# 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: $[H,K] := \langle [h,k] \mid h \in H, k \in K \rangle$

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

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 $\langle H, K \rangle$. Both $H^K$ and $K^H$ are normal inside $\langle H, K \rangle$, with $\langle H, K \rangle = KH^K = HK^H$.

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