Non cyclic group of order $p^3$ satisfies $G simeq H rtimes_{theta}K$
$begingroup$
Let $G$ be a non-cyclic group of order $p^3$ for an odd prime $p$. Prove that $G simeq H rtimes_{theta}K$, where $H$ is a normal subgroup of $G$ of order $p^2$, $K$ is a subgroup of order $p$, and $theta : K to Aut(H)$ is a homomorphism.
I managed to prove that there exists a normal subgroup $H$ of order $p^2$. Then I took some $g in G-H$. If $g$ is of order $p$, I am done since $G simeq H rtimes langle g rangle $. But what if all $g in G - H$ are of order $p^2 $?
abstract-algebra group-theory finite-groups semidirect-product
edited Dec 24 '18 at 2:11
the_fox
2,89021537
asked Dec 23 '18 at 22:08
user401516user401516
92039
$endgroup$
add a comment |
$begingroup$
Let $G$ be a non-cyclic group of order $p^3$ for an odd prime $p$. Prove that $G simeq H rtimes_{theta}K$, where $H$ is a normal subgroup of $G$ of order $p^2$, $K$ is a subgroup of order $p$, and $theta : K to Aut(H)$ is a homomorphism.
I managed to prove that there exists a normal subgroup $H$ of order $p^2$. Then I took some $g in G-H$. If $g$ is of order $p$, I am done since $G simeq H rtimes langle g rangle $. But what if all $g in G - H$ are of order $p^2 $?
abstract-algebra group-theory finite-groups semidirect-product
edited Dec 24 '18 at 2:11
the_fox
2,89021537
asked Dec 23 '18 at 22:08
user401516user401516
92039
$endgroup$
1
$begingroup$
Impossible that all the non-trivial elements in $;G;$ are of order $;p^2;$ since there exists a subgroup of order $;p;$ ...
$endgroup$
– DonAntonio
Dec 23 '18 at 22:10
$begingroup$
Sorry, I meant all $g in G - H$. I edited my question
$endgroup$
– user401516
Dec 23 '18 at 22:18
1
$begingroup$
Observe that the conclusion would be false when $p=2$. The quaternion group $Q_8$. In other words, the fact that $p$ is odd must come into play somehow.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:36
1
$begingroup$
The claim seems to follow from the observations made by the OP and the argument given in this old answer by Arturo Magidin.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 7:21
add a comment |
$begingroup$
Let $G$ be a non-cyclic group of order $p^3$ for an odd prime $p$. Prove that $G simeq H rtimes_{theta}K$, where $H$ is a normal subgroup of $G$ of order $p^2$, $K$ is a subgroup of order $p$, and $theta : K to Aut(H)$ is a homomorphism.
I managed to prove that there exists a normal subgroup $H$ of order $p^2$. Then I took some $g in G-H$. If $g$ is of order $p$, I am done since $G simeq H rtimes langle g rangle $. But what if all $g in G - H$ are of order $p^2 $?
abstract-algebra group-theory finite-groups semidirect-product
edited Dec 24 '18 at 2:11
the_fox
2,89021537
asked Dec 23 '18 at 22:08
user401516user401516
92039
$endgroup$
Let $G$ be a non-cyclic group of order $p^3$ for an odd prime $p$. Prove that $G simeq H rtimes_{theta}K$, where $H$ is a normal subgroup of $G$ of order $p^2$, $K$ is a subgroup of order $p$, and $theta : K to Aut(H)$ is a homomorphism.
I managed to prove that there exists a normal subgroup $H$ of order $p^2$. Then I took some $g in G-H$. If $g$ is of order $p$, I am done since $G simeq H rtimes langle g rangle $. But what if all $g in G - H$ are of order $p^2 $?
abstract-algebra group-theory finite-groups semidirect-product
abstract-algebra group-theory finite-groups semidirect-product
edited Dec 24 '18 at 2:11
the_fox
2,89021537
asked Dec 23 '18 at 22:08
user401516user401516
92039
edited Dec 24 '18 at 2:11
the_fox
2,89021537
asked Dec 23 '18 at 22:08
user401516user401516
92039
edited Dec 24 '18 at 2:11
the_fox
2,89021537
edited Dec 24 '18 at 2:11
the_fox
2,89021537
edited Dec 24 '18 at 2:11
the_fox
2,89021537
2,89021537
asked Dec 23 '18 at 22:08
user401516user401516
92039
asked Dec 23 '18 at 22:08
user401516user401516
92039
asked Dec 23 '18 at 22:08
user401516user401516
92039
92039
1
$begingroup$
Impossible that all the non-trivial elements in $;G;$ are of order $;p^2;$ since there exists a subgroup of order $;p;$ ...
$endgroup$
– DonAntonio
Dec 23 '18 at 22:10
$begingroup$
Sorry, I meant all $g in G - H$. I edited my question
$endgroup$
– user401516
Dec 23 '18 at 22:18
1
$begingroup$
Observe that the conclusion would be false when $p=2$. The quaternion group $Q_8$. In other words, the fact that $p$ is odd must come into play somehow.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:36
1
$begingroup$
The claim seems to follow from the observations made by the OP and the argument given in this old answer by Arturo Magidin.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 7:21
add a comment |
1
$begingroup$
Impossible that all the non-trivial elements in $;G;$ are of order $;p^2;$ since there exists a subgroup of order $;p;$ ...
$endgroup$
– DonAntonio
Dec 23 '18 at 22:10
$begingroup$
Sorry, I meant all $g in G - H$. I edited my question
$endgroup$
– user401516
Dec 23 '18 at 22:18
1
$begingroup$
Observe that the conclusion would be false when $p=2$. The quaternion group $Q_8$. In other words, the fact that $p$ is odd must come into play somehow.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:36
1
$begingroup$
The claim seems to follow from the observations made by the OP and the argument given in this old answer by Arturo Magidin.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 7:21
1
1
$begingroup$
Impossible that all the non-trivial elements in $;G;$ are of order $;p^2;$ since there exists a subgroup of order $;p;$ ...
$endgroup$
– DonAntonio
Dec 23 '18 at 22:10
$begingroup$
Impossible that all the non-trivial elements in $;G;$ are of order $;p^2;$ since there exists a subgroup of order $;p;$ ...
$endgroup$
– DonAntonio
Dec 23 '18 at 22:10
$begingroup$
Sorry, I meant all $g in G - H$. I edited my question
$endgroup$
– user401516
Dec 23 '18 at 22:18
$begingroup$
Sorry, I meant all $g in G - H$. I edited my question
$endgroup$
– user401516
Dec 23 '18 at 22:18
1
1
$begingroup$
Observe that the conclusion would be false when $p=2$. The quaternion group $Q_8$. In other words, the fact that $p$ is odd must come into play somehow.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:36
$begingroup$
Observe that the conclusion would be false when $p=2$. The quaternion group $Q_8$. In other words, the fact that $p$ is odd must come into play somehow.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:36
1
1
$begingroup$
The claim seems to follow from the observations made by the OP and the argument given in this old answer by Arturo Magidin.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 7:21
$begingroup$
The claim seems to follow from the observations made by the OP and the argument given in this old answer by Arturo Magidin.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 7:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The result follows immediately if all elements of $G$ have order $1$ or $p$, so we can assume that there exists an element of order $p^2$. This generates a cyclic subgroup of index $p$, which is normal in $G$, so we can assume that $H$ is cyclic of order $p^2$.
So let $g in G setminus H$, and assume that $g$ has order $p^2$ (if it has order $p$ then we are done). So $g^p = h^p$ for some $h in H$. We claim that $(gh^{-1})^p=1$, and hence $gh^{-1}$ has order $p$ and we are done.
This is immediate if $G$ is abelian, so suppose not. Then $[G,G]=Z(G)$ has order $p$, so commutators are central, and $(xy^{-1})^p = x^py^{-p}[y^{-1},x]^{p(p-1)/2}$ (where $[a,b]$ denotes the commutator $a^{-1}b^{-1}ab$).
Then, since $p$ is odd and $[y^{-1},x]$ has order $p$, we get $[y^{-1},x]^{p(p-1)/2}=1$ and the claim follows. (Note that this result is false when $p=2$, and the quaternion group $Q_8$ is a counterexample.)
answered Dec 24 '18 at 10:22
Derek HoltDerek Holt
53.7k53571
$endgroup$
$begingroup$
Thank you for your answer. Can you please explain why $(xy^{-1})^p=x^py^{-p}[y^{-1},x]^{p(p-1)/2}$ ?
$endgroup$
– user401516
Dec 24 '18 at 12:30
1
$begingroup$
If $[G,G] le Z(G)$ then $(ab)^k = a^kb^k[b,a]^{k(k-1)/2}$ for all $a,b in G$ and all $k ge 0$. Prove it by induction on $k$, using $ba = ab[b,a]$.
$endgroup$
– Derek Holt
Dec 24 '18 at 15:24
$begingroup$
Where did you use the fact that $H$ is cyclic?
$endgroup$
– user401516
Dec 24 '18 at 21:45
$begingroup$
The existence of $h in H$ such that $g^p=h^p$.
$endgroup$
– Derek Holt
Dec 24 '18 at 22:45
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The result follows immediately if all elements of $G$ have order $1$ or $p$, so we can assume that there exists an element of order $p^2$. This generates a cyclic subgroup of index $p$, which is normal in $G$, so we can assume that $H$ is cyclic of order $p^2$.
So let $g in G setminus H$, and assume that $g$ has order $p^2$ (if it has order $p$ then we are done). So $g^p = h^p$ for some $h in H$. We claim that $(gh^{-1})^p=1$, and hence $gh^{-1}$ has order $p$ and we are done.
This is immediate if $G$ is abelian, so suppose not. Then $[G,G]=Z(G)$ has order $p$, so commutators are central, and $(xy^{-1})^p = x^py^{-p}[y^{-1},x]^{p(p-1)/2}$ (where $[a,b]$ denotes the commutator $a^{-1}b^{-1}ab$).
Then, since $p$ is odd and $[y^{-1},x]$ has order $p$, we get $[y^{-1},x]^{p(p-1)/2}=1$ and the claim follows. (Note that this result is false when $p=2$, and the quaternion group $Q_8$ is a counterexample.)
answered Dec 24 '18 at 10:22
Derek HoltDerek Holt
53.7k53571
$endgroup$
$begingroup$
Thank you for your answer. Can you please explain why $(xy^{-1})^p=x^py^{-p}[y^{-1},x]^{p(p-1)/2}$ ?
$endgroup$
– user401516
Dec 24 '18 at 12:30
1
$begingroup$
If $[G,G] le Z(G)$ then $(ab)^k = a^kb^k[b,a]^{k(k-1)/2}$ for all $a,b in G$ and all $k ge 0$. Prove it by induction on $k$, using $ba = ab[b,a]$.
$endgroup$
– Derek Holt
Dec 24 '18 at 15:24
$begingroup$
Where did you use the fact that $H$ is cyclic?
$endgroup$
– user401516
Dec 24 '18 at 21:45
$begingroup$
The existence of $h in H$ such that $g^p=h^p$.
$endgroup$
– Derek Holt
Dec 24 '18 at 22:45
add a comment |
$begingroup$
The result follows immediately if all elements of $G$ have order $1$ or $p$, so we can assume that there exists an element of order $p^2$. This generates a cyclic subgroup of index $p$, which is normal in $G$, so we can assume that $H$ is cyclic of order $p^2$.
So let $g in G setminus H$, and assume that $g$ has order $p^2$ (if it has order $p$ then we are done). So $g^p = h^p$ for some $h in H$. We claim that $(gh^{-1})^p=1$, and hence $gh^{-1}$ has order $p$ and we are done.
This is immediate if $G$ is abelian, so suppose not. Then $[G,G]=Z(G)$ has order $p$, so commutators are central, and $(xy^{-1})^p = x^py^{-p}[y^{-1},x]^{p(p-1)/2}$ (where $[a,b]$ denotes the commutator $a^{-1}b^{-1}ab$).
Then, since $p$ is odd and $[y^{-1},x]$ has order $p$, we get $[y^{-1},x]^{p(p-1)/2}=1$ and the claim follows. (Note that this result is false when $p=2$, and the quaternion group $Q_8$ is a counterexample.)
answered Dec 24 '18 at 10:22
Derek HoltDerek Holt
53.7k53571
$endgroup$
$begingroup$
Thank you for your answer. Can you please explain why $(xy^{-1})^p=x^py^{-p}[y^{-1},x]^{p(p-1)/2}$ ?
$endgroup$
– user401516
Dec 24 '18 at 12:30
1
$begingroup$
If $[G,G] le Z(G)$ then $(ab)^k = a^kb^k[b,a]^{k(k-1)/2}$ for all $a,b in G$ and all $k ge 0$. Prove it by induction on $k$, using $ba = ab[b,a]$.
$endgroup$
– Derek Holt
Dec 24 '18 at 15:24
$begingroup$
Where did you use the fact that $H$ is cyclic?
$endgroup$
– user401516
Dec 24 '18 at 21:45
$begingroup$
The existence of $h in H$ such that $g^p=h^p$.
$endgroup$
– Derek Holt
Dec 24 '18 at 22:45
add a comment |
$begingroup$
The result follows immediately if all elements of $G$ have order $1$ or $p$, so we can assume that there exists an element of order $p^2$. This generates a cyclic subgroup of index $p$, which is normal in $G$, so we can assume that $H$ is cyclic of order $p^2$.
So let $g in G setminus H$, and assume that $g$ has order $p^2$ (if it has order $p$ then we are done). So $g^p = h^p$ for some $h in H$. We claim that $(gh^{-1})^p=1$, and hence $gh^{-1}$ has order $p$ and we are done.
This is immediate if $G$ is abelian, so suppose not. Then $[G,G]=Z(G)$ has order $p$, so commutators are central, and $(xy^{-1})^p = x^py^{-p}[y^{-1},x]^{p(p-1)/2}$ (where $[a,b]$ denotes the commutator $a^{-1}b^{-1}ab$).
Then, since $p$ is odd and $[y^{-1},x]$ has order $p$, we get $[y^{-1},x]^{p(p-1)/2}=1$ and the claim follows. (Note that this result is false when $p=2$, and the quaternion group $Q_8$ is a counterexample.)
answered Dec 24 '18 at 10:22
Derek HoltDerek Holt
53.7k53571
$endgroup$
The result follows immediately if all elements of $G$ have order $1$ or $p$, so we can assume that there exists an element of order $p^2$. This generates a cyclic subgroup of index $p$, which is normal in $G$, so we can assume that $H$ is cyclic of order $p^2$.
So let $g in G setminus H$, and assume that $g$ has order $p^2$ (if it has order $p$ then we are done). So $g^p = h^p$ for some $h in H$. We claim that $(gh^{-1})^p=1$, and hence $gh^{-1}$ has order $p$ and we are done.
This is immediate if $G$ is abelian, so suppose not. Then $[G,G]=Z(G)$ has order $p$, so commutators are central, and $(xy^{-1})^p = x^py^{-p}[y^{-1},x]^{p(p-1)/2}$ (where $[a,b]$ denotes the commutator $a^{-1}b^{-1}ab$).
Then, since $p$ is odd and $[y^{-1},x]$ has order $p$, we get $[y^{-1},x]^{p(p-1)/2}=1$ and the claim follows. (Note that this result is false when $p=2$, and the quaternion group $Q_8$ is a counterexample.)
answered Dec 24 '18 at 10:22
Derek HoltDerek Holt
53.7k53571
edited Dec 24 '18 at 11:01
answered Dec 24 '18 at 10:22
Derek HoltDerek Holt
53.7k53571
answered Dec 24 '18 at 10:22
Derek HoltDerek Holt
53.7k53571
answered Dec 24 '18 at 10:22
Derek HoltDerek Holt
53.7k53571
53.7k53571
$begingroup$
Thank you for your answer. Can you please explain why $(xy^{-1})^p=x^py^{-p}[y^{-1},x]^{p(p-1)/2}$ ?
$endgroup$
– user401516
Dec 24 '18 at 12:30
1
$begingroup$
If $[G,G] le Z(G)$ then $(ab)^k = a^kb^k[b,a]^{k(k-1)/2}$ for all $a,b in G$ and all $k ge 0$. Prove it by induction on $k$, using $ba = ab[b,a]$.
$endgroup$
– Derek Holt
Dec 24 '18 at 15:24
$begingroup$
Where did you use the fact that $H$ is cyclic?
$endgroup$
– user401516
Dec 24 '18 at 21:45
$begingroup$
The existence of $h in H$ such that $g^p=h^p$.
$endgroup$
– Derek Holt
Dec 24 '18 at 22:45
add a comment |
$begingroup$
Thank you for your answer. Can you please explain why $(xy^{-1})^p=x^py^{-p}[y^{-1},x]^{p(p-1)/2}$ ?
$endgroup$
– user401516
Dec 24 '18 at 12:30
1
$begingroup$
If $[G,G] le Z(G)$ then $(ab)^k = a^kb^k[b,a]^{k(k-1)/2}$ for all $a,b in G$ and all $k ge 0$. Prove it by induction on $k$, using $ba = ab[b,a]$.
$endgroup$
– Derek Holt
Dec 24 '18 at 15:24
$begingroup$
Where did you use the fact that $H$ is cyclic?
$endgroup$
– user401516
Dec 24 '18 at 21:45
$begingroup$
The existence of $h in H$ such that $g^p=h^p$.
$endgroup$
– Derek Holt
Dec 24 '18 at 22:45
$begingroup$
Thank you for your answer. Can you please explain why $(xy^{-1})^p=x^py^{-p}[y^{-1},x]^{p(p-1)/2}$ ?
$endgroup$
– user401516
Dec 24 '18 at 12:30
$begingroup$
Thank you for your answer. Can you please explain why $(xy^{-1})^p=x^py^{-p}[y^{-1},x]^{p(p-1)/2}$ ?
$endgroup$
– user401516
Dec 24 '18 at 12:30
1
1
$begingroup$
If $[G,G] le Z(G)$ then $(ab)^k = a^kb^k[b,a]^{k(k-1)/2}$ for all $a,b in G$ and all $k ge 0$. Prove it by induction on $k$, using $ba = ab[b,a]$.
$endgroup$
– Derek Holt
Dec 24 '18 at 15:24
$begingroup$
If $[G,G] le Z(G)$ then $(ab)^k = a^kb^k[b,a]^{k(k-1)/2}$ for all $a,b in G$ and all $k ge 0$. Prove it by induction on $k$, using $ba = ab[b,a]$.
$endgroup$
– Derek Holt
Dec 24 '18 at 15:24
$begingroup$
Where did you use the fact that $H$ is cyclic?
$endgroup$
– user401516
Dec 24 '18 at 21:45
$begingroup$
Where did you use the fact that $H$ is cyclic?
$endgroup$
– user401516
Dec 24 '18 at 21:45
$begingroup$
The existence of $h in H$ such that $g^p=h^p$.
$endgroup$
– Derek Holt
Dec 24 '18 at 22:45
$begingroup$
The existence of $h in H$ such that $g^p=h^p$.
$endgroup$
– Derek Holt
Dec 24 '18 at 22:45
add a comment |
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1
$begingroup$
Impossible that all the non-trivial elements in $;G;$ are of order $;p^2;$ since there exists a subgroup of order $;p;$ ...
$endgroup$
– DonAntonio
Dec 23 '18 at 22:10
$begingroup$
Sorry, I meant all $g in G - H$. I edited my question
$endgroup$
– user401516
Dec 23 '18 at 22:18
1
$begingroup$
Observe that the conclusion would be false when $p=2$. The quaternion group $Q_8$. In other words, the fact that $p$ is odd must come into play somehow.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 6:36
1
$begingroup$
The claim seems to follow from the observations made by the OP and the argument given in this old answer by Arturo Magidin.
$endgroup$
– Jyrki Lahtonen
Dec 24 '18 at 7:21