# Hall satisfies permuting transfer condition

This article gives the statement, and possibly proof, of a subgroup property (i.e., Hall subgroup) satisfying a subgroup metaproperty (i.e., permuting transfer condition)
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## Statement

### Statement with symbols

Suppose $H$ is a Hall subgroup of a finite group $G$. Suppose, further, that $K$ is a subgroup of $G$ such that $H$ and $K$ are permuting subgroups -- in other words, $HK = KH$. Then, $H \cap K$ is a Hall subgroup of $K$.

## Definitions used

### Hall subgroup

Further information: Hall subgroup

A subgroup $H$ of a finite group $G$ is termed a Hall subgroup if its order $|H|$ and its index $[G:H]$ are relatively prime.

## Facts used

1. Index is multiplicative
2. Lagrange's theorem
3. Product formula: This states that if $H$ and $K$ are subgroups of $G$, we have:

$|HK| = \frac{|H||K|}{|H \cap K|}$.

## Proof

Given: A finite group $G$, a Hall subgroup $H$ of $G$, a subgroup $K$ of $G$ such that $HK = KH$.

To prove: $H \cap K$ is Hall in $K$.

Proof: Rearranging the product formula (fact (3)) yields:

$\frac{|K|}{|H \cap K|} = \frac{|HK|}{|H|}$.

By Lagrange's theorem (fact (2)), and noting that $HK$ is a subgroup of $G$, we get:

$[K:H \cap K] = [HK:H]$.

By fact (1), we have:

$[HK:H][G:HK] = [G:H]$.

Thus, we get:

$[K:H \cap K][G:HK] = [G:H]$.

In particular, $[K:H \cap K]$ divides $[G:H]$. By Lagrange's theorem, we have that $|H \cap K|$ divides $|H|$. Since $|H|$ and $[G:H]$ are relatively prime, we obtain that $[K:H \cap K]$ and $|H \cap K|$ are relatively prime. Thus, $H \cap K$ is a Hall subgroup of $K$.