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?

Answer 1

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

Answer 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.