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Let $a = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$ and $b = \begin{pmatrix}0 & 1 \\ -1 & -1\end{pmatrix}$ be elements of $\mathrm{GL}_2\left(\mathbb{Q}\right)$.

Prove that the subgroup $\langle a, b \rangle$ has order 6.

I can show that $\left|a\right| = 2$, $\left|b\right| = 3$, and $\left|ab\right| = 2$. By LaGrange's theorem it follows that the order of $\langle a, b \rangle$ must be a multiple of 6.

But how can I prove that $\left|\langle a,b \rangle\right|=6$?

My best idea so far is to enumerate all possible products (i.e., $e, a, b, aa, ab, ba, bb, \ldots$) up to redundancy and show that they generate only 6 elements.

Is there a more clever way to show this?

Hints preferred, please!

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  • $\begingroup$ If $|ab| = \infty$, then $\langle a, b \rangle \supseteq \langle ab \rangle$ has infinite order too. $\endgroup$ – user296602 Jul 28 '16 at 0:56
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    $\begingroup$ Here's a hint. I wrote out a few members of your group and found, indeed, six distinct elements. Then I made a little 6 by 6 multiplication table, filled in all the products, and confirmed that the set of 6 is closed under multiplication. You should do the same. $\endgroup$ – Will Jagy Jul 28 '16 at 1:32
  • $\begingroup$ $|ab|$ is not $\infty$. $\endgroup$ – George Law Jul 28 '16 at 1:43
  • $\begingroup$ @GeorgeLaw Yes, you're right. Sorry for my error. $\endgroup$ – nimble agar Jul 28 '16 at 1:46
  • $\begingroup$ It should be easy enough to show that $\{I, a, b, b^2, ab, ab^2\}$ are in your group. Clearly if you multiply any of those on the right side by $b$ you get an element already accounted for. Multiply by $a$ and $ba = ab^2, b^2 a = ab, aba = b^2, ab^2 a = b$ $\endgroup$ – Doug M Jul 28 '16 at 1:57
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The dihedral group of degree 3 (order 6) is characterized as follows: $$D_3\ =\ \langle \rho,\sigma \mid \rho^3=\sigma^2=e,\,\sigma\rho=\rho^{-1}\sigma \rangle$$ You have shown that $b^3=a^2=I$. It remains to show that $ab=b^{-1}a$. This will mean that the mapping $a\mapsto\sigma,\,b\mapsto\rho$ is an isomorphism from $\langle a,b\rangle$ to $D_3$.

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Groups of order 6 are either $C_6$(cyclic) or $D_6$(Dihedral). $a$ and $b$ don't commute so you just need to check $aba=b^{-1}$. i.e You only have to check one relation

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Your original idea is fairly straight forward also.

Since $|a|=2$ and $|b|=3$ you can construct the set $H=\{e,a,b,b^2,ab,ab^2\}$. Notice that each of the six element must be distinct. You can show that $H$ is subgroup of $GL_2(\mathbb{Q})$, and thus a group of order $6$. Now $\langle a,b\rangle$ is the smallest subgroup generated by $a$ and $b$, and thus $\langle a,b\rangle =H$and has order $6$.

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