Show that H/K is isomorphic to $Z_2 \oplus Z_2$ We are given $H = \{(1),(13),(24),(12)(34),(13)(24),(14)(23),(1234),(1432)\}$ is a subgroup of $S_4$. Also assume $K = \{(1),(13)(24)\}$ is a normal subgroup of $H$. Show $H/K$ isomorphic to $Z_2\oplus Z_2$. 
This is just a practice question (not assignment). So I have tried finding $H/K$ explicitly.
$H/K = \{\{(1),(13)(24)\},\{(13),(24)\},\{(14)(23),(12)(34)\},\{(1234),(1423)\}\}$. We know there are only $2$ groups of order $4$. One of the elements in $H/K$ we see is $(1234)K$, doesn't this element have a order of $4$, making $H/K$ cyclic and hence not isomorphic to $Z_2\oplus Z_2$?  
 A: After Tobias comment. I realized the order of (1234)K is actually 2 and not 4. Since 
(1234)K(1234)K = (1234)(1234)K = (13)(24)K = K. We know there are only 2 groups of order 4. That is $Z_4$ and $Z_2 \bigoplus Z_2$. Since we see $H/K$ does not have any elements of order 4 it is not cyclic and cannot be isomorphic to $Z_4$, hence $H/K$ must be isomorphic to $Z_2 \bigoplus Z_2$.
A: No - be careful. $(1234)^2 = (13)(24) \in K$. Hence, $(1234)K^2 = K$, so $(1234)K$ has order 2 in $H/K$. In fact, you can just check by hand that all the elements of $H$ either square to $(1)$ or to $(13)(24)$ [You need to use the fact that disjoint cycles commute]. Hence, every element of $H/K$ has order $\leq 2$, which means $H/K$ has to be isomorphic to $\mathbb{Z}_2\oplus \mathbb{Z}_2$.
A: I think your doubt about these two being isomorphic has been solved but just in case you still don't see how to define a precise isomorphism between them note that $\mathbb{Z}_2 \bigoplus  \mathbb{Z}_2 $ has two generators, namley $(1,0)$ and $(0,1)$ also $(1,0) + (0,1) = (1,1)$. Therefore I think it's very easy for you to know what precisely your isomorphism should be and which coset should be send to which generator. You can also use the fact that there are only two groups of order 4 up to isomorphism as you already pointed out.
