Let $p$ be the smallest prime divisor of $|G|$ and $N\unlhd G $ s.t. $|N|=p$. Find ${\rm Aut}(N)$.

Question

Let $p$ be the smallest prime divisor of $|G|$ and $N\unlhd G $ s.t. $|N|=p$. Find ${\rm Aut}(N)$.

Attempt:

We have $|N|=p$ and since $p$ is a prime number, $N$ is a cyclic subgroup and so abelian, hence $Z(N)=N$.

Also ${\rm Inn}(N) \cong N/Z(N)$ can be obtained from the first isomorphism theorem.

Since $Z(N)=N\space$ we obtain ${\rm Inn}(N) \cong N/N=\{e\}$ so

$${\rm Out}(N)= {\rm Aut}(N)/{\rm Inn}(N)\implies {\rm Out}(N)={\rm Aut}(N)$$

Is my solution correct?

I will be grateful for feedback. Thanks!

Answer 1

This can be done is two steps:

1. Show that every group of prime order is cyclic

2. Show Aut $\mathbb Z_p\simeq \mathbb Z_{p-1}$