If $G$ is not abelian, then $\#\text{Inn}(G) \geq 4$

Question

I am trying to prove that if a group $G$ is non-abelian, that the inner automorphism group has four elements, so $\# \text{Inn}(G) \geq 4$.

So far I figured the following things:

Suppose $G$ is not abelian. Then $G/Z(G)$ is not cyclic, and thus $G/Z(G)$ has at least two generators. I know that automorphisms are determined by where they sent their generator. This is where I am stuck.

Any ideas?

I think you might be overcomplicating things at the end. To rephrase what you're saying: "$G/Z(G)$ cannot be generated by a single element". But all groups of order at most $3$ can, hence... — sTertooy, Jun 25 '20 at 07:42
Yes... just consider the map $\Phi$ from $G$ to $Inn(G)$ given by $g\mapsto \phi_g$ where $\phi_g(x)=gxg^{-1}$. Note that $\Phi$ is a homomorphism. What's its kernel? — WoolierThanThou, Jun 25 '20 at 07:51
The elements that commute with everything, so $Z(G)$. Thank you, it makes sense now! — bigbuddies, Jun 25 '20 at 07:55

score 4 · Accepted Answer · answered Jun 25 '20 at 14:00

4

The contrapositive is much clearer:

If $\#\text{Inn}(G) < 4$, then $G$ is abelian

The key facts are

$\text{Inn}(G) \cong G/Z(G)$
All groups of order less than $4$ are cyclic
If $G/Z(G)$ is cyclic then $G$ is abelian

answered Jun 25 '20 at 14:00

lhf

216,483

See also https://math.stackexchange.com/questions/999247/if-g-zg-is-cyclic-then-g-is-abelian-what-is-the-point – lhf Jun 25 '20 at 15:04

score 2 · Answer 2 · answered Jun 25 '20 at 13:47

2

As you identified in the comments, the key to this is the isomorphism

$$G/Z(G)\cong{\rm Inn}(G).$$

A proof of this isomorphism can be found in Gallian's "Contemporary Abstract Algebra (Eighth Edition)", page 194, Theorem 9.4.

answered Jun 25 '20 at 13:47

Shaun

44,997

If $G$ is not abelian, then $\#\text{Inn}(G) \geq 4$

2 Answers2