TITLE: Smallest possible subgroup of $\,D_6\,$ containing two elements of $\,D_6\,$
QUESTION [6 upvotes]: Let $\,G = D_6;\;$ let $\,H\,$ be the smallest subgroup containing the elements $\,r^2s\,$ and $\,sr^2.\;$ List all the elements in $\,H\,$ and explain.

My intuition leads me to $H =  \{ r^2s, \,sr^2,\, e \}$.

REPLY [4 votes]: Let me add a couple of comments. I assume the notation @MichaelJoyce has spelled out.
Both $r^2 s$ and $s r^2$ are involutions (elements of order 2), so by Lagrange's Theorem the order of $H$ must be divisible by 2. This is a helpful piece of information, signalling that your first guess had to be corrected.
Also, it is not difficult to prove that any group generated by two involutions $a$ and $b$ is dihedral, of order $2 n$, where $n$ is the order of $a b$ (or infinite, if $a b$ has infinite order).
In your case $r^2 s \cdot s r^2 = r^4$ has order 3, so you get a dihedral group of order 6 as shown in the other answers.