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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 $|yx| = p => (yx)^p=1$

Then $(Hyx)^p=H$

$Now (Hyx)^p = HyxHyx...Hyx=HyHy...Hy=Hy^p = H\ since\ |y|=p$

But I am supposed to get a contradiction in the end. I do not know what I did wrong

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  • $\begingroup$ I think you mean $\lvert G \rvert = 2 \lvert H \rvert$. $\endgroup$ – Alex Provost Dec 26 '17 at 15:59
  • $\begingroup$ yes that s how it was supposed to be.. I fixed it, thanks $\endgroup$ – CXB Dec 26 '17 at 16:01
  • $\begingroup$ For the first question, if $yx$ were the identity, then $x$ would belong to $H$, a contradiction. And if $yx$ were of order $2p = \lvert G \rvert$ then $G$ would be cyclic. $\endgroup$ – Alex Provost Dec 26 '17 at 16:01
  • $\begingroup$ May I ask why $HyxHyx...Hyx=HyHy...Hy$? $\endgroup$ – harlem Dec 27 '17 at 20:05
  • $\begingroup$ This follows since $x$ is an element of $H$, thus $Hx=H$ $\endgroup$ – CXB Dec 28 '17 at 10:31
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For the first question, if $yx$ were the identity, then $x$ would belong to $H$, a contradiction. And if $yx$ were of order $2p = \lvert G \rvert$ then $G$ would be cyclic. Therefore the order of $yx$ is either $2$ or $p$.

For the second question, note that $G/H$ has order two, so it is cyclic and consists only of two distinct cosets $H$, $Hyx$ (since $yx$ cannot belong to $H$). Since $p$ is odd, $(Hyx)^p = Hyx$ must be the nontrivial element in $G/H$, but if $yx$ were of order $p$ then we would also have $(Hyx)^p = H(yx)^p = H$, a contradiction. Therefore $yx$ has order $2$. (And in fact, it is not hard to see that $G$ has to be a dihedral group, generated by the reflections $x$ and $yx$.)

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  • $\begingroup$ why does it follow that given p is odd, then $(Hyx)^p=Hyx$ must be the nontrivial element in $G/H$ please? $\endgroup$ – CXB Dec 26 '17 at 16:38
  • $\begingroup$ @CXB Because $G/H$ is a group of order two generated by $Hyx$. So $(Hyx)^n$ is the identity $H$ when $n$ is even, and it is the generator $Hyx$ when $n$ is odd. $\endgroup$ – Alex Provost Dec 26 '17 at 16:40
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Here's another solution for the part $ yx $ can not have order $ p $, in case you are interested. By Lagrange Theorem, $ |yx| $ is either $ 1, 2, p $, or $ 2p $ where $ p $ is an odd prime, and you have shown that it can not be $ 1 $ or $ 2p $. Since $ |G| = 2p $ and $ |H| = p $, we have $ H $ is a Sylow $ p $-subgroup of $ G $. Since $ H \triangleleft G $, it is the unique Sylow $ p $-subgroup of $ G $. So if $ yx $ has order $ p $, then $ \left< yx \right> = H $, so $ x \in H $, a contradiction.

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