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4.4 Let $G$ be a finite p-group; show that if $H$ is a normal subgroup of G having order $p$, then $H$ is a subgroup of $Z(G)$.

$Z(G)=\left\{x\in G| xy=yx, \forall y\in G \right\}.$

Can you help me solve it and explain in detail? Thank you very much. Good health!

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closed as off-topic by Dietrich Burde, Javi, Derek Holt, Brian Borchers, Xander Henderson May 20 at 0:24

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  • 2
    $\begingroup$ Not following. You start by assuming that $H$ is a normal subgroup of $G$ and then you want to show that $H$ is a subgroup of $G$? $\endgroup$ – lulu May 19 at 15:01
  • $\begingroup$ "show that if 𝐻 is a normal subgroup of G having order 𝑝, then 𝐻 is a subgroup of 𝐺". What do you mean ? If $H$ is a normal subgroup of $G$, then $H$ is a subgroup... $\endgroup$ – user659895 May 19 at 15:02
  • $\begingroup$ You must have mixed words: to prove that normal subgroup is a subgroup is pretty trivial... $\endgroup$ – DonAntonio May 19 at 15:02
  • $\begingroup$ sorry, prove $H$ is a subgroup of $Z(G)$. $\endgroup$ – Trần Nam Sơn May 19 at 15:07
  • $\begingroup$ $Z(G)=\left\{x\in G| xy=yx, \forall y\in G \right\}.$ $\endgroup$ – Trần Nam Sơn May 19 at 15:11
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Use the previous exercise (4.3), which states that $K\cap Z(G)\ne 1$ whenever $K\lhd G$. Now $H\cap Z(G)\ne 1$, and so $H\cap Z(G)$ has order $p$. So $H\cap Z(G) = H$.

Hint of Exercise 4.3: $K\lhd G$ implies that $K$ is a union of some conjugacy classes of $G$.

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  • $\begingroup$ Sorry, the question seems very stupid. Can you explain why there is this line $H\cap Z(G)=Z(G)$? $\endgroup$ – Trần Nam Sơn May 19 at 15:21
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    $\begingroup$ $H\cap Z(G)$ is a subgroup of $H$, and $H$ is a group of order $p$. Thus either $H\cap Z(G) = 1$ or $H\cap Z(G) = H$. In this case it cannot be $1$, so it is $H$. It's a typo. $\endgroup$ – Hongyi Huang May 19 at 15:22
  • $\begingroup$ Thank you very much. I wish you good luck as you move on to new and different challenges. $\endgroup$ – Trần Nam Sơn May 20 at 2:49

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