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Problem 1.6.24 in D&F:

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 $n = |xy|$.

My proof: If $t = xy$, then $tx = xyx = xt^{-1}$. Thus, $t$ and $x$ satisfy the same relations as $D_{2n}$ where $n = |t|$. The elements $t$ and $x$ clearly generate $G$ because $G = \left< x, y \right>$ and $y = xt$. Summarizing, we have shown $$ G = \left< t, x \mid t^n = x^2 = e, tx = xt^{-1} \right> $$

My question: Could you please check my proof to make sure it is sound (maybe suggest other simpler proofs)?

Thanks!

Shaun
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obr
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    See also here – kabenyuk Mar 28 '22 at 16:13
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    @kabenyuk -- After a quick skim of that question, I couldn't find any proof that resembled mine. – obr Mar 28 '22 at 16:21
  • The question from the link already contains your reasoning. The most important thing I wanted to say is that you have no definition of group $D_n$. Therefore you should start with it. – kabenyuk Mar 28 '22 at 16:33
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    @kabenyuk -- Ah, yes, my mistake... $D_{2n}$ is the dihedral group of order $2n$, and has presentation $D_{2n}= \langle r,s \mid r^{n}=s^{2}=1, rs=sr^{-1} \rangle$ – obr Mar 28 '22 at 17:34

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