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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.

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Hints.

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!

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    $\begingroup$ 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

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