Definition 3

Tag 05CZ, permanent. Cite it as [Bourbaki, Tag 05CZ].

Algebra, chapter I, § 6. Extensions, solvable groups, nilpotent groups

Let $G$ be a group and $x$ and $y$ two elements of $G$ . The element $x^{-1}y^{-1}xy$ of $G$ is called the commutator of $x$ and $y$ .

(x, y) is used to denote the commutator of x and y. Then obviously

(y, x) = (x, y)^{-1}.

For x and y to commute it is necessary and sufficient that

(x, y) = e

. More generally,

xy = yx(x, y).

On the other hand we write (3)

x^y = y^{-1}xy = x(x, y) = (y, x^{-1})x.

As the mapping

x \mapsto x^y

is the inner automorphism

\operatorname{Int}(y^{-1})

(x, y)^z = (x^z, y^z)

for all

x, y, z \in G

For $x, y, z \in G$ , we prove the following relations: (4)

(x, yz) = (x, z).(x, y)^z = (x, z).(z, (y, x)).(x, y)

(4 bis)

(xy, z) = (x, z)^y.(y, z) = (x, z).((x, z), y).(y, z)

(5)

(x^y, (y, z)).(y^z, (z, y)).(z^x, (x, y)) = e

(6)

(x, yz).(z, xy).(y, zx) = e

(6 bis)

(xy, z).(yz, x).(zx, y) = e.

Now

\begin{align*} (x, yz) &= x^{-1}z^{-1}y^{-1}xyz = (x, z)z^{-1}x^{-1}y^{-1}xyz = (x, z)(x, y)^z \\ &= (x, z)(z, (x, y)^{-1})(x, y) \end{align*}

by (3), which proves (4). Formula (4 bis) follows similarly. On the other hand,

\begin{align*} (x^y, (y, z)) &= (x^y)^{-1}(z, y)(x^y)(y, z) \\ &= y^{-1}x^{-1}yz^{-1}y^{-1}zyy^{-1}xyy^{-1}z^{-1}yz \\ &= (yzy^{-1}xy)^{-1}(zxz^{-1}yz). \end{align*}

Then writing

u = yzy^{-1}xy, v = zxz^{-1}xy

and

w = xyx^{-1}zx

, we obtain

(x^y, (y, z)) = u^{-1}v.

By cyclically permuting

x, y, z

, we deduce

(y^z, (z, x)) = v^{-1}w

and

(z^x, (x, y)) = w^{-1}u,

which immediately imply (5). Finally, (6) follows by multiplying together the two sides in the three formulae obtained by cyclically permuting

x, y, z

in the formula

(x, yz) = x^{-1}z^{-1}y^{-1}xyz = (yzx)^{-1}(xyz)

, and similarly for (6 bis).

If A and B are two subgroups of G, (A, B) denotes the subgroup generated by the commutators $(a, b)$ with $a \in A$ and $b \in B$ .† Then $(A, B) = \{e\}$ if and only if A centralizes B. $(A, B) \subset A$ if and only if B normalizes A. If A and B are normal (resp. characteristic), so is $(A, B)$ .

† We here reject the notational convention made in § 1, no 1 of extending a law of composition to subsets.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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