The Subgroup of Inner Automorphisms of a Group, Inn(G)

# The Subgroup of Inner Automorphisms of a Group, Inn(G)

Recall from the Inner Automorphisms of a Group page that if $G$ is a group then an inner automorphism of $G$ is an automorphism of $G$ of the form $i_a$ where $a \in G$ and $i_a : G \to G$ is defined for all $g \in G$ by:

(1)\begin{align} \quad i_a(g) = aga^{-1} \end{align}

We will now see that the set of all inner automorphisms of a group $G$ forms a group, and that it is a subgroup of $\mathrm{Aut}(G)$.

Definition: Let $G$ be a group. The Group of Inner Automorphisms of $G$ denoted by $\mathrm{Inn}(G)$ is the set of all inner automorphisms of $G$ with the operation of composition. |

The following proposition tells us that $(\mathrm{Inn}(G), \circ)$ is a subgroup of $(\mathrm{Aut}(G), \circ)$.

Proposition 2: Let $G$ be a group. Then $\mathrm{Inn}(G)$ is a subgroup of $\mathrm{Aut}(G)$. |

**Proof:**Let $i_a, i_b \in \mathrm{Inn}(G)$. Then for all $i_a \circ i_b : G \to G$ and for all $g \in G$ we have that:

\begin{align} \quad (i_a \circ i_b)(g) = i_a(i_b(g)) = i_a(bgb^{-1}) = abgb^{-1}a^{-1} = abg(ab)^{-1} = i_{ab}(g) \in \mathrm{Inn}(G) \end{align}

- So $\mathrm{Inn}(G)$ is closed under the operation of composition.

- Now consider the identity automorphism $i : G \to G$ defined for all $g \in G$ by $i(g) = g$. Let $e$ be the identity in $G$. Then for all $i_e \in \mathrm{Inn}(G)$ is such that for all $g \in G$:

\begin{align} \quad i_e(g) = ege^{-1} = g = i(g) \end{align}

- So $i = i_e \in \mathrm{Inn}(G)$.

- Lastly, let $i_a \in \mathrm{Inn}(G)$. Then $i_{a^{-1}} \in \mathrm{Inn}(G)$ is such that for all $g \in G$:

\begin{align} \quad (i_a \circ i_{a^{-1}})(g) = i_a(i_{a^{-1}}(g)) = i_a(a^{-1}ga) = a(a^{-1}ga)a^{-1} = g = i_e(g) \end{align}

(5)
\begin{align} \quad (i_{a^{-1}} \circ i_a)(g) = i_{a^{-1}}(i_a(g)) = i_{a^{-1}}(aga^{-1}) = a^{-1}(aga^{-1})a = g = i_e(g) \end{align}

- So $\mathrm{Inn}(G)$ is a subgroup of $\mathrm{Aut}(G)$. $\blacksquare$