# An exercise in an introduction to the theory of groups [closed]

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!

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dietrich Burde, Derek Holt, Brian Borchers, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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 $G$? – lulu May 19 at 15:01
• "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... – user659895 May 19 at 15:02
• You must have mixed words: to prove that normal subgroup is a subgroup is pretty trivial... – DonAntonio May 19 at 15:02
• sorry, prove $H$ is a subgroup of $Z(G)$. – Trần Nam Sơn May 19 at 15:07
• $Z(G)=\left\{x\in G| xy=yx, \forall y\in G \right\}.$ – Trần Nam Sơn May 19 at 15:11

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 $$G$$.
• Sorry, the question seems very stupid. Can you explain why there is this line $H\cap Z(G)=Z(G)$? – Trần Nam Sơn May 19 at 15:21
• $H\cap Z(G)$ 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. – Hongyi Huang May 19 at 15:22