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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.

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.

freehumorist
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    That M's got swag. – Git Gud Jan 02 '19 at 15:41
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    @GitGud \mathscr is where it's at – Ben Grossmann Jan 02 '19 at 15:47
  • @GitGud it is not for the sake of beauty, but specially used. – freehumorist Jan 02 '19 at 15:55
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    The answer by amWhy (at the duplicate) contains the answers given below. – Dietrich Burde Jan 02 '19 at 19:54
  • Also this post has the same matrices as above in the answer by mrs. – Dietrich Burde Jan 02 '19 at 20:01
  • @DietrichBurde the question may be duplicate, because I didn't verify actually. But about the other post, maybe we had the same reference book (Graduate Algebra, by Hungerford) but the question is about linear groups this time, and they are not explained at the same stage at all here. no way I could find these to merge the posts. – freehumorist Jan 02 '19 at 20:08
  • Did you read the answer by mrs? Actually, the question on the order of a product in matrix groups is one of the most frequent duplicates for abstract algebra. Maybe I did not choose the best duplicate with respect to the title. – Dietrich Burde Jan 02 '19 at 20:09
  • @DietrichBurde You are right. Though I'm surprised he gave the exact same example as answer. Sorry about being totally impatient to dive into it. I generally do check the subject. – freehumorist Jan 02 '19 at 20:14
  • No problem. I just wanted to say that this type of example with two matrices in $GL_2(\Bbb{Z})$ is really popular... – Dietrich Burde Jan 02 '19 at 20:17

2 Answers2

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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.

Ben Grossmann
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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.

vidyarthi
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