Let $G$ be a group such that $|G| \ge 3$ then $|AutG| \ge 2$.

Question

How can I approach to this problem?

It is necessary to divide in cases? For G finite and infinite, or Abelian and non-Abelian?

The Lagrange theorem can help me?

Could you suggest some hints?

See this thread for another duplicate, and this for another near duplicate. — Jyrki Lahtonen, Aug 20 '15 at 20:35

Eugene Zhang · Answer 1 · 2015-08-20T21:31:43.393

0

For any non abelian $G, \:|G|\geqslant 3$, there is a $g, \: g\ne e, \:g\notin Z(G)$, for otherwise it would be abelian. So $\{\phi:G\to G|\:\phi(x)=g^{-1}xg,\:x\in G\}$ is a nontrivial automorphism. Plus one trivial automorphism, $|Aut(G)|\geqslant 2$.

For any abelian $G, \:|G|\geqslant 3$, there is a nontrivial permutation that is the nontrivial automorphism. So $|Aut(G)|\geqslant 2$.

edited Aug 20 '15 at 21:31

answered Aug 20 '15 at 20:34

Eugene Zhang

16,805

And if $g$ is in the center? – Jyrki Lahtonen Aug 20 '15 at 20:35
If $G$ is abelian, this won't work! – Cheerful Parsnip Aug 20 '15 at 20:35

Let $G$ be a group such that $|G| \ge 3$ then $|AutG| \ge 2$.

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