What subgroup of $S_4$ is this?

Question

What is the subgroup of $S_4$ generated by $\left(14\right)\left(23\right)$ and $\left(12\right)$?

I can see that they are both order 2, and they don't commute, but can't see where to go from here.

Answer 1

Take all "words" formed by the two elements. The order of each is $2$. The order of their product, $(1423)$, is $4$. Also, $(12)(1423)(12)=(1324)=(1423)^{-1}$. So we have $\langle a,b\mid a^4,b^2,(ba)^2\rangle $.

This is $D_4$, the dihedral group.

There are at most $8$ elements because, using $ba=a^{-1}b$, every word can be put in the form $a^mb^n$, with $0\le m\le3$ and $0\le n\le1$.

Answer 2

Where you go is to calculate products of those permutations and products of the products. When you can no longer generate anything new you will have the subgroup. (You may want to provide a proof then that you have them all.)

Here's a start: $$ (14)(23)(12) = (1423) $$ (multiplying from left to right).