How do I find the elements of $S=\langle (12),(1324)\rangle$? Let $G=S_4$, I want to find the elements of  

$$S=\langle (12),(1324)\rangle$$

How do I do this? I know that $$\langle(1324)\rangle≤S$$and I know I have to find out orders of $(12)$ and $(1324)$ and possibly use Lagrange's Theorem but I'm not sure how to . Thank you for any help.
 A: You should know that if $H$ is a subgroup of a group $G$ then $|H| \mid |G|$. So a subgroup of $S_4$ can have size only 1, 2, 3, 4, 6, 8, 12, or 24. Furthermore, as you correctly observed, $\langle (1324) \rangle \le S$, so $4\mid |S|$. So $|S|$ can only be 4, 8, 12, or 24.
Now, say $a = (1324)$, $b = (12)$ and find some useful equations. It turns out that $ba = a^{-1}b$, which tells you a very important fact: any element of $S$ can be written $a^mb^n$ for some $m$ and $n$, by using the above rule to "move all the $b$ to the right". This, together with the orders of $a$ and $b$, limits the number of elements you can have – together with the previous limitations, you will be able prove how many elements there are, and to give canonical representations for them in terms of $a$ and $b$.
(Aside: when you get used to group theory, you may be able to guess the group could be dihedral, and spot that if you label the vertices of a square 1, 3, 2, 4 clockwise, then the two elements are generators of the symmetry group $D_8$)
A: Start with the identity. Now multiply on the left by (12) and (1324) giving 2 new elements for a total of 3. Multiply each of those on the right by (12) and (1324), put those together with your old ones and remove duplicates. Repeat, alternating left and right, until this process stabilizes.
