Group theory and generalizations Proposition

Product formula for the order of HK

Builds on Product of two subgroups Intersection of subgroups is a subgroup Cosets have the same cardinality as the subgroup

Statement

Let $G$ be a group and let $H$ and $K$ be finite subgroups of $G$ . Then the product $HK$ has exactly $\dfrac{|H| \, |K|}{|H \cap K|}$ elements.

Remarks

The formula holds also when $HK$ is not a subgroup. The proof counts the pairs $(h, k)$ : two pairs $(h, k)$ and $(h', k')$ give the same product exactly when $h' = hx$ and $k' = x^{-1}k$ for some $x \in H \cap K$ . For example, two distinct subgroups of order $2$ in $S_3$ meet in $\{e\}$ , and their product has $2 \cdot 2 / 1 = 4$ elements.

Reading path

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Group concept
Subgroup definition
Left coset definition
Product of two subgroups definition
Right coset definition
Subgroup criterion statement
Cosets have the same cardinality as the subgroup statement
Intersection of subgroups is a subgroup statement
Product formula for the order of HK statement

Tag	`08PY4GNAPK`
URI	`m2://08PY4GNAPK`
Kind	statement, proposition
Writing	W0 stub
MSC	20D60
Status	This copy of the site has no build data. Run `m2 status 08PY4GNAPK`.
Uses	Product of two subgroups, Intersection of subgroups is a subgroup, Cosets have the same cardinality as the subgroup

Product formula for the order of HK

Statement

Remarks

Reading path

Further reading

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