Let $G$ be a group. Prove that $G'$ is central iff ${\rm Inn}(G)$ is abelian.

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This is Exercise 4.11 of Roman's "Fundamentals of Group Theory: An Advanced Approach." According to this search and Approach0, it is new to MSE.

The Details:

On page 33 of Roman's book,

Definition: A subgroup $H$ of a group $G$ is central if $H$ is contained in the centre of $G$.

On page 120 ibid., we have

Theorem 4.13 (1): [. . .] $${\rm Inn}(G)\cong G/Z(G).$$

The Question:

Let $G$ be a group. Prove that $G'$ is central if and only if ${\rm Inn}(G)$ is abelian.

Thoughts:

I think Theorem 4.13 (1) is important here. However, I find it hard to articulate why.


Suppose $G'\subseteq Z(G)$. Consider $\gamma_g, \gamma_h\in{\rm Inn}(G)$ for $g, h\in G$. We have for any $x\in G$ that

$$\begin{align} (\gamma_g\circ\gamma_h)(x)&=\gamma_g(\gamma_h(x))\\ &=\gamma_g(hxh^{-1})\\ &=ghxh^{-1}g^{-1}\\ &=(gh)x(gh)^{-1}\\ &=\gamma_{gh}(x). \end{align}$$

Now, I just follow my nose. Something tells me I'm heading in the right direction.

Consider $\gamma_{gh}(x)x^{-1}$; that is, $[gh, x]$. This commutes with all elements of $G$ by assumption.

Where do I go from here?

Please help :)

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This is simpler. Since $\mathrm{Inn}(G)\cong G/Z(G)$, then $\mathrm{Inn}(G)$ is abelian if and only if $G'\subseteq Z(G)$: recall that for a group $G$, a normal subgroup $N$ contains $G'$ if and only if $G/N$ is abelian.