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I've heard this fact before but wasn't sure whether it pertained to an individual inner automorphism or $Inn(G)$, the set of all inner automorphisms of G.

I'm aware that if $g \in Z(G)$, then $\phi_g$ is the identity function ($\phi_g(x) = gxg^{-1} = xgg^{-1} = x$), but wasn't sure whether I could use this fact to conclude that the kernel is therefore the center of the group. Would appreciate help as to whether I'm on the right or wrong track here.

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    $\begingroup$ I wanted to come back and make sure that my explanation made sense to you, since the answer was accepted, but there were no further comments. $\endgroup$ – Andres Mejia Nov 18 '17 at 18:54
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    $\begingroup$ Yes, it did, appreciate the explanation! $\endgroup$ – SS' Nov 18 '17 at 20:57
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The center is almost by definition elements to which the associated inner automorphism is trivial (as you mention.)

What this means is that $G/Z(G) \cong \mathrm{Inn}(G)$. This can be seen by taking the homomorphism $G \to \mathrm{Inn}(G):g \mapsto \phi_g$, where $\phi_g(x)= gxg^{-1}$, and noting that its kernel is exactly the center. You can then apply the first isomorphism theorem.


There are two concepts going on simultaneously. Any automorphism $\phi:G \to G$ has trivial kernel by assumption. However, the set of maps $\{ \phi: G \to G\}$ can be given a group structure by composition: $\phi \circ\rho$ is associative, there is an identity, and inverses since all the maps are assumed to be automorphisms. We denote this group $\mathrm{Aut}(G)$. One subgroup we care about in particular is $\mathrm{Inn}(G)$ -- These are automorphisms of the form $\phi_g:G \to G$ given by $\phi_g(x)=g x g^{-1}$ for some $g \in G$.

Now, there is another homomorphism $\Phi:G \to \mathrm{Aut}(G)$ given by $g \mapsto \phi_g$. The image of this homomorphism is precisely the subgroup $\mathrm{Inn}(g)$

$\mathrm{ker}(\Phi)=Z(G)$ since it consists of group elements, so that conjugating by them is the identity map (trivial automorphism in the group.)

In other words, $g \in \mathrm{ker}(\Phi) \iff \phi_g=id$, or in other words, $gxg^{-1}=x$ for all $x \in G$.

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Given $g\in G$, conjugation by $g$ induces an automorphism $c_g$ of $G$ which maps an arbitrary element $x$ to $gxg^{-1}$.

We then have a homomorphism $\varphi$ from $G$ into $Aut(G)$ which sends $g$ to $c_g$. The image of $G$ is the group of inner automorphisms.

So, when we speak of an inner automorphism of $G$, we mean a map $c_g$ for some $g\in G$. The kernel of any such map is trivial, being an automorphism.

However, the kernel of $\varphi$ is $Z(G)$. That is, the group of inner automorphisms is isomorphic to $G/Z(G)$.

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  • $\begingroup$ So as I understand, the kernel of an automorphism is the trivial one but the kernel of that homomorphic map shown then is $Z(G)$ $\endgroup$ – SS' Nov 17 '17 at 5:39
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    $\begingroup$ @SS' see my edits. $\endgroup$ – Andres Mejia Nov 17 '17 at 6:00

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