# Question:Inner automorphism automorphism

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A: This basically is a straightforward verification that involves some symbol manipulation. See group acts as automorphisms by conjugation, which not only shows that conjugations by elements are automorphisms, but also that the conjugation by $gh$ is the composite of conjugations by $g$ and $h$. In other words, if $c_g$ denotes conjugation by $g$, we have $c_{gh} = c_g \circ c_h$ for all $g,h \in G$.
We thus obtain a homomorphism of groups from $G$ to $\operatorname{Aut}(G)$ (the automorphism group of $G$) sending each $g \in G$ to the conjugation map $c_g$. The kernel of this is the center $Z(G)$ and the image is the inner automorphism group $\operatorname{Inn}(G)$. We thus have, from the first isomorphism theorem, that $\operatorname{Inn}(G) \cong G/Z(G)$.