Normal implies permutable

DIRECT: The fact or result stated in this article has a trivial/direct/straightforward proof provided we use the correct definitions of the terms involved
View other results with direct proofs
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property must also satisfy the second subgroup property
View all subgroup property implications | View all subgroup property non-implications
|

Property "Page" (as page type) with input value "{{{stronger}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process.Property "Page" (as page type) with input value "{{{weaker}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process.

Statement

Verbal statement

Any normal subgroup of a group is a permutable subgroup.

Symbolic statement

Let $G$ be a group and $H$ a normal subgroup of $G$. Then $H$ is a permutable, or quasinormal, subgroup of $G$. In other words $HK=KH$ for any subgroup $K$ of $G$.

Property-theoretic statement

The subgroup property of being normal is stronger than the subgroup property of being permutable.

Definitions used

Normal subgroup

Further information: Normal subgroup A subgroup $H$ of a group $G$ is a normal subgroup if for any $g\in G$, $gH=Hg$ (viz, the left cosets are the same as the right cosets).

Permutable subgroup

Further information: Permutable subgroup, permuting subgroups A subgroup $H$ of a group $G$ is a permutable subgroup if for any subgroup $K\le G$, $HK=KH$. In other words, $H$ and $K$ are permuting subgroups for every $K$, i.e.:

$\mathrm{\forall }h\in H,k\in K,\mathrm{\exists }{k}^{\prime }\in K,h^{\prime }\in H,hk=k^{\prime }{h}^{\prime }$

Proof

Let $H$ be a normal subgroup of $G$. We need to show that $H$ is permutable in $G$.

Let $K$ be any subgroup of $G$. For every $g\in K$, $Hg=gH$. Now we have:

$HK={\bigcup }_{g\in K}Hg$

and

$KH={\bigcup }_{g\in K}gH$

Since $Hg=gH\mathrm{\forall }g\in K$, we conclude that $HK=KH$.

Notice that the above proof does not anywhere use the fact that $K$ is a subgroup.