I am studying from a book about group-theory. I got the chapter of normal groups and isomorphisms. There was a question:
Let $X=\mathbb{Z}_{4}$,$X'=\{0,2\}$,$Y=\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ and $Y'=\{(0,0),(1,1)\}$.
Verify: $X'\cong Y'$ and $X/X'\cong Y/Y'$, but $X\not\cong Y$.
Where do I start? But verifying I guess it possible to prove it with one of the isomorphisms laws and I should check it manually. But How should I do it?
A group homomorphism is entirely determined by the image of the generators for the group, by the homomorphism property. For instance, if $G$ is a cyclic group with generator $g$ then the powers of $g$ determine the elements of $G$. If I now look at a homomorphism $\varphi : G \to H$ I need only specify the image of $g$ under $\varphi$ and this determines the homomorphism because $\varphi(g^k) = \varphi(g)^k$.
E.g. Let $G = \Bbb Z/(5)$ and $H = \Bbb Z/(10)$. Then $G$ is a cyclic group generated by $1$. Consider the following map
$$\varphi : G \to H : 1 \mapsto 2.$$
Then $\varphi(2) = \varphi(1 + 1) = \varphi(1) + \varphi(1) = 2 + 2 = 4$, etc.
For your question, the implication is that $X^\prime$ is considered as a subgroup of $X$ and $Y^\prime$ as a subgroup of $Y$ (which you can check of course, free exercise). How should one check that $X^\prime \cong Y^\prime$? Just write down an isomorphism (after checking that $X^\prime$ is a cyclic subgroup of $X$).
To check that $X/X^\prime \cong Y/Y^\prime$ you need to know what these quotients are, which should remind you to check that $X^\prime \triangleleft X$ and $Y^\prime \triangleleft Y$. Then write down another isomorphism $X/X^\prime \xrightarrow{\sim} Y/Y^\prime$.
For the final question, note that orders are preserved by homomorphisms (under the homomorphism property). If $X$ is cyclic then $Y$ should at least have one element with order $\lvert X \rvert$. Is this the case here?