Commutator of two subgroups

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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:

$[H, K] : = ⟨ [h, k] ∣ h \in H, k \in K ⟩$

where:

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

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

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

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

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

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

Permuting subgroups characterized in terms of commutators

Subgroups $H, K \leq G$ are permuting subgroups if and only if $[H, K] \leq H K$ ; 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