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Let $G$ be a finite group, and $H$, $K$ be subgroups.

Let $HK = \{hk: h \in H, k \in K\}$.

Let $\phi: H \times K \rightarrow HK $ be defined by $(h,k) \rightarrow hk$.

Let $|\phi^{-1}(g)| = |\{x:\phi(x)=g\}|$.

I'm looking to show $|\phi^{-1}(g)|=|H \cap G|$, and have managed to show $|\phi^{-1}(g)| \ge |H \cap G|$ (i.e. each group element in $HK$ is represented by at least $|H \cap K|$ products in ), using the following:

$$ hk = (hx)(x^{-1}k),\ \forall x \in (H \cap K) $$

However, I am struggling to see why $|\phi^{-1}(g)| \le |H \cap G|$ (i.e. each group element in $HK$ is represented by at most $|H \cap K|$ products in ), which would then imply the equality. I have seen proofs of this on other StackExchange answers (See paragraph 3 of the proof in this answer). The proof is rather terse, however, and I am struggling to follow it.

Could anyone give a very detailed, every-step-explained proof of this?

Shaun
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hegash
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    The duplicate has so many beautiful proofs. It is worth reading them. If one proof is too terse, take the next answer. There are six answers altogether. – Dietrich Burde Apr 17 '22 at 11:49
  • Fix a pair $(h,k)$. Show that each element of $\phi^{-1}(hk)$ is of the form $(ht,t^{-1}k)$, with $t\in H\cap K$. That gives you the bijection directly. – Arturo Magidin Apr 18 '22 at 02:06

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