# Homework Help: Abstract Algebra - Smallest subgroup of GL(n,R)

1. Feb 14, 2012

### Chinnu

1. The problem statement, all variables and given/known data

Let A = $\left[ \begin{array}{cccc} 1 & 1 \\ 0 & -1 \end{array} \right]$

Let B = $\left[ \begin{array}{cccc} 1 & 2 \\ 0 & -1 \end{array} \right]$

Find the smallest subgroup G of GL(n,R) that contains $A$ and $B$. Also, find the smallest subgroup H of G that contains the matrices AB and BA.

2. Relevant equations

GL(n,R) is the group of all n x n invertible matrices

Homomorphism:

$f(a*b) = f(a) * f(b)$ for some operation $*$

3. The attempt at a solution

I need a hint to start (don't solve the whole thing, please just help me start it).

So far, I can see that $A=$$A^{-1}$ and $B=$$B^{-1}$

Also, for the second part $AB\neq BA$ since $AB=\left[ \begin{array}{cccc} 1 & 1 \\ 0 & 1 \end{array} \right]$ and $BA=\left[ \begin{array}{cccc} 1 & -1 \\ 0 & 1 \end{array} \right]$, also, $\left(AB\right)^{-1}=\left[ \begin{array}{cccc} 1 & -1 \\ 0 & 1 \end{array} \right]=BA$, and, $\left(BA\right)^{-1}=\left[ \begin{array}{cccc} 1 & 1 \\ 0 & 1 \end{array} \right]=AB$

To show that something is a subgroup, I need to show that it is closed under the operation (matrix multiplication in the second part), inverses, contains the identity element, and has the homomorphism property

Can I define an operation that just gives the inverse of the matrix? (meaning I think the set {I, A, B} under that operation would be the smallest subgroup), and a similar group can be taken for the second part, just having it contain {I, AB, BA}.

Or, would the following work: The set {I, A, B, AB, BA} under matrix multiplication?
with this:
For any two elements, say $X$ and $Y$ of the set, $X*Y$ is in the set as well (given what I've shown above).

$A*(B*BA)=(A*B)*BA$, and this seems to work for all such combinations, so associativity holds.

Inverses hold given what I've shown above, and the identity is there by definition.

Last edited: Feb 14, 2012