Sufficiency of subgroup criterion

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This article gives the statement, and possibly proof, of a basic fact in group theory.
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This article gives a proof/explanation of the equivalence of multiple definitions for the term subgroup
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Statement

For a subset H of a group G, the following are equivalent:

  1. H is a subgroup, viz H is closed under the binary operation of multiplication, the inverse map, and contains the identity element
  2. H is a nonempty set closed under left quotient of elements (that is, for any a, b in H, b^{-1}a is also in H)
  3. H is a nonempty set closed under right quotient of elements (that is, for any a, b in H, ab^{-1} is also in H)

Proof

We shall here prove the equivalence of the first two conditions. Equivalence of the first and third conditions follows by analogous reasoning.

(1) implies (2)

Clearly, if H is a subgroup:

  • H is nonempty since H contains the identity element
  • Whenever a, b are in H so is b^{-1} and hence b^{-1}a

(2) implies (1)

Suppose H is a nonempty subset closed under left quotient of elements. Then, pick an element u from H. (VIDEO WARNING: In the embeddded video, the letter a is used in place of u, which is a little unwise, but the spirit of reasoning is the same).

  • e is in H: Set a = b = u to get u^{-1}u is contained in H, hence e is in H
  • g \in H \implies g^{-1} \in H: Now that e is in H, set b = g, a =e to get b^{-1}a = g^{-1}e is also in H, so g^{-1} is in H
  • x,y \in H \implies xy \in H: Set a = y, b= x^{-1}. The previous step tells us both are in H. So b^{-1}a = (x^{-1})^{-1}y is in H, which tells us that xy is in H.

Thus, H satisfies all the three conditions to be a subgroup.