# Find a group that contains elements $a$ and $b$ such that $|a|=2, |b|=11$ and $|ab|=2.$

I was initially thinking along the lines of either finding some $$S_n$$ or some $$D_n$$ based on examples I saw here.

But the question is which of these $$S_n$$ or $$D_n$$ would be correct, and what would be the value of $$n$$.

Is $$n = 11$$, or is $$n$$ a multiple of $$11$$ and $$2$$?

• $D_{11}$ (the dihedral group of the $11$-gon) will work. Of course, $S_{11}$ contains subgroups isomorphic to $D_{11}$ so it will also work.
– user169852
Commented Aug 5, 2019 at 4:50

$$|a|=2$$ means $$a^2 =1$$, so $$a=a^{-1}$$. Same for $$|ab|=2$$: $$abab=aba^{-1}b=1$$ so $$aba^{-1}=b^{-1}$$.
This is exactly (one of) the definition(s) of the dihedral group! Unluckily, you can find both notation $$D_{11}$$ and $$D_{2 \times 11} = D_{22}$$ in the literature for the* same group you are searching: check Wikipedia to understand what notation you are using.
*the dihedral group is the easiest example to find such $$a,b$$