Holt
Nov 16 at 8:05
1
Anyway, you can solve your problem by taking $M$ to be a normal subgroup of prime index, and then $MN$ is a subgroup.
– Derek Holt
Nov 16 at 8:05
|
show 2 more comments
1
$MN$ is a subgroup of $G$ and contains both $M$ and $N$. Does that help?
– user3482749
Nov 15 at 16:21
@user3482749Why $MN$ is a subgroup?If so, yes, we have $ M lneqq MN leq G $ and as $M$ is a maximal subgroup of $G$, we have $G=MN$.
– Yasmin
Nov 15 at 16:43
The result is false when $G = C_{p^infty}$, which is an infinite abelian group in which all proper subgroups are finite. And your fact is wrong when $G = C_{p^infty} times C_p$, which has the unique maximal subgroup $C_{p^inf ty}$.
– Derek Holt
Nov 15 at 21:13
1
Your original version said nothing about finiteness, so why would you expect anyone to think that you were talking about finite groups?
– Derek Holt
Nov 16 at 8:05
1
Anyway, you can solve your problem by taking $M$ to be a normal subgroup of prime index, and then $MN$ is a subgroup.
– Derek Holt
Nov 16 at 8:05
1
1
$MN$ is a subgroup of $G$ and contains both $M$ and $N$. Does that help?
– user3482749
Nov 15 at 16:21
$MN$ is a subgroup of $G$ and contains both $M$ and $N$. Does that help?
– user3482749
Nov 15 at 16:21
@user3482749Why $MN$ is a subgroup?If so, yes, we have $ M lneqq MN leq G $ and as $M$ is a maximal subgroup of $G$, we have $G=MN$.
– Yasmin
Nov 15 at 16:43
@user3482749Why $MN$ is a subgroup?If so, yes, we have $ M lneqq MN leq G $ and as $M$ is a maximal subgroup of $G$, we have $G=MN$.
– Yasmin
Nov 15 at 16:43
The result is false when $G = C_{p^infty}$, which is an infinite abelian group in which all proper subgroups are finite. And your fact is wrong when $G = C_{p^infty} times C_p$, which has the unique maximal subgroup $C_{p^inf ty}$.
– Derek Holt
Nov 15 at 21:13
The result is false when $G = C_{p^infty}$, which is an infinite abelian group in which all proper subgroups are finite. And your fact is wrong when $G = C_{p^infty} times C_p$, which has the unique maximal subgroup $C_{p^inf ty}$.
– Derek Holt
Nov 15 at 21:13
1
1
Your original version said nothing about finiteness, so why would you expect anyone to think that you were talking about finite groups?
– Derek Holt
Nov 16 at 8:05
Your original version said nothing about finiteness, so why would you expect anyone to think that you were talking about finite groups?
– Derek Holt
Nov 16 at 8:05
1
1
Anyway, you can solve your problem by taking $M$ to be a normal subgroup of prime index, and then $MN$ is a subgroup.
– Derek Holt
Nov 16 at 8:05
Anyway, you can solve your problem by taking $M$ to be a normal subgroup of prime index, and then $MN$ is a subgroup.
– Derek Holt
Nov 16 at 8:05
|
show 2 more comments
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1
$MN$ is a subgroup of $G$ and contains both $M$ and $N$. Does that help?
– user3482749
Nov 15 at 16:21
@user3482749Why $MN$ is a subgroup?If so, yes, we have $ M lneqq MN leq G $ and as $M$ is a maximal subgroup of $G$, we have $G=MN$.
– Yasmin
Nov 15 at 16:43
The result is false when $G = C_{p^infty}$, which is an infinite abelian group in which all proper subgroups are finite. And your fact is wrong when $G = C_{p^infty} times C_p$, which has the unique maximal subgroup $C_{p^inf ty}$.
– Derek Holt
Nov 15 at 21:13
1
Your original version said nothing about finiteness, so why would you expect anyone to think that you were talking about finite groups?
– Derek Holt
Nov 16 at 8:05
1
Anyway, you can solve your problem by taking $M$ to be a normal subgroup of prime index, and then $MN$ is a subgroup.
– Derek Holt
Nov 16 at 8:05