For $A,B \in \mathscr{M}_{2\times2}(\mathbb{Q}) $ of finite order, show that $AB$ has infinite order

Question

Let $G$ be the group $ ( \mathscr{M}_{2\times2}(\mathbb{Q}) , \times ) $ of nonsingular matrices.

Let $ A = \left ( \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right ) $, the order of $A$ is $4$;

Let $ B = \left ( \begin{matrix} 0 & 1 \\ -1 & -1 \end{matrix} \right ) $, the order of $B$ is $3$.

Show that $AB$ has infinite order.

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The only reasoning possible here is by contradiction as $G$ is not abelian. And so I tried, but I got stuck before any concrete development.

Any hints are welcome, Thanks.

Answer 1

We compute $$ AB = \pmatrix{1&1\\0&1} $$ We can prove (using induction, for instance) that $$ (AB)^n = \pmatrix{1&n\\0&1} $$ Since there is no positive $n$ for which $(AB)^n$ is the identity matrix, we see that $AB$ is of infinite order.

Answer 2

You have $AB=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ And $(AB)^n=\begin{pmatrix}1&n\\0&1\end{pmatrix}$ by induction. Therefore, its order in the given group is infinite.