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Problem Let $G$ be a finite group and let $x$ and $y$ be distinct elements of order $2$ in $G$ that generate $G$. Prove that $G \cong D_{2n}$, where $\vert xy \vert = n$.

Solution We have $x^2 = 1$ and $y^2=1$. If we let $t=xy$, we then have $tx = xt^{-1}$. This is because, $tx = xyx$ and $xt^{-1} = x(xy)^{-1} = xyx$, since $x^{-1} = x$ and $y^{-1} = y$. And note that $y = xt$. Hence, instead of $x$ and $y$ generating the group, we can have $x$ and $t$ to generate the group.

I proved that setting $t=xy$, we have $tx=xt^{−1}$. And since $x$ and $y$ generate G, since $y=xt$, $t$ and $x$ also generates $G$. Is this (along with the fact that $G$ is finite) sufficient to conclude that $G \cong D_{2n}$?

Also, I have another related question, can there be an infinite Dihedral group, i.e., $$D_{\infty} = \langle x,y \vert x^2=1, yx=xy^{-1}, y \text{ has infinite order}\rangle$$ Or is this isomorphic to some other group?

Mikasa
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John Smith
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  • Dor $D_{\infty}$ see http://math.stackexchange.com/questions/515512/infinite-dihedral-group, http://math.stackexchange.com/questions/215774/algebra-infinite-dihedral-group. – Dietrich Burde Dec 29 '13 at 16:41
  • @DietrichBurde Thanks. I like the other representation, i.e., $\langle x,y \vert x^2=y^2=1\rangle$, though my representation is more geometric for my liking. – John Smith Dec 29 '13 at 16:44
  • @DietrichBurde I have a couple more questions. Do you mind coming on chat? – John Smith Dec 29 '13 at 16:46
  • @DietrichBurde http://chat.stackexchange.com/rooms/36/mathematics and I am there. I will keep the problem short. – John Smith Dec 29 '13 at 16:52
  • @DietrichBurde Or I can ask you here itself. I have a question in group theory, rather a clarification. I realized somethings yesterday night and want to check if my understanding right? 1) Automorphism group is a sub-group of the permutation group. 2) Any group can be viewed as a subgroup of $S_n$. 3) The anti-automorphism i.e., the element $\sigma^2(g) = g$ is similar to the reflection element in $D_{2n}$. – John Smith Dec 29 '13 at 16:53
  • The "$;G;$ is a finite group" is vital here. Without this a group can be very infinite and non-dihedral at all... – DonAntonio Dec 29 '13 at 17:37
  • @DonAntonio Why do you say the group is non-dihedral? It is an infinite dihedral group, right? – John Smith Dec 29 '13 at 17:39
  • Well, yes indeed, @LeslieFaulkner. I was thinking of finite dihedral groups. – DonAntonio Dec 29 '13 at 17:47

2 Answers2

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Ok. I realized the answer to my own question. If $t^n$ is not $1$ for any $n \in \mathbb{Z}^+$, this would be an infinite group. Hence, $t^n$ has to be $1$ for some $n$. Taking the least such $n^*$, we get that $G \cong D_{2n^*}$. However, the other question on the infinite dihedral group still remains.

John Smith
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Yes, there is an infinite dihedral group.

Igor Rivin
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