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

2,89021537




2,89021537










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

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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

53.7k53571




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


















  • $begingroup$
    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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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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