I have a group G that is not cyclic and contains a subgroup H of order p where p is an odd prime and H has index 2.
Then $\\|G:H| = |G|/|H|=2 => |G|=2p $
Also H is a normal subgroup of G and G has an element $x$ of order 2 which is not an element of H since H has order $p$.
Now $y$ is a generator of H $i.e. <y>=H$ where y has order $p$
I also know that the order of $yx $ is not equal to $1$ or $2p$.
By considering $(Hyx)^p$ how do I prove that the order of $yx$ cannot be $p$?
This is what I have so far:
Suppose $|yx| = p => (yx)^p=1$
$Now (Hyx)^p = HyxHyx...Hyx=HyHy...Hy=Hy^p = H\ since\ |y|=p$
But I am supposed to get a contradiction in the end. I do not know what I did wrong