Non cyclic group of order $p^3$ satisfies $G simeq H rtimes_{theta}K$












7












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















7












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














7












7








7


2



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 2:11









the_fox

2,79221537




2,79221537










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














  • 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










1 Answer
1






active

oldest

votes


















3












$begingroup$

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.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









3












$begingroup$

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.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















3












$begingroup$

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.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














3












3








3





$begingroup$

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.)






share|cite|improve this answer











$endgroup$



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.)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 24 '18 at 11:01

























answered Dec 24 '18 at 10:22









Derek HoltDerek Holt

53.5k53571




53.5k53571












  • $begingroup$
    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








  • 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


















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Cabo Verde

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  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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Gyllenstierna

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