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Let $(G,\cdot)$ be a group.

If $G$ is Abelian, the commutator subgroup $[G,G]=\{ghg^{-1}h^{-1}|g,h\in G\}$ is trivial; otherwise, the commutator subgroup is not trivial and there is the Abelianisation $\mathrm{Ab}(G)=G/[G,G]$.

Also, if $G$ is Abelian, then the centre is itself or $\mathrm{Z}(G)=G$. Thus, its Inner Automorphism group $\mathrm{Inn}(G)\cong G/\mathrm{Z}(G)$ is trivial.

I can see that if $G$ is not Abelian, then its Inner Automorphism group will not be trivial, but I do not know what happens next.

Are there some interesting relationships between $[G,G]$ and $\mathrm{Inn}(G)$? Still, why are commutators and inner automorphisms both about commutativity of one group?

Thanks.

Emancipatrix
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    Inner automorphisms are "about commutativity" in so far as the inner automorphism group is isomorphic to $G/Z(G)$, so it's about the center. For the "duality" of commutator subgroups and center, see here. – Arturo Magidin Dec 16 '23 at 03:56
  • ... whatever "about" means in this situation. And without knowing what it means, I've voted to close. – Lee Mosher Dec 16 '23 at 04:39
  • Note: $[G,G]$ is not defined to be the set of elements of the form $ghg^{-1}h^{-1}$. It is the subgroup generated by all such elements. – Arturo Magidin Dec 16 '23 at 22:52

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If $[G,G]$ is trivial then $ghg^{-1}h^{-1}=e$ and so $ghg^{-1}=h$. So for every $g$ and $h$ we see that $h$ is stabilized by conjugation. Said another way, the group of inner automorphisms is trivial.

CyclotomicField
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