Proof that dihedral group $D_{2n}$ is isomorphic to the group generated by two group elements of order 2

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An exercise in Aluffi asks the reader to prove that if $a,b$ are distinct elements of order 2 in a group $G$ and $ab$ has finite order $n\geq3$, then the subgroup of $G$ generated by $a,b$ is isomorphic to the dihedral group $D_{2n}$, having already shown that $\langle x,y~|~x^2,y^2,(xy)^n\rangle$ is a presentation of $D_{2n}$.

This is intuitively obvious to me, but I'm having some trouble with a detail of the proof. My strategy is the following:

Let $H=\langle a,b\rangle$ be a subgroup of $G$ and $S=\{x,y\}$. Also let $\mathscr{R}=\{x^2,y^2,(xy)^n\}$ be a subset of $F(S)$, and let $R$ be the normal closure of $\mathscr{R}$ in $F(S)$. Define a set-function $f:S\to H$ by $x\mapsto a$ and $y\mapsto b$, and extend this to a homomorphism $\rho:F(S)\to H$. I want to prove that $\ker\rho = R$, and hence that $H\cong F(S)/R$ as desired (by the first isomorphism theorem).

Proving that $R\subseteq \ker\rho$ was easy, but I'm having some trouble showing that $\ker\rho\subseteq R$.

How should I approach this? Is there a better overall strategy for the proof?