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Let $G$ be generated by elements $a$ and $b$ where $a^n=1$ for some $n\geq 3$, $b^2=1$ and $bab^{-1}=a^{-1}$. There is a surjective homomorphism $D_{2n}\rightarrow G$, and if $G$ has order $2n$, then $G\cong D_{2n}$.

I studied the proof in here.

Since the relations of generators $a,b$ satisfy the defining relations of $G$ there is an epimorphism from $D_{2n}$ to $G$. (This can be done using theorem of free groups).

Also, it shows that $|G|\leq2n$.

My question is : Since there is an epimorphism from $D_{2n}$ to $G$, can we straight away say that $|G|\geq 2n$ and therefore $|G|=2n$?

Also if $|a|=n$ and $|b|=2$, does the argument above valid?

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    The answer to your first question is no, because the orders of $a$ and $b$ could be proper divisors of $n$ and $2$. In fact $G$ could be the trivial group. But if you know that $|a|=n$ and $|b|=2$ then the answer is yes. – 2017-01-25
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    @DerekHolt Ok I think I know where I have misunderstanding. For a surjective map $f:A\rightarrow B$, we have $|A|\geq |B|$ right? – 2017-01-25
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    @DerekHolt So for $|a|=n$ and $|b|=2$, can I use the fact that $\langle a \rangle \cap \langle b \rangle =1$ to say that $G$ contains the subset $\langle a \rangle \langle b \rangle$ of size $2n$ to complete the proof? – 2017-01-25
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    Yes you can do that! The fact that $\langle a \rangle \cap \langle b \rangle = 1$ follows from the fact that $b$ does now commute with $a$ (since you are assuming that $n \ge 3$). – 2017-01-25

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