Inner Automorphisms form Subgroup of Symmetric Group

Theorem

Let $G$ be a group.

Let $\struct {\map \Gamma G, \circ}$ be the symmetric group on $G$.

Let $\Inn G$ denote the inner automorphism group of $G$.

Then:

$\Inn G \le \struct {\map \Gamma G, \circ}$

where $\le$ denotes the relation of being a subgroup.

Proof

An inner automorphism is a permutation on $G$ by definition.

From Inner Automorphisms form Subgroup of Automorphism Group:

$\Inn G \le \Aut G$

where $\Aut G$ denotes the set of automorphisms of $G$.

From Automorphism Group is Subgroup of Symmetric Group:

$\Aut G \le \struct {\map \Gamma G, \circ}$

Thus $\Inn G \le \struct {\map \Gamma G, \circ}$ as required.

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 7.1$. Homomorphisms: Example $131$