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!

Note $$G/{ e_G}\neq G;$$ rather, $$G/ { e_G}\cong G$$ for all groups $G$. — Shaun, Jun 16 '22 at 18:16
No, you don't have a solution. You only said that $Out(N)\cong Aut(N)$. We simply have the following: if $|N|=p_1$ then $N\cong C_{p_1}$ and we have ${\rm Aut}(N)\cong C_{p_1-1}$. — Dietrich Burde, Jun 16 '22 at 18:18
I would have expected the question to be "show $N \le Z(G) $". — Andreas Caranti, Jun 16 '22 at 18:59
Most of the problem set-up is useless. That $N$ is a normal subgroup of $G$, the order of $G$, none of that matters in any way whatsoever. Your question boils down to "I have a group of order $p_1$ a prime, and I want to find its automorphism group. Why include the dross? — Arturo Magidin, Jun 16 '22 at 20:41
@Shaun Now it is better. The original post had several problems ($N$ and $G$ interchanged, a prime factorization introduced and never used etc.). Nevertheless, I still don't understand why the outer automorphism group is mentioned. I also suspect that the real question is a different one, namely to show that $N$ is normal, see this duplicate. Why? Because of the wording "smallest prime divisor" in the title. — Dietrich Burde, Jun 17 '22 at 14:56

score 0 · Accepted Answer · answered Jun 16 '22 at 18:42

0

This can be done is two steps:

answered Jun 16 '22 at 18:42

Shaun

1 Answers1