I have this problem in which i have to proof that the Dihedral Group is of even order. I've seen some proofs which are somehow complicated, I proved it in a way that seems to me simple, but I'm new at proving and i want to know if my proof is correct.
The Dihedral group is the set of symbols $x^iy^j$ such that: $$ x^i | i= 0, 1. y^j | j = 0, 1, ... , n-1 $$ $$ x^2 = e = y^n $$ $$ xy = y^{-1}x$$
First we prove that $xy^j = y^{-j}x$ proceeding by induction. The case for 1 comes from the definition, let us now proof for k+1. $$ xy^{k+1} = (xy)y^k = y^{-1}xy^k = y^{-1}(xy^k) = y^{-1}(y^{-k}x) = y^{-(k+1)}x.$$ Which is what we wanted to prove.
Now, we prove that xy^j is of order 2. $$ (xy^j)^2 = (xy^j)(xy^j) = (xy^j)(y^{-j}x) = x(y^jy^{-j})x = xx = e $$
Since $xy^j$ is of order 2 then by the Langrange Theorem, the order of the group must be a multiple of 2, which is what we wanted to proof.
Is my proof correct? Seems to me like it is. Thank you for your help.