Question

Yet more group theory..

Find the isometries of R^2 which map the subset \[\left\{ (x,y) \in \mathbb{R} ^{2} : y=x\right\}\] to
itself. In particular, identify the transformations, show that they preserve the subset and state the associated symmetry group.

I thought you could just reflect in y=x twice and get back to the original point, but apparently I'm not allowed...could someone please explain how I would go about doing this question?


Thank you

Answer 1

@Mertsj

Answer 2

nevermind … >.<

Answer 3

for isometries that map \(x=y\) to itself... hmm... \((x,y)\mapsto(-x,-y)\) is one ("inverting" the line), also \((x,y)\mapsto(x+k,y+k)\) i.e. translation along the line

Answer 4

that seems like it otherwise the maps would either not map \(x=y\) to itself or would not be isometries. I forgot to mention the trivial identity isometry \((x,y)\mapsto(x,y)\) (this you can get from the involution I described earlier though)

Answer 5

OMG oldrin is a genius

Answer 6

my guess is that the symmetry group can be generated with the following four maps:$$(x,y)\mapsto(-x,-y)\\(x,y)\mapsto(x-1,y-1)\\(x,y)\mapsto(x+1,y+1)$$

Answer 7

three* maps

Answer 8

thanks @nincompoop I forgot the \((x,y)\mapsto(y,x)\) map... :-p

Answer 9

lawllllllllllllllllllllllllllllllllllllllllllllllllll

Answer 10

@oldrin.bataku I honestly can't thank you enough! Thanks for the help and clear explanation!

Answer 11

np it's a shame I couldn't give these groups proper names... also I should've just written \((x,y)\mapsto(x+k,y+k)\text{ for }k\in\mathbb{R}\) like before

Answer 12

@satellite73
i've attached the original question paper
the solution link is:
http://www.mth.kcl.ac.uk/~bdoyon/symmetries2013/solu2012.pdf

It was question B5