# 2x2 matrices and groups under multiplication

Let n be a positive integer.

(a) Let G be a set of real 2x2 matrices $A$ such that the $detA$ is a rational number of the form $m/n^t$ where $m$ and $t$ are nonnegative integers and $m\ne0$. Is G a group under multiplication?

(b) Same for $H$ of all real 2x2 matrices A such that $detA$ is of the form $n^t$ where $t\in \mathbb{Z}$?

I know the four defining properties of a group are closure, associativity, identity element, and inverse:

1. Closure - For every pair of elements $a$ and $b$, $a*b$ must be an element of the group. Should I use $GL(2,{\Bbb{R}})$ as a mental reference (it's a group under multiplication) to see if the properties of a group hold for both $G$ and $H$ in same way?

2. Associativity - For any three elements $a, b,$ and $c$, the equality $a*(b*c)=(a*b)*c$ must hold. I know multiplication of 2x2 matrices is always associative so this holds.

3. Identity element - There is a group element $e$, the identity element, such that $a*e=e*a=a$ for any $a$ in the group.

4. Inverse - For every group element $a$, there is an element $b$, the inverse, such that $a*b=b*a=e$, where $e$ is the identity. Looks like the determinants for $G$ and $H$ are not equal to $0$, so they have an inverse.

The determinant forms are throwing me off a bit...can someone provide some guidance? Or a more formal way of proving this?

• Is $t$ a fixed integer? – Omnomnomnom Nov 3 '15 at 19:09
• And what do you mean by using $GL(2,R)$ "as an example"? – Omnomnomnom Nov 3 '15 at 19:11
• You cannot use just one example to prove that a property holds for all examples! – The Chaz 2.0 Nov 3 '15 at 19:16
• The formal way to prove this is to start off with "Let A, B be in G. That means, by definition, that detA is [...] and detB is [...]. Then their product AB is also in G because its determinant is..." – The Chaz 2.0 Nov 3 '15 at 19:17
• No no I understand that, I'm just saying would it be helpful to look at GL(2,R) as an example of a group under multiplication where these properties hold (just for my own reference)...I wasn't going to use this group to actually prove these necessary properties of G and H here. – user0990 Nov 3 '15 at 19:20

Hint: In determining closure and inverses, note that $$\det(AB) = \det(A)\det(B)\\ \det(A^{-1}) = 1/\det(A)$$
• For G's closure, would showing that $\frac{mx}{n^ty^s}=\frac{m}{n^t}\frac{x}{y^s}$ where $x$ is nonnegative, $y$ and $s$ are positive integers suffice? – user0990 Nov 4 '15 at 16:59
• Also, can't we simply say that inversion holds for G since the $detA$ is nonzero? (Using the fact that an n × n matrix has an inverse if and only if its determinant is nonzero) – user0990 Nov 4 '15 at 17:36
• Just because a matrix in $G$ has an inverse, doesn't mean that inverse is itself an element of $G$ – Omnomnomnom Nov 4 '15 at 17:54
• For closure: what you need to check is whether that product can itself be written in the form $m/n^t$ for some choice of integers $m,n,t$ – Omnomnomnom Nov 4 '15 at 17:56