# Let $P$ be a finite p-group, $A\vartriangleleft P$, with $|A|= p$. Show that $A \subseteq Z(P)$

Let $P$ be a finite p-group, $A\vartriangleleft P$, with $|A|= p$. Show that $A \subseteq Z(P)$

I have to show that if $x \in A \Rightarrow x \in Z(P)$.

My attempt:

Since A is a normal subgroup of P then for all $p \in P$, $pA = Ap$

If $x \in A$, and $A\vartriangleleft P$ then by the definition of normal subgroup,

$$pxp^{-1} \in A, \forall p \in P.$$

we also have that $|A| = p$ and P is a p-group, then the order of each element in P is a power of $p$.

$$Z(P) = \{z \in P ∣ \forall p \in P, zp = pz\}$$

I know that the center is a normal subgroup of P.

Since $|A| = p$, then $A$ is also a finite abelian p-group. We must show that $A$ must be contained in $Z(P)$ or be equal.

If $x \in A \Rightarrow x \in P$, now pick an arbitrary $p \in P$ then $pxp^{-1} \in A$ we must show that $px = xp$.

Using these definitions I can not complete the proof. I do not see where I should use the fact that P is a p-group and that the order of A is p. Any ideas? Alternative definition? Alternative approach by contradiction ?

Let $\varphi: P \rightarrow \text{Aut}(A)$ be the group homomorphism that sends an element $p \in P$ to the automorphism of $A$ given by $$a \mapsto p a p^{-1}.$$ Now observe that $|A| = p$ implies that $A \cong \mathbb{Z}/p\mathbb{Z}$ and hence $\text{Aut}(A) \cong \mathbb{Z}/{(p -1)\mathbb{Z}}$. Since the order of $P$ is a power of $p$ and the order of $\text{Aut}(A)$ is coprime to $p$, it follows that the image of $\varphi$ only contains the identity. But this is equivalent to showing that $A$ is in the center.