Math Is Fun Forum

AlexGo

Guest

Groups

Let G be the special linear group SL(2,5) of 2 × 2 matrices of determinant 1 over the field F_5 of integers modulo 5, so that the arithmetic in G is modulo 5. Show that G is a group of order 120.

I'm having trouble with this. I've let g be the matrix with entries a, b, c and d in the obvious places. Am I right in saying that a, b, c and d can only have the values 0, 1, 2, 3 or 4? I've then drawn out a multiplication table (not modulo 5... we wouldn't calculate the determinant modulo 5 would we?), and considered the number of ways of getting ad-bc = 1.

We could have:

9 - 8: 2 ways
4 - 3: 6 ways
3 - 2: 4 ways
2 - 1: 2 ways
1 - 0: 9 ways

giving a total of 23 ways.

23 doesn't relate to 120 in any obvious way. What am I missing?

Thanks

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Groups

Am I right in saying that a, b, c and d can only have the values 0, 1, 2, 3 or 4?

Yes.

we wouldn't calculate the determinant modulo 5 would we?

Yes, you would.  In general you can consider SL(n, R), where R is any ring.  The determinate here is the same as the determinant in the real case, but the elements add and multiply up to 1 in R.

First note that a and d may be switched without affecting the determinant, and the same with b and c.  So we may assume a < d and b < c, and then factor in the extra ways later.  This trick is an extremely important concept in group theory: Symmetries make calculations easier, but only when you use them.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

AlexGo

Guest

Re: Groups

Though I do note that 5(23 + 1) = 120...

I've also realised that we can swap a with d, and b with c.

alexGo

Guest

Re: Groups

Thanks for that Ricky

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Groups

Whoops, just be careful with the case where a = d or b = c (or both!) of course.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

alexGo

Guest

Re: Groups

Okay, I've now got that out fine. The question goes on to ask me to prove that the only element of G of order to is -I, which I've done. It then asks

Find a subgroup of G isomorphic to Q_8.

I'm not sure how to go about doing this? Obviously it contains -I, and I need 3 more elements of order 4 to generate the group ...

Thanks

alexGo

Guest

Re: Groups

Okay, I've now got that out fine. The question goes on to ask me to prove that the only element of G of order to is -I, which I've done.

The above should say "of order two "

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Groups

I'm not sure how to go about doing this?

There is no way (that I know of) to just look at the group and say, "Ok, here is what the elements are going to be."  Remember G has only 120 elements, which really isn't all that many.  The task at hand is to reduce this down to a more feasible number of candidates.

The trick to doing this is to assume that the elements exist, and then figure out what properties they would satisfy.  So, what properties must these elements satisfy, and how do those properties eliminate elements from G?

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."