# matrix groups

• Sep 24th 2011, 08:08 PM
burak100
matrix groups
$A= \left( \begin{matrix} i & 0 \\ 0 & -i \end{matrix} \right)$
, $B= \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)$
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Show that $\langle A, B \rangle$ is subgroup of $GL_2(\mathbb{C})$. And Show that $\langle A, B \rangle$ generated by $A$ and $B$, and order of $\langle A, B \rangle$ is 8 ?
• Sep 24th 2011, 08:16 PM
Drexel28
Re: matrix groups
Quote:

Originally Posted by burak100
$A= \left( \begin{matrix} i & 0 \\ 0 & -i \end{matrix} \right)$
, $B= \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)$
\\
Show that $\langle A, B \rangle$ is subgroup of $GL_2(\mathbb{C})$. And Show that $\langle A, B \rangle$ generated by $A$ and $B$, and order of $\langle A, B \rangle$ is 8 ?

Note that $A,B$ satisfy the relations $A^4=B^4=1$ and $AB=A^{-2}BA$ and so you can easily verify that the map $f:Q_8\to\text{GL}_2(\mathbb{C})$ (where $Q_8$ is the Quaternion group) defined by $i\mapsto A$ and $j\mapsto B$ is a homomorphism, and since it maps generators to generators we see that it surjects onto $\langle A,B\rangle$, but clearly $\text{im }f\leqslant AB$ and so $\text{im }f=\langle A,B\rangle$. Now, since $\langle A,B\rangle$ can be covered by $Q_8$ we know that $\left|\langle A,B\rangle\right|\leqslant 8$ but since $\left|\langle A,B\rangle\right|\mid |Q_8|=8$ you must have that $\left|\langle A,B\rangle\right\=1,2,4,8$. But, since (as can easily be checked) $I,-I,A,-A,B$ are $5$ distinct elements of $\langle A,B\rangle$ we may conclude that $\left|\langle A,B\rangle\right|=8$ and moreover that $f$ is an isomorphism (since a finite surjection between two sets of equal cardinality must be a bijection).
• Sep 24th 2011, 08:22 PM
burak100
Re: 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, 08:25 PM
Drexel28
Re: matrix groups
Well, I don't know what notation you are using but the conventional notation is $\langle S\rangle$ is the subgroup generated by the set $S$. So, by DEFINITION $A,B$ generate $\langle A,B\rangle$.
• Sep 24th 2011, 08:42 PM
burak100
Re: 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>= $\left( \begin{matrix} 0 & i \\ -i & 0 \end{matrix}\right)$ and it is just an element of GL_2(C) right? not a subgroup....?
• Sep 24th 2011, 08:44 PM
SlipEternal
Re: matrix groups
Perhaps try to just generate the subgroup with $A$ and not $B$ in order to arrive at the contradiction that you need both in order to generate the subgroup? Then try just $B$ and not $A$?
• Sep 24th 2011, 08:52 PM
Drexel28
Re: matrix groups
Quote:

Originally Posted by burak100
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>= $\left( \begin{matrix} 0 & i \\ -i & 0 \end{matrix}\right)$ and it is just an element of GL_2(C) right? not a subgroup....?

This doesn't make sense. It's conceivable that they have defined the group $\langle S\rangle$ to be the set of all words (I assume you know what that means) in $S$ but this is equivalent to generated.
• Sep 24th 2011, 09:43 PM
burak100
Re: 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, 06:12 AM
Deveno
Re: 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.