# 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.

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If you think you have to find out the orders of those elements, why haven't you done so? Do you know what Lagrange's Theorem is? – Ben Millwood Jan 4 '13 at 14:57

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 $D_8$)