Part A:
Given [tex]f:Z \rightarrow Z, [/tex] defined by [tex]f(x)=-x[/tex]
[tex]f(x+y)=-(x+y)=-x-y \\ \\ f(x)+f(y)=-x+(-y)=-x-y[/tex]
but
[tex]f(xy)=-xy \\ \\ f(x)\cdot f(y)=-x\cdot-y=xy[/tex]
Since, f(xy) ≠ f(x)f(y)
Therefore, the function is not a homomorphism.
Part B:
Given [tex]f:Z_2 \rightarrow Z_2, [/tex] defined by [tex]f(x)=-x[/tex]
Note that in [tex]Z_2[/tex], -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular [tex]f(x)=x[/tex]
[tex]f(x+y)=x+y \\ \\ f(x)+f(y)=x+y[/tex]
and
[tex]f(xy)=xy \\ \\ f(x)\cdot f(y)=xy[/tex]
Therefore, the function is a homomorphism.
Part C:
Given [tex]g:Q\rightarrow Q[/tex], defined by [tex]g(x)= \frac{1}{x^2+1} [/tex]
[tex]g(x+y)= \frac{1}{(x+y)^2+1} = \frac{1}{x^2+2xy+y^2+1} \\ \\ g(x)+g(y)= \frac{1}{x^2+1} + \frac{1}{y^2+1} = \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)} = \frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1} [/tex]
Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.
Part D:
Given [tex]h:R\rightarrow M(R)[/tex], defined by [tex]h(a)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right) [/tex]
[tex]h(a+b)= \left(\begin{array}{cc}-(a+b)&0\\a+b&0\end{array}\right)= \left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right) \\ \\ h(a)+h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)+ \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)=\left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right)[/tex]
but
[tex]h(ab)= \left(\begin{array}{cc}-ab&0\\ab&0\end{array}\right) \\ \\ h(a)\cdot h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)\cdot \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)= \left(\begin{array}{cc}ab&0\\-ab&0\end{array}\right)[/tex]
Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.
Part E:
Given [tex]f:Z_{12}\rightarrow Z_4[/tex], defined by [tex]\left([x_{12}]\right)=[x_4][/tex], where [tex][u_n][/tex] denotes the lass of the integer [tex]u[/tex] in [tex]Z_n[/tex].
Then, for any [tex][a_{12}],[b_{12}]\in Z_{12}[/tex], we have
[tex]f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)[/tex]
and
[tex]f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)[/tex]
Therefore, the function is a homomorphism.
