Commutator of a group and a subgroup implies normal

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This article describes a computation relating the result of the commutator operator on two known subgroup properties or properties of subsets of groups: (i.e., improper subgroup and subgroup), to another known subgroup property (i.e., normal subgroup)
View a complete list of commutator computations
This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., subgroup realizable as the commutator of the whole group and a subgroup) must also satisfy the second subgroup property (i.e., normal subgroup)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about subgroup realizable as the commutator of the whole group and a subgroup|Get more facts about normal subgroup


Statement with symbols

Suppose G is a group and H is any subgroup. Then, the commutator [G,H], defined as:

[G,H] := \langle [g,h] \mid g \in G, h \in H \rangle

is a normal subgroup of G.

Related facts

Stronger facts

Other related facts

Breakdown for Lie rings

Facts used

  1. Subgroup normalizes its commutator with any subset: If K \le G and A is a subset of G, then K normalizes the commutator [A,K] = [K,A].


Given: A group G, a subgroup H.

To prove: The subgroup [G,H] is normal in G.

Proof: Apply fact (1) to K = G and A = H. We get that G normalizes [G,H]. Hence, [G,H] is normal in G.