If $H$ is a normal subgroup of $p$-group $G$ then why is $H\cap Z(G)\neq \{e\}$?

Question

If $H\neq\{e\}$ is a normal subgroup of the $p$-group $G$

Why is it that $H\cap Z(G)\neq \{e\}$?

Also if the last statement is true why is it that if $|H|=p$ then

$H\subseteq Z(G)$?

Answer 1

Here is a significantly more detailed proof than the one in the link provided by manthanomen. Let me know if you need any further details.

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Observe that we can let $G$ act on $H$ by conjugation. This is a valid group action because if $g \in G$ and $h \in H$, we have $ghg^{-1} \in H$ since $H \lhd G$.

Every group action partitions the target set into orbits. In this case, $H$ is partitioned into orbits (conjugacy classes), where the orbit containing $h \in H$ is $\{ghg^{-1} : g \in G\} \subset H$. Let $\mathcal O$ denote the set of orbits. Then $|H|$ is the sum of the sizes of the orbits: $$|H| = \sum_{X \in \mathcal O} |X|$$ As usual, we are interested in the fixed points of the action. A point $h \in H$ is fixed if its orbit is just $\{h\}$. For this action, $h \in H$ is fixed if and only if $ghg^{-1} = h$ for all $g \in G$, if and only if $gh = hg$ for all $g \in G$, if and only if $h \in H \cap Z(G)$. So, the number of fixed points is $|H \cap Z(G)|$, which means that $$|H| = |H \cap Z(G)| + \sum |X|$$ where the sum is now taken over only the non-singleton orbits. Now, the size of each orbit is a divisor of $|G|$, which is a power of $p$, which means that each $|X|$ in this equation is divisible by $p$. Since $1 < H \leq G$, we see that $|H|$ is also divisible by $p$, so if we take the above equation modulo $p$, we get $$0 \equiv |H \cap Z(G)| \mod p$$ Since $H \cap Z(G)$ is nonempty (all subgroups contain the identity), this means that $|H \cap Z(G)|$ is a positive integer divisible by $p$, and in particular, $|H \cap Z(G)| > 1$. (Indeed, $|H \cap Z(G)| \geq p$.)

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Note that if we take $H = G$, we get the important special case that $|Z(G)| > 1$, in other words, every $p$-group $G$ has a non-trivial center.

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To answer the last question, if we further assume that $|H| = p$, then $|H \cap Z(G)| \leq p$ since $H \cap Z(G) \leq H$. But we just showed that $|H \cap Z(G)| \geq p$, so this forces $|H \cap Z(G)| = p = |H|$, and therefore $H \cap Z(G) = H$, which means that $H \leq Z(G)$.