I have a proof that I must produce, but I'm a little unsure of how to structure it, as I don't have a great deal of practice in forming rigorous proofs.

We have, G the group of 2x2 invertible matrices over $\mathbb{R}$ (with matrix multiplication), and $H\leq G$ consisting of only those matrices with determinant 1.

I must show that, for $g,g' \epsilon\ G$, then $gH=g'H$ if and only if det(g) = det(g')

What I know is that cosets are either equivalent or disjoint - that is, if there is any element in both cosets, then those cosets must be equivalent. How would I form the most rigorous proof of this "if and only if" statement?

Note: I definitely don't want to ask people to do my homework for me, however I've been struggling with proving this rigorously for days.



First suppose that $gH=g'H$. Remember that this is an equality of sets: every matrix in the LHS is also in the RHS, and conversely. So if $h_1$ is a matrix in $H$, then $gh_1$ is in $gH$; so $gh_1$ is in $g'H$; so $gh_1=g'h_2$ for some $h_2$ in $H$. See if you can use this to prove that $\det(g)=\det(g')$.

Conversely, suppose that $\det(g)=\det(g')$. We have to show that $gH\subseteq g'H$ and $g'H\subseteq gH$. So, consider any element of $gH$; it can be written in the form $gh_1$, where $h_1$ is in $H$. You need to find a matrix $h_2$ such that $gh_1=g'h_2$, then confirm that your matrix $h_2$ is an element of $H$.

Good luck!

  • 1
    $\begingroup$ Thank you! I'll give it a try and email the proof to my lecturer. The theory behind all of this is sound, its just my proof writing that needs work. $\endgroup$ – Yoshi Mar 24 '14 at 6:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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