Re: Questions around GL(2,C)

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Originally Posted by

**shmounal** Attachment 22414
Ok so I've been set these questions as revision for new course on Algebra and after a summer of no maths my brain has turned to mush!!

For part 1 to 3 I feel reasonably confident I have the correct answers as all pretty simple matrix multiplication but really struggling with part 4.

Is it enough to just show the members of the groups by working out various multiplications of a, a^2, a^3 and b, b^2, b^3, b^4 then doing a multiplication table with all members of group to show that they can be self contained if you like and that each ab,... b^4a^3 etc are members of GL(2,C). This feels a bit laborious and I know how rigorous my lecturer is. just wondering if there is a nicer simpler way to do it without tons of boring matrix multiplication!! And for finding the order I'm at a bit of a loss!!

Any help appreciated!!

HINT: Your generators commute-ish. What I mean is that if A is the generator with the i-terms, and B is the generator with the w-terms, then $\displaystyle AB=B^{-1}A$. To prove this, show that $\displaystyle BAB=A$.

Can you see why this proves that your group is finite?

Re: Questions around GL(2,C)

personally, i would be inclined to use the result of problem 3 to show that we can write $\displaystyle ba = a^2b$, and then to use this to show we can write any product of a's and b's in the form $\displaystyle a^jb^k$. this, together with proving that a and b are of finte orders, should yield the order of the group.

since the set of $\displaystyle a^jb^k$ is finite, and the above suggestion also shows closure, we have a group.

if i am correct, and i could be wrong, this is a group i have not encounted much, but appears to be rather lovely.

Re: Questions around GL(2,C)

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Originally Posted by

**Deveno** personally, i would be inclined to use the result of problem 3 to show that we can write $\displaystyle ba = a^2b$, and then to use this to show we can write any product of a's and b's in the form $\displaystyle a^jb^k$. this, together with proving that a and b are of finte orders, should yield the order of the group.

since the set of $\displaystyle a^jb^k$ is finite, and the above suggestion also shows closure, we have a group.

if i am correct, and i could be wrong, this is a group i have not encounted much, but appears to be rather lovely.

This is, essentially, the same way as I suggested, I just hadn't read the previous questions...

Also, it is the dihedral group of order 6...because $\displaystyle a^3=1$, $\displaystyle b^2=1$ and $\displaystyle ba=a^{-1}b$...and it has 6 elements so it isn't a homomorphic image of the dihedral group of order 6 and so it must be it...

Re: Questions around GL(2,C)

are you quite sure about this? my calculations indicate b is of order 4, and the dihedral group of order 6 has no elements of order 4. as i indicated in my previous post, this is an "unusual" group.

swlabr, this is for you: the 3-sylow subgroup is normal, the 2-sylow subgroups are not.

Re: Questions around GL(2,C)

Quote:

Originally Posted by

**Deveno** are you quite sure about this? my calculations indicate b is of order 4, and the dihedral group of order 6 has no elements of order 4. as i indicated in my previous post, this is an "unusual" group.

swlabr, this is for you: the 3-sylow subgroup is normal, the 2-sylow subgroups are not.

Sorry, of course, I wasn't quite thinking straight (I was doing it in my head, which is always a bad start!).

If my memory serves me correctly, there are only three non-abelian groups of order 12.

It is not $\displaystyle A_4$ as $\displaystyle A_4$ has more than one cyclic subgroup of order 3 (and so the Sylow 3-subgroups are not normal - take (123) and (124)).

It is not the Dihedral group of order 12 as this contains no elements of order 4.

This only leaves one group...$\displaystyle T$. The only thing I know about it is that it is the third non-abelian group of order 12. It sticks in my memory, because in my honours group theory course it was mentioned as an aside in the very last lecture, as the last lecture was used to show that we could classify all groups of order up to...15?

Re: Questions around GL(2,C)

you win a prize! also known as Q3, also known as the 3rd dicyclic group, or Z3 x| Z4 (we have the non-trivial homomorphism Z4-->Aut(Z3) given by x-->(-1)^x).

it can't be D6, because the 6 elements of the rotation subgroup all have orders of 1,2,3 or 6, and the other 6 elements all have order 2...so there are no elements of order 4.

as i said before, rather lovely.

Re: Questions around GL(2,C)

Quote:

Originally Posted by

**Deveno**

it can't be D6, because the 6 elements of the rotation subgroup all have orders of 1,2,3 or 6, and the other 6 elements all have order 2...so there are no elements of order 4.

Yeah - I got that edit in before your reply ;)

Re: Questions around GL(2,C)

Cheers for all that help everyone, seems as though I was on the right lines and you've all classified it for me!! Thanks for all the help!