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Every order $p$ subgroup is normal in a $p$ group ($p$ here denotes a prime number)?

I think that it's not true, I'm but unable to construct an example. Any hints?

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    $\begingroup$ I assume $p$ is a prime? $\endgroup$ – Luiz Cordeiro Sep 19 '16 at 19:01
  • $\begingroup$ @LuizCordeiro Yes. Btw,do people use $p$ group terminilogy even if $p$ is not prime? $\endgroup$ – Math Lover Sep 19 '16 at 19:04
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    $\begingroup$ Hint: If there is an example, it must be nonabelian. What is the smallest nonabelian $p$-group you can think of? $\endgroup$ – Wojowu Sep 19 '16 at 19:07
  • $\begingroup$ @Wojowu Quaternion group of order $8$,but every subgroup is abelian. $\endgroup$ – Math Lover Sep 19 '16 at 19:08
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    $\begingroup$ Abelianness of a subgroup doesn't imply it's normal. Unfortunately, in this example, there is a unique order $2$ subgroup which is normal. But there is another group of this order: $D_4$. $\endgroup$ – Wojowu Sep 19 '16 at 19:19
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A direct counter example: take the dihedral group

$$D_{2\cdot4}=\langle\,r,s\;/\;s^2=r^4=1\,,\,\,srs=r^{-1}\,\rangle$$

and now, take for example $\;H=\left\{\,1,\,\,r^2s\,\right\}\;$ . This is a subgroup since

$$r^2s\cdot r^2s=r^2\left(srs\right)^2=r^2\cdot r^{-2}=1\in H$$

but it is not normal since

$$r^{-1}(r^2s)r=rsr\neq\begin{cases}1\iff r=r^{-1}s=r^3s\implies r^2s=1\,,\,\text{contradiction}\\{}\\r^2s\iff sr=rs\iff D_{2\cdot3}\;\text{is abelian, contradiction}\end{cases}$$

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This is not necessarily true.

Exercise: Let $G,H$ be groups and $H\curvearrowright G$ an action by automorphisms. Then $H$ is normal in the semidirect product $G\rtimes H$ if and only if the action of $H$ on $G$ is trivial.

Let $C_p=\langle x:x^p=1\rangle$ be the cyclic group of order $p$, $G=(C_p)^p$ and let $H=C_p$ act on $G$ by permuting entries: $x\cdot(g_1,g_2,\ldots,g_p)=(g_p,g_1,g_2,\ldots,g_{p-1})$. Then $H$ is a non-normal subgroup of order $p$ of the $p$-group $G\rtimes H$.

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