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Let $G$ be a group of order $p^4$ for $p$ prime. Prove that $G'$ is abelian
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Let $G$ be a group of order $p^4$ for $p$ prime. Prove that $G'$ is abelian where $G'$ is the commutator subgroup of $G$.
group-theory finite-groups abelian-groups
edited Jul 24 at 12:36
Daniel Fischer♦
171k16154274
asked Jul 24 at 10:31
Rimon ben david
244
 |Â
show 1 more comment
up vote
4
down vote
favorite
Let $G$ be a group of order $p^4$ for $p$ prime. Prove that $G'$ is abelian where $G'$ is the commutator subgroup of $G$.
group-theory finite-groups abelian-groups
edited Jul 24 at 12:36
Daniel Fischer♦
171k16154274
asked Jul 24 at 10:31
Rimon ben david
244
2
Are you familiar with the result that any group of order $p^2$ is abelian?
– Tobias Kildetoft
Jul 24 at 10:33
yes but I don't understand how I can use it here. I thought about the possible order of G' but I got stuck in the middle. The possible order is 1,p,p^2,p^3,p^4 but I don't know how to proceed
– Rimon ben david
Jul 24 at 10:37
Are you also familiar with the fact that $G$ will have a normal subgroup of any of those orders?
– Tobias Kildetoft
Jul 24 at 10:38
No :(, and how I can use it?
– Rimon ben david
Jul 24 at 10:39
2
Then show that and then think about how you can use it.
– Tobias Kildetoft
Jul 24 at 10:42
 |Â
show 1 more comment
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $G$ be a group of order $p^4$ for $p$ prime. Prove that $G'$ is abelian where $G'$ is the commutator subgroup of $G$.
group-theory finite-groups abelian-groups
edited Jul 24 at 12:36
Daniel Fischer♦
171k16154274
asked Jul 24 at 10:31
Rimon ben david
244
Let $G$ be a group of order $p^4$ for $p$ prime. Prove that $G'$ is abelian where $G'$ is the commutator subgroup of $G$.
group-theory finite-groups abelian-groups
edited Jul 24 at 12:36
Daniel Fischer♦
171k16154274
asked Jul 24 at 10:31
Rimon ben david
244
edited Jul 24 at 12:36
Daniel Fischer♦
171k16154274
edited Jul 24 at 12:36
Daniel Fischer♦
171k16154274
edited Jul 24 at 12:36
Daniel Fischer♦
171k16154274
171k16154274
asked Jul 24 at 10:31
Rimon ben david
244
asked Jul 24 at 10:31
Rimon ben david
244
asked Jul 24 at 10:31
Rimon ben david
244
244
2
Are you familiar with the result that any group of order $p^2$ is abelian?
– Tobias Kildetoft
Jul 24 at 10:33
yes but I don't understand how I can use it here. I thought about the possible order of G' but I got stuck in the middle. The possible order is 1,p,p^2,p^3,p^4 but I don't know how to proceed
– Rimon ben david
Jul 24 at 10:37
Are you also familiar with the fact that $G$ will have a normal subgroup of any of those orders?
– Tobias Kildetoft
Jul 24 at 10:38
No :(, and how I can use it?
– Rimon ben david
Jul 24 at 10:39
2
Then show that and then think about how you can use it.
– Tobias Kildetoft
Jul 24 at 10:42
 |Â
show 1 more comment
2
Are you familiar with the result that any group of order $p^2$ is abelian?
– Tobias Kildetoft
Jul 24 at 10:33
yes but I don't understand how I can use it here. I thought about the possible order of G' but I got stuck in the middle. The possible order is 1,p,p^2,p^3,p^4 but I don't know how to proceed
– Rimon ben david
Jul 24 at 10:37
Are you also familiar with the fact that $G$ will have a normal subgroup of any of those orders?
– Tobias Kildetoft
Jul 24 at 10:38
No :(, and how I can use it?
– Rimon ben david
Jul 24 at 10:39
2
Then show that and then think about how you can use it.
– Tobias Kildetoft
Jul 24 at 10:42
2
2
Are you familiar with the result that any group of order $p^2$ is abelian?
– Tobias Kildetoft
Jul 24 at 10:33
Are you familiar with the result that any group of order $p^2$ is abelian?
– Tobias Kildetoft
Jul 24 at 10:33
yes but I don't understand how I can use it here. I thought about the possible order of G' but I got stuck in the middle. The possible order is 1,p,p^2,p^3,p^4 but I don't know how to proceed
– Rimon ben david
Jul 24 at 10:37
yes but I don't understand how I can use it here. I thought about the possible order of G' but I got stuck in the middle. The possible order is 1,p,p^2,p^3,p^4 but I don't know how to proceed
– Rimon ben david
Jul 24 at 10:37
Are you also familiar with the fact that $G$ will have a normal subgroup of any of those orders?
– Tobias Kildetoft
Jul 24 at 10:38
Are you also familiar with the fact that $G$ will have a normal subgroup of any of those orders?
– Tobias Kildetoft
Jul 24 at 10:38
No :(, and how I can use it?
– Rimon ben david
Jul 24 at 10:39
No :(, and how I can use it?
– Rimon ben david
Jul 24 at 10:39
2
2
Then show that and then think about how you can use it.
– Tobias Kildetoft
Jul 24 at 10:42
Then show that and then think about how you can use it.
– Tobias Kildetoft
Jul 24 at 10:42
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
10
down vote
Since the center of a $p$-group is non-trivial then it follows that every group of order $p^2$ is abelian. And so is every group of order $1$ or $p$. So it is enough to show that $G'$ is of order $1$, $p$ or $p^2$.
Since $G$ has a normal subgroup $N$ of order $p^2$ (actually any $p^k$) then $G/N$ is of order $p^2$ as well. But we've already established that groups of order $p^2$ are abelian. Since $G/N$ is abelian then $G'subseteq N$ meaning $|G'|leq p^2$. $Box$
answered Jul 24 at 11:19
freakish
8,4771524
add a comment |Â
up vote
1
down vote
Using some finite $p$-group theory: since $G$ is solvable $G' subsetneq G$, hence $|G:G'| geq p$. But $|G:G'|=p$ implies that $G/G'$ is cyclic and hence $G$ would be cyclic. (Proof: look at the Frattini subgroup $Phi(G)=G'G^p$. $G/G'$ cyclic implies $G/Phi(G)$ being cyclic, say $G/Phi(G)=langle bar gPhi(G) rangle$. And remember that $Phi(G)$ are non-generators. This implies that $G=langle g ranglePhi(G)$, and since $Phi(G)$ are non-generators, $G=langle g rangle$ is cyclic.). Hence we may assume $|G'| leq p^2$ and groups of order dividing $p^2$ are abelian.
answered Jul 24 at 14:54
Nicky Hekster
26.9k53052
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
Since the center of a $p$-group is non-trivial then it follows that every group of order $p^2$ is abelian. And so is every group of order $1$ or $p$. So it is enough to show that $G'$ is of order $1$, $p$ or $p^2$.
Since $G$ has a normal subgroup $N$ of order $p^2$ (actually any $p^k$) then $G/N$ is of order $p^2$ as well. But we've already established that groups of order $p^2$ are abelian. Since $G/N$ is abelian then $G'subseteq N$ meaning $|G'|leq p^2$. $Box$
answered Jul 24 at 11:19
freakish
8,4771524
add a comment |Â
up vote
10
down vote
Since the center of a $p$-group is non-trivial then it follows that every group of order $p^2$ is abelian. And so is every group of order $1$ or $p$. So it is enough to show that $G'$ is of order $1$, $p$ or $p^2$.
Since $G$ has a normal subgroup $N$ of order $p^2$ (actually any $p^k$) then $G/N$ is of order $p^2$ as well. But we've already established that groups of order $p^2$ are abelian. Since $G/N$ is abelian then $G'subseteq N$ meaning $|G'|leq p^2$. $Box$
answered Jul 24 at 11:19
freakish
8,4771524
add a comment |Â
up vote
10
down vote
up vote
10
down vote
Since the center of a $p$-group is non-trivial then it follows that every group of order $p^2$ is abelian. And so is every group of order $1$ or $p$. So it is enough to show that $G'$ is of order $1$, $p$ or $p^2$.
Since $G$ has a normal subgroup $N$ of order $p^2$ (actually any $p^k$) then $G/N$ is of order $p^2$ as well. But we've already established that groups of order $p^2$ are abelian. Since $G/N$ is abelian then $G'subseteq N$ meaning $|G'|leq p^2$. $Box$
answered Jul 24 at 11:19
freakish
8,4771524
Since the center of a $p$-group is non-trivial then it follows that every group of order $p^2$ is abelian. And so is every group of order $1$ or $p$. So it is enough to show that $G'$ is of order $1$, $p$ or $p^2$.
Since $G$ has a normal subgroup $N$ of order $p^2$ (actually any $p^k$) then $G/N$ is of order $p^2$ as well. But we've already established that groups of order $p^2$ are abelian. Since $G/N$ is abelian then $G'subseteq N$ meaning $|G'|leq p^2$. $Box$
answered Jul 24 at 11:19
freakish
8,4771524
answered Jul 24 at 11:19
freakish
8,4771524
answered Jul 24 at 11:19
freakish
8,4771524
answered Jul 24 at 11:19
freakish
8,4771524
8,4771524
add a comment |Â
add a comment |Â
up vote
1
down vote
Using some finite $p$-group theory: since $G$ is solvable $G' subsetneq G$, hence $|G:G'| geq p$. But $|G:G'|=p$ implies that $G/G'$ is cyclic and hence $G$ would be cyclic. (Proof: look at the Frattini subgroup $Phi(G)=G'G^p$. $G/G'$ cyclic implies $G/Phi(G)$ being cyclic, say $G/Phi(G)=langle bar gPhi(G) rangle$. And remember that $Phi(G)$ are non-generators. This implies that $G=langle g ranglePhi(G)$, and since $Phi(G)$ are non-generators, $G=langle g rangle$ is cyclic.). Hence we may assume $|G'| leq p^2$ and groups of order dividing $p^2$ are abelian.
answered Jul 24 at 14:54
Nicky Hekster
26.9k53052
add a comment |Â
up vote
1
down vote
Using some finite $p$-group theory: since $G$ is solvable $G' subsetneq G$, hence $|G:G'| geq p$. But $|G:G'|=p$ implies that $G/G'$ is cyclic and hence $G$ would be cyclic. (Proof: look at the Frattini subgroup $Phi(G)=G'G^p$. $G/G'$ cyclic implies $G/Phi(G)$ being cyclic, say $G/Phi(G)=langle bar gPhi(G) rangle$. And remember that $Phi(G)$ are non-generators. This implies that $G=langle g ranglePhi(G)$, and since $Phi(G)$ are non-generators, $G=langle g rangle$ is cyclic.). Hence we may assume $|G'| leq p^2$ and groups of order dividing $p^2$ are abelian.
answered Jul 24 at 14:54
Nicky Hekster
26.9k53052
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Using some finite $p$-group theory: since $G$ is solvable $G' subsetneq G$, hence $|G:G'| geq p$. But $|G:G'|=p$ implies that $G/G'$ is cyclic and hence $G$ would be cyclic. (Proof: look at the Frattini subgroup $Phi(G)=G'G^p$. $G/G'$ cyclic implies $G/Phi(G)$ being cyclic, say $G/Phi(G)=langle bar gPhi(G) rangle$. And remember that $Phi(G)$ are non-generators. This implies that $G=langle g ranglePhi(G)$, and since $Phi(G)$ are non-generators, $G=langle g rangle$ is cyclic.). Hence we may assume $|G'| leq p^2$ and groups of order dividing $p^2$ are abelian.
answered Jul 24 at 14:54
Nicky Hekster
26.9k53052
Using some finite $p$-group theory: since $G$ is solvable $G' subsetneq G$, hence $|G:G'| geq p$. But $|G:G'|=p$ implies that $G/G'$ is cyclic and hence $G$ would be cyclic. (Proof: look at the Frattini subgroup $Phi(G)=G'G^p$. $G/G'$ cyclic implies $G/Phi(G)$ being cyclic, say $G/Phi(G)=langle bar gPhi(G) rangle$. And remember that $Phi(G)$ are non-generators. This implies that $G=langle g ranglePhi(G)$, and since $Phi(G)$ are non-generators, $G=langle g rangle$ is cyclic.). Hence we may assume $|G'| leq p^2$ and groups of order dividing $p^2$ are abelian.
answered Jul 24 at 14:54
Nicky Hekster
26.9k53052
answered Jul 24 at 14:54
Nicky Hekster
26.9k53052
answered Jul 24 at 14:54
Nicky Hekster
26.9k53052
answered Jul 24 at 14:54
Nicky Hekster
26.9k53052
26.9k53052
add a comment |Â
add a comment |Â
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2
Are you familiar with the result that any group of order $p^2$ is abelian?
– Tobias Kildetoft
Jul 24 at 10:33
yes but I don't understand how I can use it here. I thought about the possible order of G' but I got stuck in the middle. The possible order is 1,p,p^2,p^3,p^4 but I don't know how to proceed
– Rimon ben david
Jul 24 at 10:37
Are you also familiar with the fact that $G$ will have a normal subgroup of any of those orders?
– Tobias Kildetoft
Jul 24 at 10:38
No :(, and how I can use it?
– Rimon ben david
Jul 24 at 10:39
2
Then show that and then think about how you can use it.
– Tobias Kildetoft
Jul 24 at 10:42