$|\langle a,b \mid a^2=b^2=(ab)^n=1\rangle|\leq 2n$

Question

I'm trying to prove that group $G=\langle a,b \mid a^2=b^2=(ab)^n=1\rangle\cong D_n$ of order $2n$.

My proof is very similar to the one given here, but I'm struggling to show that $|G|\leq 2n$.

Could you, please, help me with this.

Answer 1

If a word in $G$ contains some $a^k$ or $b^k$ then $k< 2$. So a word looks like $$bababab \text{ or } ababab \text{ or } abababa \text{ etc}$$

But there could only be $n-1$ times of $ab$ and $n$ times of $ba$ since $(ab)^n=1$.

So how many element can we produce?

EDIT

Since $(ab)^n=1\Rightarrow (ab)^{n-1}=ba,\ (ab)^{n-2}=baba,\ (ab)^{n-3}=bababa$ and so on. So the length of each word can not be bigger than $n$.

For length $1$ we have $2$ words : $a,b$

For length $2$ we have $2$ words : $ab,ba$

For length $3$ we have $2$ words : $aba,bab$

and so on

Hence when we reach length $n$ we have a total of $$2+2+...+2=2n$$ words and it is $2n\leq 2n$

Answer 2

Let $c=ab$. Then $a=cb^{-1}$. The presentation in question is then

$$\langle a,b,c\mid (cb^{-1})^2=b^2=c^n=1, a=cb^{-1}\rangle,$$

which is then, by Tietze transformations, equivalent to

$$\langle b,c\mid b^2=c^n=1, bcb=c^{-1}\rangle,$$

which is a standard presentation for $D_n$.