3
$\begingroup$

It is clear to me that using semidirect products we can get the groups of order $2p$ for $p=5,7$. I know that there is only one order 2 automorphism of $\mathbb{Z}_q$ for $p=5,7$. How can I use this information to see that $D_{2q}$ and $\mathbb{Z}_{2q}$ are the only groups we can construct?

$\endgroup$
  • $\begingroup$ Semidirect products are classified by the twisting automorphism. Since it is clear to you that the groups have to be semidirect products, and you know there are only $2$ automorphisms, you are done. $\endgroup$ – Cheerful Parsnip Mar 22 '18 at 18:50
  • $\begingroup$ This is true for any odd prime $p$. $\endgroup$ – Lord Shark the Unknown Mar 22 '18 at 18:50
  • $\begingroup$ @LordSharktheUnknown: True for any prime, odd or even :) $\endgroup$ – Steve D Mar 22 '18 at 18:59
5
$\begingroup$

This is because there are then just two homomorphisms from $\Bbb Z_2$ to $\textrm{Aut}(\Bbb Z_p)\cong\Bbb Z_p^*$. That is the trivial automorphism, and the one mapping $1$ to $-1$. These give the cyclic group and the dihedral groups of orders $2p$ respectively.

$\endgroup$
4
$\begingroup$

I suspect some people very new to group theory might want an answer to this so I'll give a possibly more full answer. I'll be assuming a little knowledge of group theory so feel free to ask me to expand:

Let $G$ be a group of order $2p$ for any prime $p$.

$G$ then has some element of order $p$, say $x$, and some element of order $2$, say $y$. Let $N=\langle x\rangle$ (the subgroup generated by $x$).

Since $[G:N]=2$, $N$ is normal in $G$, so $yxy^{-1}=x^i$ for some $i\in\{1,\ldots,p-1\}$.

Repeating we get $x=y^2xy^{-2}=yx^iy^{-1}=(yxy^{-1})^i=x^{i^2}$.

In particular $i^2\equiv 1$ (mod $p$) so $i\equiv\pm 1$ (mod $p$). If $i=1$ then $x$ and $y$ commute so $G\cong \mathbb{Z}_{2p}$. If $i=-1$ then $G\cong\langle x,y|x^p=y^2=1,yxy^{-1}=x^{-1}\rangle\cong D_{2p}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.