The $i$-th center $Z_{i}(G)$

Question

Let $H$ be a normal subgroup of a $p$-group $G$, $H$ is of order $p^i$. Prove that $H$ is contained in the $i$-th center $Z_{i}(G)$.

Recall that we define $Z_{0}(G)=1$, and for $i>0$, $Z_{i}$ is the subgroup of $G$ corresponding to $Z(G/Z_{i-1})$ by the Correspondence Theorem: $Z_{i}/Z_{i-1}=Z(G/Z_{i-1})$

The sequence of subgroups $Z_{0}\subset Z_{1}\subset Z_{2}\subset\ldots$ is called the upper central series of $G$

I use induction on $i$ and consider $G/Z(G)$. The case $i=0$ is trivial ($H=1$ and $Z_{0}(G)=1$). How should I continue the proof?

Thanks for any insight.

Answer 1

That is exercise $9$, page $222$ of the book Groups: An Introduction to Ideas and Methods of the Theory of Groups, Antonio Machi. A detailed hint can be found on that book.

P/s: that exercise of that book is a stronger version of this problem.

Answer 2

If a subgroup $H$ is normal, then the conjugacy class of every $h\in H$ is contained in $H.$ The size of a conjugacy class is a power of $p$ (since it is the quotient of the order of $G$ by the order of the centralizer of an element). Since the identity has conjugacy class of size $1,$ that means that there are at least another $p-1$ elements in $H$ whose conjugacy classes have size one. But that is the same as being in the center of $G.$ So, $H\cap Z(G) \neq \{e\}.$ Now, mod out by $Z(G),$ repeat.