I have the following excercise given:
Define a surj. homo. $f: \mathbb{Z}_{2} \times \mathbb{Z}_{4} -> \mathbb{Z}_{2}$ such that $f((0,1))=0$ and determine its Kernel $N=\ker(f)$. List the cosets of $N$.
I already managed to find the homomorphism. But i can not get out the cosets. As $\ker(f)=N$ I defined:
${(0,0), (0,1), (0,2), (0,3)}$
because these get mapped to 0 in $\mathbb{Z}_{2}$.
But how do i calculate the cosets now? And what is the index?
Consider the map $\varphi : \Bbb Z_2 \times \Bbb Z_4 \to \Bbb Z_2$ defined by $\varphi((0,x)) = 0$ and $\varphi((1,x)) = 1.$ We note that $\varphi((x,y) + (u,v)) = \varphi((x+u, y+v)) = 0$ if $x + u \equiv 0$ (mod 2) and 1 if $x + u \equiv 1$ (mod 2). Observe that $x + u \equiv 0$ (mod 2) if and only if $x = u$ so that $\varphi((x+u, y+v)) = 0 + 0$ or $1 + 1.$ Either way, we have that $\varphi((x,u) + (y,v)) = \varphi(x,y) + \varphi(u,v).$ On the other hand, $x + u \equiv 1$ (mod 2) if and only if $x = 1$ and $u = 0$ or vice-versa, hence either $\varphi(x,y) = 1$ and $\varphi(u,v) = 0$ or vice-versa. Either way, $\varphi((x,y) + (u,v)) = \varphi(x,y) + \varphi(u,v),$ so $\varphi$ is a group homomorphism. Certainly, $\varphi$ is surjective, and $\varphi((0,1)) = 0,$ as desired. Observe that $\ker \varphi = \{(0,0), (0,1), (0,2), (0,3) \} \cong \Bbb Z_4.$ Our only cosets are $(0,0) + \ker \varphi$ and $(1,0) + \ker \varphi$ so that $(\Bbb Z_2 \times \Bbb Z_4)/ \ker \varphi \cong \Bbb Z_2.$ But this is no surprise since this is guaranteed by the First Isomorphism Theorem. Calculating the orders at this step is straightforward: either use the isomorphism we now have, or use Lagrange's Theorem. Ultimately, we have that $[\Bbb Z_2 \times \Bbb Z_4 : \ker \varphi] = |\Bbb Z_2 \times \Bbb Z_4|/|\ker \varphi| = 8/4 = 2.$