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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 attempt : This question already asked here.But i have different approach

Im thinking about different answer

My answer : Since every element of $G$ can be written as $x^a (xy)^b$ for $ 0 \le a\le 2$ and $0 \le b \le n$

let define a function $f : D_{2n} \to G $ by $$f(s^ar^b)=x^a(xy)^b$$

Now we have to prove that $f$ is homomorphism

proof : Take $ s^ar^b$ and $s^c r^d \in D_{2n}$

$f(s^ar^b.s^cr^d)= f(s^{a+c}r^{b+d})= x^{a+c}(xy)^{b+d}$

After that im not able to proceed further

Shaun
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jasmine
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    Why does $rs=sr$? – lhf Feb 26 '21 at 16:46
  • that will represent abelian group @lhf but i have not written $rs=sr$? – jasmine Feb 26 '21 at 16:49
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    Why does $f(s^ar^b.s^cr^d)= f(s^{a+c}r^{b+d})$ ? – lhf Feb 26 '21 at 16:51
  • @lhf $(s^a.s^c )=s^{a+c}$ and $ r^b.r^d=r^{b+d}$ – jasmine Feb 26 '21 at 16:52
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    So you have used $rs=sr$. – ancient mathematician Feb 26 '21 at 16:53
  • okss got its now $rs =sr ^{-1}$ so we have $f(s^ar^b.s^cr^d)= f(s^{a+c}r^{d-b})= x^{a+c}(xy)^{d-b}$ – jasmine Feb 26 '21 at 17:07
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    And what if $c=0$? Then you are claiming that $s^ar^br^d$, which is clearly equal to $s^ar^{b+d}$, is also equal to $s^ar^{d-b}$. So you are claiming that $r^{b+d}=r^{d-b}$ for all $b$ and $d$. That seems unlikely. You are going to have to consider different cases depending on whether $c=0$ or $c=1$. – Arturo Magidin Feb 26 '21 at 17:08
  • @ArturoMagidin if $c=0$ , then $f(s^ar^{d-b})=x^a(xy)^{d-b}= x^a(xy)^bx^0(xy)^d=f(s^ar^b)f(s^0r^d)$

    if $c=1$ , then $f(s^{a+1}r^{d-b})=x^{a+1}(xy)^{d-b}= x^{a}(xy)^bx^1(xy)^d=f(s^ar^b)f(s^1r^d)$

     – jasmine Feb 26 '21 at 17:20
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    If $c=0$, then $(s^ar^b)(s^cr^d)$ is not equal to $s^ar^{d-b}$ (except in the special case that $n$ is even and $b=\frac{n}{2}$). Your entire reply is based on an incorrect assertion, and as such is worthless. – Arturo Magidin Feb 26 '21 at 17:29

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