Inner automorphism

of a group $G$

An automorphism $\phi$ such that $$ \phi(x) = g^{-1} x g $$

for a certain fixed element $g \in G$: that is, $\phi$ is conjugation by $g$. The set of all inner automorphisms of $G$ forms a normal subgroup $\mathrm{Inn}(G)$ in the group $\mathrm{Aut}(G)$ of all automorphisms of $G$; this subgroup is isomorphic to $G / Z(G)$, where $Z(G)$ is the centre of $G$ (cf. Centre of a group). Automorphisms that are not inner are called outer automorphisms. The outer automorphism group is the quotient $\mathrm{Out}(G) = \mathrm{Aut}(G)/\mathrm{Inn}(G)$.

Other relevant concepts include those of an inner automorphism of a monoid (a semi-group with a unit element) and an inner automorphism of a ring (associative with a unit element), which are introduced in a similar way using invertible elements.

#### Comments[edit]

Let $\mathfrak{g}$ be a Lie algebra and $x \in \mathfrak{g}$ an element of $\mathfrak{g}$ for which $\mathrm{ad}(x) : y \mapsto [x,y]$ is nilpotent. Then $$ \exp(\mathrm{ad}(x)) = \mathrm{id} + \mathrm{ad}(x) + \frac{1}{2!}\mathrm{ad}(x)^2 + \cdots $$ defines an automorphism of $\mathfrak{g}$. Such an automorphism is called an inner automorphism of $\mathfrak{g}$. More generally, the elements in the group $\mathrm{Int}(\mathfrak{g})$ generated by them are called inner automorphisms. It is a normal subgroup of $\mathrm{Aut}(\mathfrak{g})$.

If $G$ is a real or complex Lie group with semi-simple Lie algebra, then the inner automorphisms constitute precisely the identity component of the group $\mathrm{Aut}(\mathfrak{g})$ of automorphisms of $\mathfrak{g}$.

#### References[edit]

[a1] |  M. Hall jr., "The theory of groups" , Macmillan (1959)

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[a2] |  J.E. Humphreys, "Introduction to Lie algebras and representation theory" , Springer (1972) pp. §5.4
[a3] |  J.-P. Serre, "Algèbres de Lie semi-simples complexes" , Benjamin (1966)