Question

Which of the following functions are homomorphisms?

Answer 1

Part A:

Given
`f:Z \rightarrow Z,` defined by
`f(x)=-x`


`f(x+y)=-(x+y)=-x-y \\ \\ f(x)+f(y)=-x+(-y)=-x-y`

but


`f(xy)=-xy \\ \\ f(x)\cdot f(y)=-x\cdot-y=xy`

Since, f(xy) ≠ f(x)f(y)

Therefore, the function is not a homomorphism.



Part B:

Given
`f:Z_2 \rightarrow Z_2,` defined by
`f(x)=-x`

Note that in
`Z_2`, -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular
`f(x)=x`


`f(x+y)=x+y \\ \\ f(x)+f(y)=x+y`

and


`f(xy)=xy \\ \\ f(x)\cdot f(y)=xy`

Therefore, the function is a homomorphism.



Part C:

Given
`g:Q\rightarrow Q`, defined by
`g(x)= (1)/(x^2+1)`


`g(x+y)= (1)/((x+y)^2+1) = (1)/(x^2+2xy+y^2+1) \\ \\ g(x)+g(y)= (1)/(x^2+1) + (1)/(y^2+1) = (y^2+1+x^2+1)/((x^2+1)(y^2+1)) = (x^2+y^2+2)/(x^2y^2+x^2+y^2+1)`

Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.



Part D:

Given
`h:R\rightarrow M(R)`, defined by
`h(a)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)`


`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)`

but


`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)`

Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.



Part E:

Given
`f:Z_(12)\rightarrow Z_4`, defined by
`\left([x_(12)]\right)=[x_4]`, where
`[u_n]` denotes the lass of the integer
`u` in
`Z_n`.

Then, for any
`[a_(12)],[b_(12)]\in Z_(12)`, we have


`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)`

and


`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)`

Therefore, the function is a homomorphism.