Commutator of two subgroups

Definition

Symbol-free definition

The commutator of two subgroups of a group is defined as the subgroup generated by commutators between elements in the two subgroups.

Definition with symbols

Suppose $G$ is a group and $H$ and $K$ are subgroups of $G$. The commutator of the subgroups $H$ and $K$, denoted $[H,K]$, is defined as:

$[H,K]:=⟨[h,k]\mid h\in H,k\in K⟩$

where:

$[h,k]=h^{-1}{k}^{-1}hk$

is the commutator of the elements $h$ and $k$.

Note that there are two conventions for commutators; in some other conventions:

$[h,k]=hk{h}^{-1}{k}^{-1}$.

Whatever the convention, the set of commutators is the same; the commutator of $h$ and $k$ in the former convention equals the commutator of $h^{-1}$ and $k^{-1}$ in the latter convention.

Facts

Commutator, closure and join

If $H,K\le G$ are subgroups, let $H^K$ denote the closure of $H$ under the action of $K$. Define $K^H$ analogously. We then have:

• $[H,K]$ is a normal subgroup inside $H^K$. In fact, $H^K=H[H,K]$, where $H$ normalizes $[H,K]$.
• $[H,K]$ is a normal subgroup inside $K^H$. In fact, $K^H=K[H,K]$ where $K$ normalizes $[H,K]$.
• $[H,K]$ is a normal subgroup inside $⟨H,K⟩$. Both $H^K$ and $K^H$ are normal inside $⟨H,K⟩$, with $⟨H,K⟩=K{H}^K=H{K}^H$.

For full proof, refer: Commutator of two subgroups is normal in join

Normalizing characterized in terms of commutators

For subgroups $H,K\le G$, $K$ is contained in the normalizer of $H$ if and only if $[H,K]\le H$. (In particular, $H$ is normal if and only if $[H,G]\le H$).

Similarly, $H$ is contained in the normalizer of $K$ if and only if $[H,K]\le K$. Thus, the subgroups $H$ and $K$ normalize each other iff $[H,K]\le H\cap K$. In particular, if both subgroups are normal, their commutator is contained in their intersection.

Permuting subgroups characterized in terms of commutators

Subgroups $H,K\le G$ are permuting subgroups if and only if $[H,K]\le HK$; in other words, the commutator of the subgroups is contained in their product.

Normal closure and quotient

The commutator of two subgroups need not, in general, be a normal subgroup. The normal closure of the commutator of two subgroups is of greater interest. If $L$ denotes the normal closure of $[H,K]$ for $H,K$ subgroups of $G$, then the images of $H$ and $K$ in $G/L$ commute element-wise. Conversely, any normal subgroup for which the images of $H$ and $K$ commute element-wise in the quotient, must be contained in $L$.

However, in the special case when both $H$ and $K$ are normal, the commutator of the subgroups is also normal. Further information: Commutator of normal subgroups is normal