1
$\begingroup$

So I have a group of order $2m$ with these elements: $$(\overline{0},\overline{0}),(\overline{1},\overline{0})...(\overline{m-1},\overline{0})$$ $$(\overline{0},\overline{1})(\overline{-1},\overline{1})...(\overline{-(m-1)},\overline{1})$$

Dihedral group $D_m$ has these elements: $$(r^0s^0)(r^1s^0)...(r^{m-1}s^0)$$ $$(r^0s^1)(r^1s^1)...(r^{m-1}s^1)$$

I want write that it is isomorphic to the dihedral group of order m. It is so obvious to me that I'm not sure how to prove it.

Thanks for the help

$\endgroup$
  • $\begingroup$ What is the group operation on your $2m$ elements? $\endgroup$ – Servaes Dec 2 '18 at 20:40
  • 1
    $\begingroup$ Group G of order 2m is: $$\mathbb{Z}_m \rtimes_{\varphi} \mathbb{Z}_2$$ and I am given that $\varphi_{1+2\mathbb{Z}}(1+m\mathbb{Z}) = m-1+m\mathbb{Z}$ $\endgroup$ – JoeyF Dec 2 '18 at 20:40
0
$\begingroup$

As you want to show an isomorphism with a dihedral group, how about constructing a faithful action of your group on the regular $m$-gon? This yields an injection into $D_m$, and because they are both groups of order $2m$ this must be an isomorphism.

Given that the group is constructed as a semidirect product, another approach is to take subgroups $T,S\subset D_m$ such that $T\cong\Bbb{Z}_m$ and $S\cong\Bbb{Z}_2$. And show that $D_m=T\rtimes_{\psi}S$ where $\psi:\ S\ \longrightarrow\ \operatorname{Aut}T$ is the map obtained from $\varphi$ through the isomorphisms $T\cong\Bbb{Z}_m$ and $S\cong\Bbb{Z}_2$. Of course you want $\psi$ to turn out to be the conjugation action of $S$ on $T$.

$\endgroup$
  • $\begingroup$ I haven't seen group actions yet in my class. But thanks for the help! $\endgroup$ – JoeyF Dec 2 '18 at 20:52
0
$\begingroup$

Probably the easiest way to see is to show you have 2 generators, $r$ and $s$ with orders n and 2 respectively. Furthermore, $sr^{-1}=rs$. Hence, it is the dihedral group with 2n elements.

$\endgroup$

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.