Find cosets of $H$ in $G$ 
Let $G$ = $\operatorname{Sym}(\{1,2,3,4\})$ and let $ H = ⟨(1,2,3,4),(2,4)⟩$. Write out all the cosets of $H$ in $G$

So, I know that $G$ contains $4!= 24$ elements, because it's the permutation group. 
I don't understand how to get the generated group $H$, so my first question is: Is $H$ a group generated by 2 elements?, the next question is: How to get that group ( I saw in another question that $H=⟨(1,2)⟩=\{(1,2),(2,4),(0,0)\}$, I think that it depends based in the group operation).
My first approach to solve the problem is find the $H$ group and then, I can find the cosets (and that is another problem: How to define that coset) of $H$. 
So, thank you very much for your responses. 
 A: If $H=<\alpha,\beta>$ and $|\alpha|=4 ,|\beta|=2$ then $H=\{(\alpha^4=) I,
\alpha, \alpha ^2 ,\alpha ^3, \beta , \alpha \beta , \alpha ^2\beta , \alpha^3 \beta \}.$
This is how you compute the subgroup generated by two elements $\alpha $ and $\beta$.
EDIT :
In generally if $H = <\alpha, \beta>$ and $\alpha, \beta$ do not have any numbers in common i.e. disjoint cycles then $\alpha^m\beta^n = \beta^n\alpha^m$ for all $m,n \in \mathbb{Z}$, because disjoint cycles commute. Therefore in that case $\{\alpha^m\beta^n : m,n \in \mathbb{Z}\}$ is the complete list of elements of $H$.
However, if the cycles aren't disjoint then we need to compute products of the form $\beta^j\alpha^i$, $\alpha^i\beta^j\alpha^k$, $\beta^i\alpha^j\beta^k\alpha^l$,... etc and develop a relation like $\alpha^s\beta^r= \beta^t\alpha^u$ etc. for some $r,s,t,u \in \mathbb{Z}$ and use Lagrange's theorem to get the order of the subgroup because in generally $\beta^j\alpha^i$ may be distinct from the elements listed above.
Now, in this example the cycles are not disjoint. So I computed  some other elements, basically $\beta\alpha, \beta\alpha^2, \beta \alpha^3$ and found the relations $\beta\alpha=\alpha^3 \beta , \beta\alpha^2 = \alpha ^2\beta,
\beta \alpha^3= \alpha \beta$. And since $|\alpha|=4 ,|\beta|=2$, using associative property and the relations we can compute any combination of products of $\alpha$ and $\beta$ and see that they all belong to the complete list that I stated for $H$.
