# counterexample, function help

• Oct 14th 2011, 03:58 AM
Juneu436
counterexample, function help
Hey, I can't seem to get anywhere with this big question so if someone could work through this question with me so I understand that would be greatly appreciated.

a) If $\displaystyle g \in M_n$ give a counterexample to $\displaystyle g(x+y)=g(x)+g(y)$

b) Prove $\displaystyle g(\frac{1}{a}x_1 +...+ \frac{1}{a}x_a)=\frac{1}{a}g(x_1)+...+\frac{1}{a}g (x_a)$

c) Using part b, prove that if $\displaystyle G \le M_n$ is finite then $\displaystyle G$ fixes at least one element $\displaystyle k \in R^n$

I think for part c that if $\displaystyle |G|=a$, I can then pick some $\displaystyle x \in R^n$ and consider $\displaystyle x_i = h_ix$ for each element $\displaystyle h_1,...,h_a \in G$.
But I can't get any of part a,b or c anyway.

• Oct 14th 2011, 04:23 AM
Deveno
Re: counterexample, function help
what is $\displaystyle M_n$?
• Oct 14th 2011, 04:35 AM
Juneu436
Re: counterexample, function help
M is the group of all rigid motions of the plane, n is dimension.
• Oct 14th 2011, 06:14 AM
Deveno
Re: counterexample, function help
do these rigid motions have to be origin-preserving? if not, then we can let g(x) = x + a, where a is any (non-origin) point in R^n.

then g(x+y) = x + y + a, whereas g(x) + g(y) = x + y + 2a.
• Oct 15th 2011, 03:55 AM
Juneu436
Re: counterexample, function help
Quote:

Originally Posted by Deveno
do these rigid motions have to be origin-preserving? if not, then we can let g(x) = x + a, where a is any (non-origin) point in R^n.

then g(x+y) = x + y + a, whereas g(x) + g(y) = x + y + 2a.

well M is the coarsest classification of orientation-preserving and orientation-reversing motion, and g(x)=x+a is a translation and thus, orientation-preserving. So you answer makes perfect sense, thanks.

Any thoughts on part b and c?
• Oct 15th 2011, 05:56 PM
Juneu436
Re: counterexample, function help
I have thought of an approach for part c, but it does not use part b.

Any help here.

Thanks
• Oct 16th 2011, 04:32 PM
Juneu436
Re: counterexample, function help
I have thought of doing it like [tex]g(\frac{1}{a}
sorry gys my computer is stuffing up and my working is wrong, any help
• Oct 16th 2011, 06:50 PM
Juneu436
Re: counterexample, function help
If G is finite then every element of G must be a rotation or a reflection. So I get the rough idea behind part c, but I can't get part b.

Any help at all?
Thanks