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.