Finite index in finite double coset index implies finite double coset index

This article describes a computation relating the result of the Composition operator (?) on two known subgroup properties (i.e., Subgroup of finite index (?) and Subgroup of finite double coset index (?)), to another known subgroup property (i.e., Subgroup of finite double coset index (?))
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Statement

Statement with symbols

Suppose $H \le K \le G$ are groups such that $H$ has finite index $r$ in $K$, and $K$ has finite Double coset index (?) $s$ in $G$: there are $s$ double cosets of $K$ in $G$. Then, $H$ has finite double coset index in $G$, and the double coset index of $H$ in $G$ is bounded by:

$r + r^2(s - 1)$.

Related facts

Other facts about index and double coset index

• Index is multiplicative: If $H \le K \le G$, then $[G:K][K:H] = [G:H]$ in the sense of possibly infinite cardinals. In particular, if $K$ has finite index in $G$ and $H$ has finite index in $K$, then $H$ has finite index in $G$.
• Finite double coset index is not transitive: We can have a situation of groups $H \le K \le G$ such that $H$ has finite double coset index in $K$ and $K$ has finite double coset index in $G$, but $H$ does not have finite double coset index in $G$.

Tightness of the result

For $G$ equal to the dihedral group of order $2p$ for an odd prime $p$, $K$ of order two, and $H$ trivial, the bound is tight. More generally, the bound is tight when $G$ is a Frobenius group, $K$ is a Frobenius complement in $G$, and $H$ is the trivial subgroup.

Proof

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