Given that $H$ and $K$ are finite subgroups of $G$ of order $o(H)$ and $o(K)$, prove that $$o(HK) = \frac{o(H)\,o(K)}{o(H\cap K)}$$
I have proved for specific case when $H$ and $K$ have only $e$ in common, but how do I prove the general case? Hints?
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user26857
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Taylor Ted
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4See also this post and other posts linked there. – Martin Sleziak Mar 21 '16 at 13:35
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Every element of $HK$ is of the form $hk$, where $h\in H$ and $k\in K$; further, there are $o(H) o(K)$ such representations. However, these representations are not necessarily unique. If $hk = h'k'$ for some $h,h'\in H$ and $k,k'\in K$, then $(h')^{-1}h = k'k^{-1}$. Since $H$ is a subgroup of $G$, $(h')^{-1}h\in H$; since $K$ is a subgroup of $G$, $k'k^{-1}\in K$. Therefore, if $x = (h')^{-1}h = k'k^{-1} $, then $x \in H\cap K$, $h = h'x$ and $k' = xk$. It follows that the number of repetitions of given element $hk \in HK$ is $o(H\cap K)$. Thus $o(HK)o(H\cap K) = o(H)o(K)$, or $o(HK) = o(H)o(K)/o(H\cap K)$.
kobe
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I had a typo which I will fix. Right-multiply by $k^{-1}$ to get $h = h'k'k^{-1}$. Then left-multiply by $(h')^{-1}$ to get $(h')^{-1}h = k'k^{-1}$. – kobe Mar 24 '15 at 19:11
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no i mean you have written h=h'x and k'=xk .it follows that .... how do it follow? – Taylor Ted Mar 24 '15 at 19:16
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There are $o(H\cap K)$ choices for $x$. So there are $o(H\cap K)$ ways to represent $hk$. – kobe Mar 24 '15 at 19:20
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