# A question about dihedral group

The presentation of the dihedral group is $$D_{2n}= \langle r,s \mid r^{n}=s^{2}=1, (sr)^2=1 \rangle.$$ Now, let $$G$$ be a group of order $$2n$$. Is it true that if $$G$$ contains two elements $$a,b$$ such that $$ord(a)=n$$, $$ord(b)=2$$ and $$ord(ba)=2$$, then $$G\cong D_{2n}$$ ?

Edit: If not, what are the conditions that $$G$$ has to fulfill in order to be isomorphic to $$D_{2n}$$ ?

• In fact any group generated by two elements $a$ and $b$ of order 2 is dihedral of order $2\cdot o(ab)$. So we have a dihedral group of order 2n as a subgroup of our group. But the order of the group is 2n so we get the result. – nobody Sep 3 '19 at 6:46

YES, it is. Let's call $$G_0$$ your presentation of $$D_{2n}$$. The universal property of $$G_0$$ tells you that there is a morphism $$G_0\to G$$ which sends $$r$$ to $$a$$ dand $$s$$ to $$b$$.

Claim. $$G$$ is generated by $$a$$ and $$b$$.

For, it is enough to show that the various elements $$a^k, ba^\ell$$ are pairwise distinct (where $$0\leq k,\ell\leq n-1$$).

The only non trivial thing to prove is that $$a^k= ba^\ell$$ for some $$k,\ell$$ cannot happen. Assume the contrary, so that $$b=a^m, m=k-\ell$$, so $$-(n-1)\leq m\leq n-1$$. Taking inverse and using the fact that $$b$$ has order $$2$$, one may assume that $$0\leq m\leq n-1$$.

The order of $$b$$ is $$2$$, so if $$n$$ is odd it cannot happen. Write $$n=2r$$. Then $$b=a^r$$, the unique element of order of the cyclic group generated by $$a$$. Now $$ba=a^{r+1}$$, which has order $$2r/gcd(2r,r+1)$$. Now $$gcd (2r,r+1)$$ is $$1$$ or $$2$$ since $$2(r+1)-2r=2$$. Consequently, $$o(ba)=2r$$ or $$r$$.

Therefore, if $$r\geq 3$$ you have a contradiction. If $$r=2$$, o$$(a^3)=4$$ and you have a contradiction again.

To sum up, $$G$$ is generated by $$a$$ and $$b$$. Hence the morphism above is surjective, hence bijective.

Note that $$[G:\langle a\rangle]=2$$ so $$\langle a \rangle$$ is normal in $$G$$.
We also have $$baba=1$$ so $$b^{-1}ab=a^{-1}$$.
This means $$G\cong C_n\rtimes C_2=D_{2n}$$
Since in $$G$$, the relation $$a^n=b^2=1, (ab)^2=1$$ hold (i.e., the ones defining $$D_{2n}$$), it follows that $$H:=\langle a,b\rangle$$ is a subgroup of $$G$$ that is isomorphic to a quotient of $$D_{2n}$$ (via $$r\mapsto a$$, $$s\mapsto b$$). From $$a\in H$$, clearly $$|H|\ge n$$, so either $$H=G$$ or $$H=\langle a\rangle$$. In the first case, $$H$$ has the same (finite) order as $$D_{2n}$$, hence the quotient $$H$$ (and so $$G$$) must be isomorphic to $$D_{2n}$$ itself. In the second case, $$H$$ is cyclic. A cyclic group has at most one element of order $$2$$, hence $$b=ab$$, so $$a=1$$, contradicting $$b\in\langle a\rangle$$.