- #1

- 240

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter teleport
- Start date

- #1

- 240

- 0

- #2

- 240

- 0

- #3

morphism

Science Advisor

Homework Helper

- 2,015

- 4

Why is |GL(2, Z_3)| odd? Isn't it just 48? In fact, the order of GL(2, Z_p) is even for all primes p.

But anyway, why would you think that GL and the dihedral group are isomorphic? The latter is much nicer, e.g. it has a presentation with 2 generators.

Ironically, a proof that GL(2, Z_p) isn't going to be isomorphic to the dihedral group having the same order follows from the fact that of how simple the center (which I believe you're calling the "commutant") of D_n is: it's either isomorphic to Z_1 or Z_2 (the cyclic groups of order 1 or 2, resp.), depending on whether n is odd or even, resp., while the center of GL(2, Z_p) is bigger when p>3. (This doesn't rule out the case when p=3, however; it turns out |C(GL(2, Z_3))| = |C(D_24)| = 2.)

So I suggest you abandon this particular approach. It's not hard to "just do it": what matrices commute with all the other ones?

But anyway, why would you think that GL and the dihedral group are isomorphic? The latter is much nicer, e.g. it has a presentation with 2 generators.

Ironically, a proof that GL(2, Z_p) isn't going to be isomorphic to the dihedral group having the same order follows from the fact that of how simple the center (which I believe you're calling the "commutant") of D_n is: it's either isomorphic to Z_1 or Z_2 (the cyclic groups of order 1 or 2, resp.), depending on whether n is odd or even, resp., while the center of GL(2, Z_p) is bigger when p>3. (This doesn't rule out the case when p=3, however; it turns out |C(GL(2, Z_3))| = |C(D_24)| = 2.)

So I suggest you abandon this particular approach. It's not hard to "just do it": what matrices commute with all the other ones?

Last edited:

- #4

- 240

- 0

It was a wild passionate fantasy of the moment I didn't really take the time to think about it before briging it up. The definition for commutant of a group G that we are using in class is the set of all [itex]xyx^{-1}y^{-1}[/itex] with for all x, y in G. But I actually solved it. Really simple! The det of any any element in the commutator of GL is 1, so the element is in SL. But any element in SL is invertible, and hence can be written as a commutator of two invertible matrices in GL. Hence the commutant of GL(2,Z_p) is SL(2,Z_p).

Last edited:

- #5

morphism

Science Advisor

Homework Helper

- 2,015

- 4

And you did get it right: the commutator subgroup of GL(2, Z_p) is SL(2, Z_p). Well, almost. If p=2, then since the GL(2, Z_2) =~ D_3, it follows that the comm subgrp of GL in this case is that of D_3, which is (isomorphic to) Z_3, while |SL(2, Z_2)| = 6.

Also, I'm not very convinced by your proof of the converse, i.e. that every element of SL(2, Z_p) arises as a commutator of GL(2, Z_p). You said that: "Also any X in SL(n = 2, Z_p) is invertible and hence can be written as a commutator of two invertible matrices over Z_p." But doesn't this apply to every X in GL as well? Thus we're lead to conclude that the commutator subgroup of GL is GL itself...

- #6

- 240

- 0

I just saw your post in the Homework forum (isn't it against the rules to post the same thread across forums?)

oops, didn't know this. Sorry to the moderators :)

And you did get it right: the commutator subgroup of GL(2, Z_p) is SL(2, Z_p). Well, almost. If p=2, then since the GL(2, Z_2) =~ D_3, it follows that the comm subgrp of GL in this case is that of D_3, which is (isomorphic to) Z_3, while |SL(2, Z_2)| = 6.

Nice. But you mean it works for all other p?

Also, I'm not very convinced by your proof of the converse, i.e. that every element of SL(2, Z_p) arises as a commutator of GL(2, Z_p). You said that: "Also any X in SL(n = 2, Z_p) is invertible and hence can be written as a commutator of two invertible matrices over Z_p." But doesn't this apply to every X in GL as well? Thus we're lead to conclude that the commutator subgroup of GL is GL itself...

You are totally right, that was care free. I am thinking of writing X as a commutator of two matrices with det 1 which would seem to do it. But I don't know how to get around the p = 2 singularity (maybe even 3).

- #7

morphism

Science Advisor

Homework Helper

- 2,015

- 4

Yeah - for all odd primes, the commutator subgroup of GL(2, Z_p) is SL(2, Z_p).Nice. But you mean it works for all other p?

- #8

- 240

- 0

why?

- #9

morphism

Science Advisor

Homework Helper

- 2,015

- 4

- #10

- 240

- 0

- #11

- 240

- 0

Also, I'm not very convinced by your proof of the converse, i.e. that every element of SL(2, Z_p) arises as a commutator of GL(2, Z_p). You said that: "Also any X in SL(n = 2, Z_p) is invertible and hence can be written as a commutator of two invertible matrices over Z_p." But doesn't this apply to every X in GL as well? Thus we're lead to conclude that the commutator subgroup of GL is GL itself...

Hmm, I don't think it can apply to any X in GL because not every X in GL has det 1, while any commutator does.

- #12

morphism

Science Advisor

Homework Helper

- 2,015

- 4

- #13

- 240

- 0

- #14

morphism

Science Advisor

Homework Helper

- 2,015

- 4

I don't know! At any rate, the proof I have in mind is tedious.Is the actual proof that complicated to not even dream of reading it this year?

Did you speak to your professor?

Share: