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:

- Closure – For any a, b\in H we have that ab\in H
- 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.