Proposition 3

Let G be a group. For every element x of G, the mapping Int(x): $y \mapsto xyx^{-1}$ of G into itself is an automorphism of G. The mapping Int: $x \mapsto \operatorname{Int}(x)$ of G into Aut(G) is a group homomorphism, whose kernel is the centre of G and whose image is a normal subgroup of Aut(G).

If $x, y$ and $z$ are elements of G, then $(xyx^{-1})(xz x^{-1}) = xyz x^{-1}$ and hence Int(x) is an endomorphism of G. For $x$ and $y$ elements of G,

for all $z \in G$, $x(yzy^{-1})x^{-1} = (xy)z(xy)^{-1}$. On the other hand, $\operatorname{Int}(e)$ is the identity mapping of G. The mapping Int is therefore a monoid homomorphism from G to the monoid End(G) of endomorphisms of the group G. As the elements of G are invertible, the mapping Int takes its values in the set Aut(G) of invertible elements of End(G) (§ 2, no. 3). Now $xyx^{-1} = y$ if and only if $x$ and $y$ commute and hence $\operatorname{Int}(x)$ is the identity mapping of G if and only if $x$ is a central element. Finally, let $\alpha$ be an automorphism of G and let $x \in G$; then

For $y \in G$,

Hence $\alpha \cdot \operatorname{Int}(G) \cdot \alpha^{-1} \subset \operatorname{Int}(G)$.