Isomorphisms

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.

If $\displaystyle Aut \, \mathbb{Z}_6$ is isomorphic to $\displaystyle \mathbb{Z}_2$ then there must exist at least one bijection $\displaystyle F:Aut \, \mathbb{Z}_6 \rightarrow \mathbb{Z}_2$. But the two sets have a different number of members, so we can't define a bijection between them!

Needless to say my argument must have a flaw in it somewhere. But where?

I am having a similar problem with the first part using $\displaystyle Aut \, \mathbb{Z}$. I presume whatever I got wrong above is similar to what I've got wrong with this one.

(This is what happens when a Physicist tries to teach himself Mathematics! :))

-Dan

Quote:

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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.

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$\displaystyle Aut \, \mathbb{Z}_6$ is the set of all bijective maps $\displaystyle \{f: \mathbb{Z}_6 \rightarrow \mathbb{Z}_6\}$
which preserve the structure of $\displaystyle \mathbb{Z}_6$. So it is not equivalent to
the set of all permutations of {0,1,2,3,4,5}, as such a
permutation need not preserve the required structure
(persumably the field structure, or maybe the additve
group structure?).

RonL

Quote:

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Originally Posted by CaptainBlack

$\displaystyle Aut \, \mathbb{Z}_6$ is the set of all bijective maps $\displaystyle \{f: \mathbb{Z}_6 \rightarrow \mathbb{Z}_6\}$
which preserve the structure of $\displaystyle \mathbb{Z}_6$. So it is not equivalent to
the set of all permutations of {0,1,2,3,4,5}, as such a
permutation need not preserve the required structure
(persumably the field structure, or maybe the additve
group structure?).

RonL

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THERE we go! I knew I had to be forgetting something! Thanks!

-Dan