The Whitehead lemma , named after John Henry Constantine Whitehead , is a statement from the mathematical field of ring theory . The lemma describes the commutator group of the linear group over a ring with one element .

## The linear group

Let it be a ring with a single element. Then is the matrix ring , that is the amount of matrices with components from , a ring with identity. Here is the group of invertible elements, the so-called general linear group -th degree. The image ${\ displaystyle R}$${\ displaystyle M_ {n} (R)}$${\ displaystyle n \ times n}$${\ displaystyle R}$${\ displaystyle GL (n, R)}$ ${\ displaystyle n}$

${\ displaystyle GL (n, R) \ rightarrow GL (n + 1, R), \, {\ begin {pmatrix} a_ {1,1} & \ ldots & a_ {1, n} \\\ vdots & \ ddots & \ vdots \\ a_ {n, 1} & \ ldots & a_ {n, n} \ end {pmatrix}} \ mapsto {\ begin {pmatrix} a_ {1,1} & \ ldots & a_ {1, n} & 0 \\\ vdots & \ ddots & \ vdots & \ vdots \\ a_ {n, 1} & \ ldots & a_ {n, n} & 0 \\ 0 & \ ldots & 0 & 1 \ end {pmatrix}}}$

is obviously an injective group homomorphism , with which one can understand as a subgroup of . The union is called a linear group , sometimes also a stable linear group , according to the construction it is the group of all invertible matrices which, with a finite number of exceptions, agree with the infinite identity matrix . ${\ displaystyle GL (n, R)}$${\ displaystyle GL (n + 1, R)}$${\ displaystyle \ textstyle GL (R): = \ bigcup _ {n \ in \ mathbb {N}} GL (n, R)}$${\ displaystyle \ infty \ times \ infty}$

In each group the elementary matrices of type 1 are contained, they create a subgroup and due to the above homomorphism one can understand as a subgroup of and form the union again . Apparently is a sub. ${\ displaystyle GL (n, R)}$${\ displaystyle E (n, R)}$${\ displaystyle E (n, R)}$${\ displaystyle E (n + 1, R)}$${\ displaystyle \ textstyle E (R) = \ bigcup _ {n \ in \ mathbb {N}} E (n, R)}$${\ displaystyle E (R) \ subset GL (R)}$

## Statement of the Whitehead Lemma

Let it be a ring with a single element. Then is , that is, is the commutator group of . In addition, is , that is, is a perfect group . ${\ displaystyle R}$${\ displaystyle E (R) = [GL (R), GL (R)]}$${\ displaystyle E (R)}$${\ displaystyle GL (R)}$${\ displaystyle E (R) = [E (R), E (R)]}$${\ displaystyle E (R)}$

## Remarks

${\ displaystyle E (R)}$is as a commutator group a normal divisor in , that means you can form the factor group. This is of great importance in the algebraic K-theory and is referred to there with . There is the Abelization of , in particular, it is an Abelian group . ${\ displaystyle GL (R)}$ ${\ displaystyle GL (R) / E (R)}$${\ displaystyle K_ {1} (R)}$${\ displaystyle K_ {1} (R) = GL (R) / E (R) = GL (R) / [GL (R), GL (R)]}$${\ displaystyle K_ {1} (R)}$${\ displaystyle GL (R)}$

If there is a body , it is well known that one has a determinant mapping into the group of invertible elements of the body. One can show that exactly the core of the determinant mapping is and the determinant mapping therefore induces an isomorphism . ${\ displaystyle R}$ ${\ displaystyle \ det: GL (R) \ rightarrow R ^ {*} = R \ setminus \ {0 \}}$${\ displaystyle E (R)}$${\ displaystyle K_ {1} (R) = GL (R) / E (R) \ rightarrow R ^ {*}}$

The simplest field is the remainder class field and according to the above is one element and therefore . It is ${\ displaystyle R = \ mathbb {Z} / 2 = \ {0.1 \}}$${\ displaystyle GL (\ mathbb {Z} / 2) / E (\ mathbb {Z} / 2) \ cong (\ mathbb {Z} / 2) ^ {*} = \ {1 \}}$${\ displaystyle GL (\ mathbb {Z} / 2) = E (\ mathbb {Z} / 2)}$

${\ displaystyle GL (2, \ mathbb {Z} / 2) = \ {{\ begin {pmatrix} 1 & 0 \\ 0 & 1 \ end {pmatrix}}, {\ begin {pmatrix} 1 & 1 \\ 0 & 1 \ end {pmatrix} }, {\ begin {pmatrix} 1 & 0 \\ 1 & 1 \ end {pmatrix}}, {\ begin {pmatrix} 0 & 1 \\ 1 & 0 \ end {pmatrix}}, {\ begin {pmatrix} 1 & 1 \\ 1 & 0 \ end {pmatrix }}, {\ begin {pmatrix} 0 & 1 \\ 1 & 1 \ end {pmatrix}} \}}$

an alternating element, non-commutative group, which must therefore be isomorphic to S 3 . Their commutator group is three-element, more precisely

${\ displaystyle [GL (2, \ mathbb {Z} / 2), GL (2, \ mathbb {Z} / 2)] = \ {{\ begin {pmatrix} 1 & 0 \\ 0 & 1 \ end {pmatrix}}, {\ begin {pmatrix} 1 & 1 \\ 1 & 0 \ end {pmatrix}}, {\ begin {pmatrix} 0 & 1 \\ 1 & 1 \ end {pmatrix}} \}}$,

but is generated by the elementary matrices, that is, for degree 2 applies . This example shows that Whitehead's lemma does not hold for finite dimensions. So one cannot do without the transition to infinite-dimensional matrices. ${\ displaystyle GL (2, \ mathbb {Z} / 2)}$${\ displaystyle E (2, \ mathbb {Z} / 2) \ neq [GL (2, \ mathbb {Z} / 2), GL (2, \ mathbb {Z} / 2)]}$

## Individual evidence

1. Jonathan Rosenberg: Algebraic K-Theory and Its Applications , Springer Verlag 1994, ISBN 3-540-94248-3 , sentence 2.1.4
2. ^ John Milnor: Introduction to algebraic K-theory , Annals of Mathematics Studies 72, Princeton University Press, 1971. Section 3.1
3. Jonathan Rosenberg: Algebraic K-Theory and Its Applications , Springer Verlag 1994, ISBN 3-540-94248-3 , sentence 2.2.2