To compute the factor group $\mathbb{Z}_4/\mathbb{Z}_2$

Question

how to compute the factor group $\mathbb{Z}_4/\mathbb{Z}_2$?

I made this example on my mind, so I don't know if it even makes sense.

To be more precise, I want to find all cosets that are elements of this factor group. I know it is of order 2 by Theorem of Lagrange, and that it is isomorphic to $\mathbb{Z}_2$. But when I try to calculate the cosets, I get {0,1}, {1,2}, {3,0}.

Answer 1

You have the group $\Bbb Z/(4) = \lbrace 0, 1, 2, 3\rbrace$ and the subgroup $H = \lbrace 0, 2\rbrace$. Since $\Bbb Z/(4)$ is Abelian, $H$ is obviously normal. You have the cosets $0 + H, 1 + H, 2 + H$ and $3 + H$. Since $0, 2 \in H$, you have $0 + H = 2 + H = H$. The other cosets are $1 + H$ and $3 + H$ and these are equal. Hence,

$$(\Bbb Z/(4))/H = \lbrace H, 1 + H\rbrace.$$

Now $H + (1 + H) = 1 + H$ and $(1 + H) + (1 + H) = 2 + H = H$, so you have a group isomorphic to $\Bbb Z/(2)$ (under the isomorphism $1 \mapsto 1 +H)$.

Answer 2

We know that up to isomorphism, it has to be $\mathbb Z_2$, since the index must be $4/2=2$ and there is only one group of that order.

On the other hand, there is only one subgroup $\mathbb Z_2 \hookrightarrow \mathbb Z_4$, where $\mathbb Z_2=\{0,2\}$, so you can directly compute $\mathbb Z_4/\mathbb Z_2$.