Let's say $G$ is a $p$-group so that $|G|=p^{r}$ for $p$ prime and $r>1$. Let say $N\subset G$ a subgroup with $|N|=p$. I'm trying to prove that $N$ is contained in the center ($Z(G)=\{a\in G\mid xa=ax, \forall x\in G\}$) of $G$ if and only if $N$ is a normal subgroup of $G$.
I have a proof but I got the hint that I needed to look at the conjugacy classes of $N$ and I didn't use that, but I can't find my mistake.
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So first assume that $N \subset Z(G)$. Let's take $g\in G$ random and $n \in N$ random. Now will follow because $N\subset Z(G)$ that $gn=ng$. So $gng^{-1}=n$ $\in N$ This implies that $N$ is a normal subgroup.
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Now I assume that $N$ is a normal subgroup. I wanted to do exact the same thing as above but in the other direction. I think I forget something or do it wrong.
Can someone maybe help and tell me why I need to use that $G$ is a $p$-group.
EDIT: It's wasn't clear what i meant with that I wanted to do exact the same thing for the other direction.
So i thougt that if we know that $N$ is a normal subgroup of $G$ That we could say: $\forall g\in G, \forall n\in N: gng^{-1}\in N$.
I took a random $n\in N$. Because $N$ is an normal subgroup of $G$ is $n\in G$. Now we need to prove that $\forall x\in G: xn=nx$.
So $xnx^{-1}\in N$ because that $N$ is a normal subgroup of $G$ So there exist an $n_{0}\in N $ so that $xnx^{1}=n_{0}$. So $xn$=$n_{0}x$. I thought this was enough?
Best Answer
Every normal subgroup of a nilpotent group has a nontrivial intersection with the center (induction on the length of the upper central series). Since a group of prime order has exactly two subgroups, the statement follows.
Now using conjugacy classes. Take an element $x\in N$, then the conjugacy class $C_x$ of $x$ in the whole group $G$ must have order dividing $p^n$, so it is a power of $p$. On the other hand $|C_x|\le p-1$ since $N$ is normal. Hence $|C_x|=1$ and $N$ is in the center.