Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
Josh I and @joriki: Sorry for the confusion! –  amWhy Nov 28 '12 at 21:02

1 Answer 1

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.

share|improve this answer
    
+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. –  Mike Nov 28 '12 at 21:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.