Sufficiency of subgroup criterion

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This article gives a proof/explanation of the equivalence of multiple definitions for the term subgroup
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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)


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.