Prove that J is a subgroup of G Could somebody please help me with the following question?
Let $G$ be the group consisting of all invertible $2\times2$ matrices with coefficients in $\mathbb Z/3\mathbb Z$. Here the group is the usual matrix multiplication, and the unit element is $e:=\begin{bmatrix}1&0\\0&1\end{bmatrix} $
Consider the matrices $g:=\begin{bmatrix}0&1\\1&1\end{bmatrix} $
$h:=\begin{bmatrix}1&1\\0&2\end{bmatrix} $
Show that $J:=\{h^n* g^m : m,n \in \mathbb{Z}_{>0} \}$  is a subgroup of $G$.
I have determined that the unit element is in $J$, but I don't know how to show the other axioms of the subgroup criterion as the group is not commutative.
Thank you!
 A: Here is an idea. We want to verify the properties of a subgroup. $J$ is a subgroup of $G$, by definition iff


*

*$a, b \in J \Rightarrow ab \in J$

*$e_G \in J$ (the identity of $G$ has to belong to $J$)

*If $a \in J$, then $a^{-1}\in J$.


The identity element is 
$$e_G:=h^2 = g^8 = \begin{pmatrix}1&0\\0&1\end{pmatrix}$$
so the second property is satisfied. For the third property. Let's take $j\in J$, then there exist $m,n \in \Bbb{Z}_{>0}$ such that
$$j = h^{m}g^{n}$$
Notice that $h^2 = g^8 = e_G.$ We then can find numbers $N, M$ with
$$n + N = 8k, \quad \text{for some $k\in \Bbb{Z}_{>0}$}$$
and
$$m + M = 2\tilde{k}, \quad \text{for some $\tilde{k}\in \Bbb{Z}_{>0}$}$$
Then define $j^{-1}$ as 
$$j^{-1} := g^{N}h^{M},$$
which is the product of the two elemts $g^{N} = h^2g^{N} \in J$ and $h^{M}\in J$. We get
$$\begin{align}
j * j^{-1} &= h^mg^n * g^N h^M \\
&= h^m g^{n+M}h^M\\
&=h^M (g^8)^kh^M \\
&=h^mh^M\\
& = h^{m+M}\\
& = (h^2)^\tilde{k} \\
&=e_G.
\end{align}$$ 
For the first property of the definition, (which you should prove before the third), take $j_{1,2,3}$ as 
$$j_1 = h^{m_1}g^{n_1}, \quad j_2 = h^{m_2}g^{n_2}, \quad j_3 = h^{m_3}g^{n_3}$$
and use the following relations
$$h^2 = g^8 =e_G, \qquad hgh = g^3, \qquad hg^2h = g^6$$
to reduce $j_1*j_2*j_3$ to an element of the form $h^{m_4}g^{n_4}$ for some $m_4,n_4 \in \Bbb{Z}_{> 0}$.
