Three subgroups lemma

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title: "Three subgroups lemma" type: doc version: 1 created: 2026-02-28 author: "Wikipedia contributors" status: active scope: public tags: ["lemmas-in-group-theory", "articles-containing-proofs"] topic_path: "general/lemmas-in-group-theory" source: "https://en.wikipedia.org/wiki/Three_subgroups_lemma" license: "CC BY-SA 4.0" wikipedia_page_id: 0 wikipedia_revision_id: 0

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 CG(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.

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.

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.

Notes

References

• {{cite book | author = I. Martin Isaacs | author-link = Martin Isaacs | year = 1993 | title = Algebra, a graduate course | edition = 1st | publisher = Brooks/Cole Publishing Company | isbn = 0-534-19002-2

References

1. Isaacs, Lemma 8.27, p. 111
2. Isaacs, Corollary 8.28, p. 111

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