## ganeshie8 one year ago #38 http://i.gyazo.com/46f5bc52baecd530ffb8cc50c6c02b79.png

1. anonymous

if nobody answers the first question check this link out. looks like a guy figured out : https://answers.yahoo.com/question/index?qid=20090921143403AA7tpJg

2. zzr0ck3r

Suppose $$x^5=e$$. If $$x^k=e$$ for $$k=1,2,3,4$$ then we get that $$\gcd(k,5)=1$$ for $$k=1,2,3,4$$ implies there exists $$m,n$$ s.t. $$km+5n=1$$ so that $$x^{km}x^{5n}=x\implies x^{km}=x\implies x=e$$. Since we are saying that $$x$$ is not the identity we have that $$5$$ is the order of $$x$$. Now surely we have $$(x^2)^5=e$$, $$(x^3)^5=e$$, and $$(x^4)^5=e$$, and in general $$(x^{2+l})^5=e, (x^{3+l})^5=e, (x^{4+l})^5=e$$ and $$x^{5l}$$ for $$l=0,1,2,3,4,5,...$$ so the solutions come if $$4's$$ in the finite case.

3. zzr0ck3r

The only thing I can think they want for the infinite case is to note that there will be infinite solutions?

4. misty1212

$\left( \begin{array}{cc} a & a\\ a & a \end{array} \right) \left( \begin{array}{cc} x_{11} & x_{12} \\ x_{21} & x_{22} \end{array} \right) =\left( \begin{array}{cc} a & a\\ a & a \end{array} \right)$$x_{11}+x_{21}=1,x_{12}+x_{22}=1$is all that is needed i think

5. ganeshie8

Ahh thanks! $$\phi(5)=4$$. From your proof, looks it is related the number of integers less than $$5$$ that are relatively prime to $$5$$. so, in a finite group, is it always the case that number of nonidentity elements of order $$k$$ equals $$\phi(k)$$ ? @zzr0ck3r

6. ganeshie8

$\left( \begin{array}{cc} a & a\\ a & a \end{array} \right) \left( \begin{array}{cc} x_{11} & x_{12} \\ x_{21} & x_{22} \end{array} \right) =\left( \begin{array}{cc} a & a\\ a & a \end{array} \right)$$x_{11}+x_{21}=1,x_{12}+x_{22}=1$ thnks @misty1212 , i think that will do. since all the elements of the identity matrix must be equal, we get $\left( \begin{array}{cc} a & a\\ a & a \end{array} \right) \left( \begin{array}{cc} x & x \\ x & x \end{array} \right) =\left( \begin{array}{cc} a & a\\ a & a \end{array} \right)$$x+x=1 \implies x=1/2$ so the identity must be $$\left( \begin{array}{cc} 1/2 & 1/2\\ 1/2 & 1/2 \end{array} \right)$$