Subgroup normalizes its commutator with any subset

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Statement

Statement with the right-action convention

Suppose G is a group, H is a subgroup of G, and A is a subset of G. Then, H normalizes the following subgroup of G:

[A,H] := \langle [a,h] \mid a \in A, h \in H \rangle.

Here, [a,h] := a^{-1}h^{-1}ah denotes the commutator of the two elements.

Note that since [a,h]^{-1} = [h,a], the subgroup [A,H] equals the subgroup [H,A].

Statement with the left-action convention

Suppose G is a group, H is a subgroup of G, and A is a subset of G. Then, H normalizes the following subgroup of G:

[A,H] := \langle [a,h] \mid a \in A, h \in H \rangle.

Here, [a,h] := aha^{-1}h^{-1} denotes the commutator of the two elements.

Note that since [a,h]^{-1} = [h,a], the subgroup [A,H] equals the subgroup [H,A].

Difference between the two statements

Note also that the subgroup [A,H] defined in the right-action convention is the subgroup [A^{-1},H] as per the left-action convention. However, the truth of the two statements is still equivalent since the quantification is over all subsets of G.

Related facts

Applications

Facts used

  1. Formula for commutator of element and product of two elements: With the right-action convention, this is expressed as:

\! [x,yz] = [x,z][x,y]^z.

where a^g = g^{-1}ag. and [a,b] = a^{-1}b^{-1}ab.

With the left-action convention, it is expressed as:

\! [x,yz] = [x,y]c_y([x,z])

where [g,h] := ghg^{-1}h^{-1} and c_g(h) := ghg^{-1}.

Proof

With the right-action convention

Given: A group G, a subgroup H, a subset A of G.

To prove: H normalizes [A,H].

Proof: Since [a,h] (with a \in A, h \in H) form a generating set for H, it suffices to show that [a,h]^k \in [A,H] for any k \in H. Let's do this. By fact (1), we have:

\! [a,hk] = [a,k][a,h]^k.

This rearranges to give:

\! [a,h]^k = [a,k]^{-1}[a,hk].

Note that since k,h \in H, hk \in H. Thus, the right side is a left quotient of two elements in [A,H], hence it is in [A,H], completing the proof.

With the left-action convention

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