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Show that is subgroup of . And Show that generated by and , and order of is 8 ?

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- Sep 24th 2011, 09:08 PMburak100matrix groups

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Show that is subgroup of . And Show that generated by and , and order of is 8 ? - Sep 24th 2011, 09:16 PMDrexel28Re: matrix groups
Note that satisfy the relations and and so you can easily verify that the map (where is the Quaternion group) defined by and is a homomorphism, and since it maps generators to generators we see that it surjects onto , but clearly and so . Now, since can be covered by we know that but since you must have that . But, since (as can easily be checked) are distinct elements of we may conclude that and moreover that is an isomorphism (since a finite surjection between two sets of equal cardinality must be a bijection).

- Sep 24th 2011, 09:22 PMburak100Re: matrix groups
this is for order of <A, B> right?

How can I show that <A, B> is generated by A and B - Sep 24th 2011, 09:25 PMDrexel28Re: matrix groups
Well, I don't know what notation you are using but the conventional notation is is the subgroup generated by the set . So, by DEFINITION generate .

- Sep 24th 2011, 09:42 PMburak100Re: matrix groups
Actually , in the question doesn't define what is <A, B> but , I think that means; check that the group generated by all possible products of A and B is a subgroup of order 8????

because if we think that <A, B> is just an inner product then <A, B>= and it is just an element of GL_2(C) right? not a subgroup....? - Sep 24th 2011, 09:44 PMSlipEternalRe: matrix groups
Perhaps try to just generate the subgroup with and not in order to arrive at the contradiction that you need both in order to generate the subgroup? Then try just and not ?

- Sep 24th 2011, 09:52 PMDrexel28Re: matrix groups
- Sep 24th 2011, 10:43 PMburak100Re: matrix groups
I, A, B, AB, BA, AAB, AAA, BAA these are the possibilities with using the fact A^4=I and B^2=I and AB=-BA

- Sep 25th 2011, 07:12 AMDevenoRe: matrix groups
i am confused when reading this thread. for example i get that B^2 = I, so that B is of order 2. furthermore i agree with the OP, and disagree with drexel28 that: AB = -BA, rather than AB = A^-2BA.

first, let's agree on a standard way to write the possible products:

since AB = -BA, whenever we get a B before an A, let's replace the BA with -AB. this means we can always put the A's first.

also, note that A^2 = -I, so A^3 = -A. so let's agree to write all instances of -A as A^3, and when the powers of A get greater than 4,

we can reduce them, since A^4 = I.

now, we know I, A, A^2 and A^3 are all different, so that's 4 elements. since B is of order 2, the only possible products left are:

B, AB, A^2B and A^3B. this tells us we have 8 elements in the set of all possible products of A's and B's.

to show this is a subgroup of GL2(C), we need only show closure, as this is a finite set.

but it should be clear that we can write [(A^j1)(B^k1)][(A^j2)(B^k2)], j1,j2 = 0,1,2 or 3 and k1,k2 = 0 or 1 (here i am using the convention that A^0 = B^0 = I).

as one of our 8 products above by pulling the A^j2 term in front of the B^k1 term, using BA = A^3B (repeatedly, if necessary).

for example: (A^3B)(A^2B) = (A^3)(BA)(AB) = (A^3)(A^3B)(AB) = (A^6B)AB = (A^2B)AB

= A^2(BA)B = A^2(A^3B)B = A^5B^2 = A.

alternatively, we can note that the rules A^4 = B^2 = I, BA = A^3B = A^-1B are the same generating relations as the dihedral group of order 8,

with <A> being the rotation group, and B a generating reflection.