Normal subgroup of order $p$ is in the center

This type of exercises are difficult to judge in a forum like this, because we need to know precisely, what has been covered up to that point. Many posters are then inclined to use extra bits they have learned (later in a similar course).

I get the feeling that the following is what might have been expected. This is just fleshing out the hint in Steve D's comment, so I make it a CW.

Let $H$ be a normal subgroup of order $p$. We know (Lagrange's theorem) that $H$ is cyclic. Let $g$ be a generator. Let $x$ be any element of $G$. By normality of $H$ we know that $$ xgx^{-1}=g^k, $$ for some integer $k, 0<k<p$. Conjugation by $x$ is an automorphism of $H$, so $xg^{t}x^{-1}=g^{tk}$ for all integers $t$. In particular we get that $$ x^2g x^{-2}=x(xgx^{-1})x^{-1}=xg^kx^{-1}=g^{k^2}. $$ An obvious induction then proves that $$ x^tgx^{-t}=g^{k^t} $$ for all natural numbers $t$. But, again by Lagrange's theorem $x^{p^2}=1$. Therefore $$ g=1g1^{-1}=g^{k^{p^2}}. $$ As $g$ is of order $p$, this means that $1\equiv k^{p^2}\pmod p$. But two applications of Little Fermat tell us that $$ k\equiv k^p \equiv k^{p^2}\equiv 1\pmod p. $$ Recalling the constraint $0<k<p$ we can conclude that $k=1$. Therefore $x$ and $g$ commute. Obviously then $x$ commutes with all the powers of $g$. As $x$ was arbitray, we have shown that $H\le Z(G)$.