// Workers AI · dad joke modeWhat did the Three subgroups lemma say to its friend? "Let's group up

In mathematics, more specifically group theory, the three subgroups lemma is a result concerning commutators. It is a consequence of Philip Hall and Ernst Witt's eponymous identity.

Notation

In what follows, the following notation will be employed:

If H and K are subgroups of a group G, the commutator of H and K, denoted by [H, K], is defined as the subgroup of G generated by commutators between elements in the two subgroups. If L is a third subgroup, the convention that [H,K,L] = [[H,K],L] will be followed.
If x and y are elements of a group G, the conjugate of x by y will be denoted by $x^{y}$ .
If H is a subgroup of a group G, then the centralizer of H in G will be denoted by C_G(H).

Statement

Let X, Y and Z be subgroups of a group G, and assume

[X,Y,Z]=1

and

[Y,Z,X]=1.

Then $[Z,X,Y]=1$ .^[1]

More generally, for a normal subgroup $N$ of $G$ , if $[X,Y,Z]\subseteq N$ and $[Y,Z,X]\subseteq N$ , then $[Z,X,Y]\subseteq N$ .^[2]

Proof and the Hall–Witt identity

Hall–Witt identity

If $x,y,z\in G$ , then

[x,y^{-1},z]^{y}\cdot [y,z^{-1},x]^{z}\cdot [z,x^{-1},y]^{x}=1.

Proof of the three subgroups lemma

Let $x\in X$ , $y\in Y$ , and $z\in Z$ . Then $[x,y^{-1},z]=1=[y,z^{-1},x]$ , and by the Hall–Witt identity above, it follows that $[z,x^{-1},y]^{x}=1$ and so $[z,x^{-1},y]=1$ . Therefore, $[z,x^{-1}]\in \mathbf {C} _{G}(Y)$ for all $z\in Z$ and $x\in X$ . Since these elements generate $[Z,X]$ , we conclude that $[Z,X]\subseteq \mathbf {C} _{G}(Y)$ and hence $[Z,X,Y]=1$ .