When Center of group is a subset of Normalizer($Z(G) \subset N(a)$).

Question

DEFINITION:

If $a \in G$, then $N(a)$, the normalizer of $a$ in $G$, is the set $N (a) = \{ x \in G | xa = ax \} $.

$Z(G)$ is the center of the group.

I found the following proof -

Lemma: If $|G| = p^2$ where p is a prime number, then G is abelian.

Proof: The aim is to show that $Z(G) = G$.

At any rate, we know that $Z(G) \neq e$ is a subgroup of G so that $|Z(G)| = p $ or $ p^2 $ .

If $|Z(G)|= p^2$ , then $Z(G) = G$ and we are done.

Suppose that $|Z(G)| = p$; let $a \in G, a \notin Z(G)$. Thus $N(a)$ is a subgroup of $G, Z(G) \subset N(a)$,

$ a \in N(a)$ , so that $| N(a)| > p$, yet by Lagrange's theorem $|N(a)| \mid |G| = p^2$ . The only way out is for $|N(a)| = p^ 2 $, implying that $a \in Z(G)$, a contradiction. Thus $|Z(G)|$ =p is not an actual possibility.$\blacksquare$

-Topics in Algebra(2nd-edition-1975) by Herstein, page 84.

Question: Why $Z(G) \subset N(a)$ ? or how it is derived ?

Answer 1

$Z(G) = \{x \in G : gx = xg \textrm{ } \forall g \in G \}$. Hence, if $x \in Z(G)$ then $xa = ax$ i.e. $x \in N(a)$.