Originally Posted by **topsquark**

Have you ever thought you knew something well and talked about it for years, only to suddenly find out one day you were wrong? (Shaking my head in despair!) I thought I knew without a shadow of a doubt what an isomorphism was...

The problem: Let G be a group and Aut G the set of all automorphisms of G.

a) Show Aut G is a group. (Easy. I did this.)

b) Show $\displaystyle Aut \, \mathbb{Z} \cong \mathbb{Z}_2$ and $\displaystyle Aut \, \mathbb{Z}_6 \cong \mathbb{Z}_2$.

Part b) is what's bothering me. Let's take the second case first since it typifies the problem I'm having.

The automorphism is the set of all bijective maps $\displaystyle \{f: \mathbb{Z}_6 \rightarrow \mathbb{Z}_6\}$. Now, we may think of the general element of this set essentially as an element of the set of permutations of {0,1,2,3,4,5}. This means the set $\displaystyle Aut \, \mathbb{Z}_6$ has 6! elements. The set $\displaystyle \mathbb{Z}_2$ has 2 elements.