Prove that $G \cong\mathrm{Inn}(G)$ if and only if $Z(G)$ is trivial

Claim: Let $G$ be a group. Prove that $G \cong\mathrm{Inn}(G)$ if and only if $Z(G)$ is trivial.

Could anyone offer a hint on proving this claim just using simple properties of group isomorphisms? (i.e., not using the fact that the quotient group $G / Z(G)$ is isomorphic to the group of inner automorphisms of $G$.)

EDIT:

It turns out that this Claim is false as stated. There are several counterexamples, several of which are provided in answers below, for the case of G being infinite. However, the claim holds for finite groups.

Solution 1:

The claim is false as stated. groupprops has a counterexample.

The correct claim is specifically about the natural map $G \to \text{Inn}(G)$, and as I mentioned in my comment above you don't need to know anything about quotient groups to prove the correct claim: you just need to determine when this map is injective.

Solution 2:

Hints: if $\,\phi_a\,$ denotes the inner isomorphism determined by $\,a\,$ , i.e. $\,\phi_a(x):=axa^{-1}\,$ , then

(1) Check that $\,F:G\to\operatorname{Aut}(G)\;,\;\;F(a):=\phi_a\,$ , is a homomorphism ;

(2) Prove $\,Z(G)=\ker(F)\,\,,\,\,\,\operatorname{Inn}(G)=\operatorname{Im}(F)$

(3) Apply the first isomorphism theorem

Added: As already noted in the comments, this only shows that $\,Z(G)=1\Longrightarrow G\cong Inn(G)\,$

Solution 3:

A somewhat simpler and better known (if maybe not for this property) counterexample against the claim than the one cited by Qiaochu (but it is very similar). Take $G=O(2,\mathbf R)$, the orthogonal group of the plane. Choose a line $L$ though the origin, then there is a surjective morphism of groups $G\to G$ that sends every rotation $\rho$ to $\rho^2$, and fixes the reflection $\sigma_L$ with respect to $L$. Every other reflection is of the form $\sigma'=\rho\sigma_L$ maps to $\rho^2\sigma_L$, which is the reflection wit respect to the line $\sigma'(L)$ (which shows that all reflections are in the image); it is easily checked that this is a morphism of groups. The kernel of this morphism has two elements, the identity and the half-turn rotation, which also form the center of $G$. Therefore by the first isomorphism theorem $G\cong G/Z(G)\cong\operatorname{Inn}(G)$ even though $Z(G)$ is not trivial.