Automorphism group of Klein four is S3.

[Kreizhn]

Homework Statement

Show that \text{Aut}_{\text{Grp}}(\mathbb Z /2 \mathbb Z \times \mathbb Z /2 \mathbb Z) \cong S_3. That is, show that the automorphism group of the Klein four group is the symmetric group on 3-letters.

The Attempt at a Solution

I think the argument here is pretty simple. First we need to analyze the automorphism group. Since automorphisms preserve the identity, there are only 27 possible set mappings. Since automorphisms preserve order and are bijective, there are only 6 possible mappings which correspond to permutations of the 3 non-identity elements, since they all have order 2.

Now one can show by hand that these permutations are all automorphisms. What I am hoping for is a more elegant way of showing this other than doing it by hand. The only thing I can possibly think of is that any two non-identity elements generate the group. Hence so long as we map two distinct elements to any other two distinct elements, we get the entire group. Is this correct? Is there another way of claiming this?

 

[micromass]

Hi Kreizhn!

The trick is to use linear algebra here. Indeed, \mathbb{Z}_2\times\mathbb{Z}_2 can be seen as a vector space over \mathbb{Z}_2. Furthermore, every group homomorphism is actually a \mathbb{Z}_2-linear map. Thus all the automorphisms can be described by invertible matrices. For example:

\left(\begin{array}{cc} a & b\\ c & d\\ \end{array}\right)

where (1,0) is being sent to (a,c) and (0,1) is being sent to (b,d). So the only thing you need to do is figure out how many matrices are invertible (i.e. have nonzero determinant).

Very analogously, one can show that the automorphism group of \mathbb{Z}_p^n is in fact GL_n(\mathbb{Z}_p), and now you know where it comes from!

 

[Kreizhn]

Thanks!

There seems to be some representation theory in here, though I'm not sufficiently familiar with it to comment on it further. It seems to me that the automorphism group will (almost by definition) be given by the representation G \to GL(G). Can we do better than that? Namely, when can we explicitly say that \text{Aut}(G) = GL_n(G) for some finite n?

 

[Kreizhn]

Oh! Also, if we're giving the automorphism group in terms of GL_2(\mathbb Z_2 ), and we give S_3 as 3x3 permutation matrices, is there an obvious way of jumping from one to the other?

 

[micromass]

Thanks!

There seems to be some representation theory in here, though I'm not sufficiently familiar with it to comment on it further. It seems to me that the automorphism group will (almost by definition) be given by the representation G \to GL(G).
Can we do better than that? Namely, when can we explicitly say that \text{Aut}(G) = GL_n(G) for some finite n?

Do you mean for every group G here? It's not clear to me how you would see any group G as a vector space (and thus how you would define GL(G)). In particular, if G is nonabelian, then I don't see what GL(G) means. The only way I really see it work is for \mathbb{Z}_n^p, but I might be missing something/

Oh! Also, if we're giving the automorphism group in terms of GL_2(\mathbb Z_2 ), and we give S_3 as 3x3 permutation matrices, is there an obvious way of jumping from one to the other?

Uh, depend how you would see S_3 as permutation matrices. What you ask is if we represent S_3 as representation matrices over the field \mathbb{Z}_2 and if you represent S_3 as matrices over \mathbb{C}, does there exist a connection?
Well, I don't know, but I'll look into it.

 

[Kreizhn]

Sorry for the ambiguity. I meant any particular G. In such an instance, I believe that GL(G) is actually defined as the automorphism group of G. Maybe just weird notation? I think I have seen that used in a representation theory book, just because it works out well for finite dimensional representations.

If memory serves, we can view any permutation group Sn as a set of nxn permutation matrices (with precisely one 1 in every row and column). I believe this is a representation of Sn over \mathbb Z_2. Like I said though, my memory and experience with rep.theory is shaky.

Edit: Nevermind, looked it up. The notation GL(V) is still specific to vector space automorphisms.

 

[Kreizhn]

Anyway, I wouldn't worry about it too much. There probably is an answer but I wouldn't know enough rep.theory to understand it well. Thanks.

 

[micromass]

If memory serves, we can view any permutation group Sn as a set of nxn permutation matrices (with precisely one 1 in every row and column). I believe this is a representation of Sn over \mathbb Z_2. Like I said though, my memory and experience with rep.theory is shaky.

Yes, this representation is also a representation over \mathbb{Z}_2. But note that we do not have this representation here. The representation you describe is one of 3x3-matrices. While the representation in this problem is one of 2x2-matrices!

 

[Kreizhn]

Yes, this representation is also a representation over \mathbb{Z}_2. But note that we do not have this representation here. The representation you describe is one of 3x3-matrices. While the representation in this problem is one of 2x2-matrices!

Yes, but ultimately we say that \text{Aut}(\mathbb Z_2 \times \mathbb Z_2 ) \cong GL_2(\mathbb Z_2) and that \text{Aut}(\mathbb Z_2 \times \mathbb Z_2 ) \cong S_3. So if that permutation matrix representation S_3 \to GL_3(\mathbb Z_2) is in fact an isomorphism onto some subset \hat S_3 \subseteq GL_3(\mathbb Z_2 ) then shouldn't we be able to map GL_2(\mathbb Z_2) \cong \hat S_3?

I can probably pump it out by hand, so if there isn't an elegant answer then don't worry about it. I was just curious if there is indeed an elegant answer. Or maybe I've just confused myself

 

[micromass]

Yes, but ultimately we say that \text{Aut}(\mathbb Z_2 \times \mathbb Z_2 ) \cong GL_2(\mathbb Z_2) and that \text{Aut}(\mathbb Z_2 \times \mathbb Z_2 ) \cong S_3. So if that permutation matrix representation S_3 \to GL_3(\mathbb Z_2) is in fact an isomorphism onto some subset \hat S_3 \subseteq GL_3(\mathbb Z_2 ) then shouldn't we be able to map GL_2(\mathbb Z_2) \cong \hat S_3?

I can probably pump it out by hand, so if there isn't an elegant answer then don't worry about it. I was just curious if there is indeed an elegant answer. Or maybe I've just confused myself

Yes, you are correct, of course. There will be a bijection between those two representations. But I don't know if this bijection has an interesting form or is canonical in some way. It is a nice question, though...

 

[Kreizhn]

I will think about it more. Thank you for your help.