# Suppose that $G$ is nonabelian. Must $|\mathrm{Out}(G)| = |\mathrm{Aut}(G)|/|\mathrm{Inn}(G)|$?

I would say not. Suppose that $G = D_4$. By the article I found, $D_4$ has 4 outer automorphisms. I understand how reflection comes to play, but $|\mathrm{Out}(G)| \neq |\mathrm{Aut}(G)|/|\mathrm{Inn}(G)|$. Is this true?

• You've already noted in a previous question the isomorphism $\operatorname{Aut}(G)/\operatorname{Inn}(G)=\operatorname{Out}(G)$. If $\operatorname{Aut}(G)$ is finite, your result follows from Lagrange's theorem. – Jared May 21 '13 at 18:08
• How could this not be the case? The outer automorphisms are precisely the quotient by the inner ones – Tobias Kildetoft May 21 '13 at 18:08
• ${\rm Out}(G)={\rm Aut}(G)/{\rm Inn}(G)$ is always true. What makes you think that $|{\rm Out}(G)|$ does not equal $|{\rm Aut}(G)|/|{\rm Inn}(G)|$ in the case of $G=D_4$? – anon May 21 '13 at 18:09
• This earlier post might help clarify things for you, if you're confused by your computations: Inner and Outer automorphisms of $D_4$ – amWhy May 21 '13 at 18:15
• What is your definition of $\operatorname{Out}(G)$? – Hagen von Eitzen May 21 '13 at 18:42

$\operatorname{Out}(G)$ is defined as the quotient $\operatorname{Aut}(G)/\operatorname{Inn}(G)$, so $$\left|\operatorname{Out}(G)\right|=\left[\operatorname{Aut}(G):\operatorname{Inn}(G)\right]=\frac{\left|\operatorname{Aut}(G)\right|}{\left|\operatorname{Inn}(G)\right|}$$ no matter what $G$ is. Your confusion seems to be that you think that $\operatorname{Out}(G)$ denotes the set of otuer automorphisms of $G$, which is not true.
• It might be a good idea to include the count of outer automorphisms $\{ \alpha \in \operatorname{Aut}(G) : \alpha \text{ is outer } \} = \operatorname{Aut}(G) - \operatorname{Inn}(G)$ has size $|\operatorname{Aut}(G)| - |\operatorname{Inn}(G)|$ – Jack Schmidt May 22 '13 at 11:50