Coset Conjoinment and Theorem

coset conjoinment

If $H, K$ are Subgroups of $G$ . Define:

H K = {h k | h \in H, k \in K}

Theorem

If $H, K \leq G$ are finite then:

| H K | = \frac{| H | | K |}{| H \cap K |}

Proof

Notice that $H K$ is a union of left cosets of $K$ . Namely:

H K = ⋃_{h \in H} h K

Since each of $K$ has $| K |$ elements it suffices to find the number of distinct left cosets of the form $h K$ . But for $h_{1} K = h_{2} K \forall h_{1}, h_{2} \in H ⟺ h_{2}^{- 1} h_{1} \in K$ via Left (and Right) Cosets form a Partition in G. Then:

h_{1} K = h_{2} K ⟺ h_{2}^{- 1} h_{1} \in H \cap K ⟺ h_{1} (H \cap K) = h_{2} (H \cap K)

Thus the number of distinct cosets of the form $h K$ for $h \in H$ is the number of distinct cosets $h (H \cap K)$ for $h \in H$ . The latter number, by Lagrange's Divisibility Theorem of Order of Subgroups equals $\frac{| H |}{| H \cap K |}$ . Thus $H K$ consists of that number of distinct cosets for $K$ , each of which has $| K |$ elements, giving the formula above.

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