
Homomorphism
http://i453.photobucket.com/albums/q...188/math21.jpg
So I want to show f(g1 + g2) = f(g1) + f(g2)?
I'm confused as to what exactly it is i'm adding, I would guess a generall 2x2 matrix i.e ((a b c d) + (e f g h)) Is this right? I don't understand exactly where O2 comes into it...
I know it's not an Isomorphism as it's not injective, due to a rotation of 2*pi, which means it's not bijective and thus not an isomorphism.
I'd guess that reflection on the other hand is an isomorphism as no single reflection gives itself, i.e you'd have to reflect through the same line twice.
Is this corect? Help would be appreciated mainly for the first part though.

Hi
Take care! The law for matrices in $\displaystyle O_2$ will be multiplication $\displaystyle (O_2\subset GL(2,\mathbb{R}).$ Addition can't be used: $\displaystyle I_2\in O_2$ but $\displaystyle 2I_2\notin O_2$ (since its determinant is $\displaystyle 4$)
Do you now see why $\displaystyle \theta\mapsto rot_{\theta}$ is a homomorphism?
About isomorphisms, you're right for rotations.
But before thinking wether the second function is an isomorphism or not, is it a homomorphism?

Not sure really. Do I want to show that two matrices added together, then apply the function f, but i don't know where to apply the fact it's in I_2...

We have $\displaystyle rot_{\theta}=\begin{pmatrix}cos\theta & sin\theta \\ sin\theta & cos\theta\end{pmatrix}$
To prove that $\displaystyle f$ is a morphism, what you have to show is that, if $\displaystyle a,b$ are two reals, $\displaystyle rot_{a+b}=:f(a+b)=f(a)f(b):=rot_arot_b,$ i.e.
$\displaystyle \begin{pmatrix}cos(a+b) & sin(a+b) \\ sin(a+b) & cos(a+b)\end{pmatrix}=\begin{pmatrix}cosa & sina \\ sina & cosa\end{pmatrix}\begin{pmatrix}cosb & sinb \\ sinb & cosb\end{pmatrix}$

Thanks for the help so far, so now i've expanded the cos(a+b) and sin(a+b) into the two relevant trig identities, but I can't see how to get it into the final form without cheating as such. Is there a factor i can take out?
Also i've managed to prove that refθ is not a HM, I did this by using "Z's" do you reckon this is ok, or do I have to use the reflection matrix again of (cosθ sinθ sinθ cosθ)?

The easiest way I think to prove the matrix equality is to compute the product and check that the matrix you obtain is the one you got by expanding $\displaystyle cos(a+b)$ and $\displaystyle sin(a+b)$.
To prove that ref is not a homomorphism, you can show that the composition of two reflections over a line in $\displaystyle \mathbb{R}^2$ is not always a reflection over a line, hence you just need one example. So there are a lot of good answers, but I'm sorry I don't understand "Z's" ...