7

I am completely stuck at the question

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 $n = |xy|.$

I have proved that

Let $x$ and $y$ be elements of order 2 in any group $G$. If $t = xy$ then $tx = xt^{-1}$.

Can I get some hints?

Shaun
  • 44,997
Tumbleweed
  • 1,001
  • 2
    It seems that the title became different! – Mikasa Sep 08 '13 at 06:25
  • The same question was already asked here. – Orat Sep 08 '13 at 06:27
  • @BabakS. I don't think so though the title of the above link would be more appropriate. Anyway, someone would re-edit the title properly if necessary. – Orat Sep 08 '13 at 06:34
  • 1
    $\langle x, y \mid x^2 = y^2 = 1 \rangle$ is easily seen to be an infinite group. If you do not know though the use of presentations please do not use them! – PVAL-inactive Sep 08 '13 at 06:51
  • ~>_<~ @PVAL I don't know though the use of presentations, but I didn't use them neither! – Tumbleweed Sep 08 '13 at 06:54
  • @BabakS. The group in the title now is not the Klein four group. There are irreducible words of arbitrary length of the form $xyxyxyxy\dots$. The group is infinite and usually referred to as the infinite dihedral group, with $xy$ and $y$ to be taken as the generators in the usual presentation of the infinite dihedral group. – PVAL-inactive Sep 08 '13 at 06:55
  • @PVAL: What I commented was that the earlier title is good to think appropriately about the group. At the first sight, one might be confused seeing this edited one with Klien four group, if he/she didn't follow the rest. – Mikasa Sep 08 '13 at 07:13
  • Also related: http://math.stackexchange.com/questions/160168/prove-a-group-generated-by-two-involutions-is-dihedral – user1729 Jan 16 '14 at 13:40

4 Answers4

7

Hint: Take the homomorphism that sends $xy$ to a rotation, and $y$ to a reflection. Show that this homomorphism is an isomorphism.

PVAL-inactive
  • 8,551
6

By introducing the letter $t$, you have proven that you group can be written as a quotient of the dihedral group of order $2n$, $D_{2n}=\langle x, t; x^2, t^n, x^{-1}tx=t^{-1}\rangle$ (can you see why?). It also proves that every element has the form $xt^i$ or $t^i$. If the quotient was non-trivial then there would exist some integer $i$ with $0<i<n$ such that either $xt^i=1$ or $t^i=1$. However, $t^i\neq 1$ (why?). So $xt^i=1$, and so $x=t^i$. Can you see the required contradiction which completes the proof?

user1729
  • 31,015
5

Personally, I like the following presentation for $D_{2n}$:

$$\langle x,y\mid x^n=y^2=(xy)^2=1\rangle$$ Now assume that the following presentation is given: $$G=\langle x,y\mid x^2=y^2=(xy)^n=1\rangle$$ So we have: $$ \begin{align*} G=\langle x,y&\mid x^2=y^2=(xy)^n=1\rangle\\ \cong \langle x,y,a&\mid x^2=y^2=(xy)^n=1,a=xy\rangle\\ \cong \langle x,y,a&\mid x^2=y^2=(xy)^n=a^n=1,a=xy\rangle\\ \cong \langle x,y,a&\mid x^2=y^2=a^n=1,a=xy\rangle\\ \cong \langle x,y,a&\mid x^2=y^2=a^n=1,a=xy,(ay^{-1})^2=1\rangle\\ \cong \langle x,y,a&\mid y^2=a^n=1,a=xy,(ay^{-1})^2=1\rangle\\ \cong \langle y,a&\mid y^2=a^n=1,(ay^{-1})^2=1\rangle\\ \cong \langle y,a,b&\mid y^2=a^n=1,(ay^{-1})^2=1,b=y^{-1}\rangle\\ \cong \langle y,a,b&\mid y^2=a^n=1,(ay^{-1})^2=1,b=y^{-1},b^2=1\rangle\\ \cong \langle y,a,b&\mid a^n=1,(ay^{-1})^2=1,b=y^{-1},b^2=1\rangle\\ \cong \langle y,a,b&\mid a^n=b^2=(ab)^2=(ay^{-1})^2=1,b=y^{-1}\rangle\\ \cong \langle y,a,b&\mid a^n=b^2=(ab)^2=1,b=y^{-1}\rangle\\ \cong \langle a,b&\mid a^n=b^2=(ab)^2=1\rangle \end{align*}$$ Which is $D_{2n}$.

user1729
  • 31,015
Mikasa
  • 67,374
  • Wow, impressive, thanks Babak! But I have a question at the very beginning - how could you rule out $yx, xyx$ as elements of $G$? – Tumbleweed Sep 08 '13 at 06:49
  • $(xy)^2=1$ so $xyxy=1$ so $xyx=y^{-1}$ ans since $y=y^{-1}$ so $xyx=y$. – Mikasa Sep 08 '13 at 06:54
  • Sorry I couldn't follow - I only know that $x^2 = y^2 = xxyy = 1$. How can I get $xyxy = 1$? – Tumbleweed Sep 08 '13 at 06:57
  • @Tumbleweed: In fact we should know the following points before. Using these points, the presentation, I wrote at the beginning does make senses. That is: $$x=\begin{pmatrix} 1 & 2 &\cdots &n \end{pmatrix}$$ and $$y=\begin{pmatrix} 1 & 2 & 3 & \cdots & n-2 &n-1 & n\ 1 & n & n-1 & \cdots &4 &3 &2\ \end{pmatrix}$$ – Mikasa Sep 08 '13 at 07:07
  • 1
    Yikes! +1 for all the formatting! – amWhy Sep 08 '13 at 14:15
  • You need to prove that every quotient is also a dihedral group (or is cyclic)... – user1729 Sep 08 '13 at 16:24
  • (And I hope you don't mind my formatting edit...) – user1729 Sep 08 '13 at 16:29
  • @user1729: Thanks for editing, but I thought that if I could make the group $G$ to have the same presentation as $D_{2n}$, then the problem is solved. – Mikasa Sep 08 '13 at 16:40
  • @BabakS. I think the best way to understand why you need this additional step is to consider the following question: Prove that every group which is generated by two elements $a$ and $b$ which each have order four and whose product has finite order at least three is infinite. This is not true, but your logic would prove that it was (the problem being homomorphic images). This is because these groups have presentation $\langle a, b; a^4, b^4, (ab)^n, \mathbf{r}\rangle$ for $n\geq 3$. You need to think about what is contained in the set r. (Note: If r is empty these groups are infinite). – user1729 Sep 08 '13 at 16:44
  • @user1729: you mean that we need to show all conditions in von Dyck's substitution theorem are satisfied? – Mikasa Sep 08 '13 at 17:05
  • @BabakS. I am not sure what von Dyck's substitution theorem is, but my point is that the conditions this question gives you also hold for all homomorphic images of your group. It is a quirk of dihedral groups that their quotients are either dihedral or cyclic. But you need to prove this quirk... – user1729 Sep 08 '13 at 17:11
1

We call the two elements $s$, $t$ instead, so $G=\langle s,t\rangle$.

Note that $G=\langle st, t\rangle$ since $(st)t=s$. Since $G$ is finite, $st$ has a finite order, say $m$, so that $(st)^m=1_G$. We also have $[(st)t]^2=s^2=1$.

We claim that there are no other relations, other than $(st)^m=t^2=[(st)t]^2=1$.

Suppose to the contrary $sts=1$. Then $sstss=ss$, i.e.\ $t=1$, a contradiction. Similarly if $ststs=1$, $tsststsst=tsst$ implies $s=1$, a contradiction. Inductively, $(st)^ks\neq 1$ and $(ts)^kt\neq 1$ for any $k\geq 1$.

Thus $$G\cong D_{2m}=\langle a,b|a^m=b^2=(ab)^2=1\rangle.$$

yoyostein
  • 19,608