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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if nobody answers the first question check this link out. looks like a guy figured out : https://answers.yahoo.com/question/index?qid=20090921143403AA7tpJg

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Suppose \(x^5=e\). 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.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2The only thing I can think they want for the infinite case is to note that there will be infinite solutions?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1\[\left( \begin{array}{cc} a & a\\ a & a \end{array} \right) \left( \begin{array}{cc} x_{11} & x_{12} \\ x_{21} & x_{22} \end{array} \right) =\left( \begin{array}{cc} a & a\\ a & a \end{array} \right)\]\[x_{11}+x_{21}=1,x_{12}+x_{22}=1\]is all that is needed i think

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Ahh 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\left( \begin{array}{cc} a & a\\ a & a \end{array} \right) \left( \begin{array}{cc} x_{11} & x_{12} \\ x_{21} & x_{22} \end{array} \right) =\left( \begin{array}{cc} a & a\\ a & a \end{array} \right)\]\[x_{11}+x_{21}=1,x_{12}+x_{22}=1\] 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 \(\left( \begin{array}{cc} 1/2 & 1/2\\ 1/2 & 1/2 \end{array} \right)\)
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