The inner automorphism group is a normal subgroup of the full automorphism group

If \sigma \in \mathsf{Aut}(G) and \varphi_g is conjugation by g, prove that \sigma \varphi_g \sigma^{-1} = \varphi_{\sigma(g)} Deduce that \mathsf{Inn}(G) \vartriangleleft \mathsf{Aut}(G). (The quotient \mathsf{Aut}(G)/\mathsf{Inn}(G) is called the outer automorphism group of G.)


Note that (\sigma \varphi_g \sigma^{-1})(x) = \sigma(\varphi_g(\sigma^{-1}(x))) = \sigma(g \sigma^{-1}(x) g^{-1}) = \sigma(g) x \sigma(g)^{-1} = \varphi_{\sigma(g)}(x). Hence \sigma \varphi_g \sigma^{-1} = \varphi_{\sigma(g)} \in \mathsf{Inn}(G).

More generally, this implies that \sigma \mathsf{Inn}(G) \sigma^{-1} \subseteq \mathsf{Inn}(G) for all \sigma \in \mathsf{Aut}(G); hence \mathsf{Inn}(G) \vartriangleleft \mathsf{Aut}(G).

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: