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I have found the six subgroups that I know that $\mathrm{GL}_2(\Bbb Z_2)$ has, but now I want to prove that this is all. How can I do this? I am currently thinking that I should argue using the fact that if $H \le G$ is a subgroup, then $|H|$ must divide $|G|$ which would imply that I can have no subgroup of order $4$ or $5$. However I don't think my reasoning is going down a conclusive route. Any help appreciated, thanks!

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Josh I and @joriki: Sorry for the confusion! – amWhy Nov 28 '12 at 21:02
up vote 5 down vote accepted

As you wrote in a comment, you're aware that $\mathrm{GL}_2(\Bbb Z_2) \cong S_3$. Subgroups of order $2$ or $3$ must be cyclic and thus generated by one non-identity element; there are five non-identity elements, two of which generate the same subgroup, so that makes $4$ subgroups. The only other possibilities are order $1$ for the trivial subgroup and order $6$ for the group itself, for a total of $6$ subgroups.

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+1 I didn't intend to delete your comment, only my (misguided) answer! – amWhy Nov 28 '12 at 21:05
Aha! I knew I had forgotten that we knew that subgroups of order 2 or 3 must be cyclic. – Moderat Nov 28 '12 at 21:10

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