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In mathematics, a subset of a group that itself forms a group is called a subgroup. So a subgroup satisfies all of the four group axioms.

How do you verify whether a subset of a group is a subgroup?

Suppose that we want to check whether a subset H of a group G is a subgroup or not. Going by the above definition we would have to check if H satisfies the four group axioms – closure, associativity, the existence of an identity, and the existence of inverses.

However, since the elements of H come from a group it is clear that associativity is already satisfied. So we need not check for associativity.

We claim that the closure property and existence of inverses automatically implies the existence of identity.

Proof: Pick an element a\in H. Since the subgroup contains inverses we see that a^{-1}\in H.

By the closure property, aa^{-1} = e\in H

Hence the identity element e belongs to H.

So we see that in order to verify whether a subset H is a subgroup or not we only need to verify two things.

Two step subgroup criterion:

A subset H of G forms a subgroup if:

  1. Closure – For any a, b\in H we have that ab\in H
  2. Existence of Inverses – For any a\in H we have that a^{-1}\in H

The two step criterion can be converted to a one step criterion by simply verifying that if a, b\in H then ab^{-1}\in H.

Examples of Subgroups:

A) We know that the set of integers Z along with the addition operation forms a group. Let H denote the set of even integers. We claim that the set of even integers forms a subgroup.

Proof: 1. We know that the sum of two even integers is even, so the axiom of closure is satisfied.

2. If a is an even integer then clearly (-a) is also an even integer. Hence, H is closed under taking inverses.

By the two step subgroup criterion we conclude that the set of even integers forms a subgroup.

B) We know that the set of nonzero rationals Q* along with the multiplication operation forms a group. We claim that H = {+1,-1} forms a subgroup

Proof: 1. We can clearly verify by hand that H is closed under multiplication. So the closure property is satisfied.

2. The multiplicative inverse of 1 is 1 and that of (-1) is (-1). So the set H is closed under taking inverses.

By the two step subgroup criterion we conclude that the set {+1,-1} forms a subgroup.

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