Suppose that G is an abelian group and H is subset of G then, prove that the subset H = {a∈ G: `a^3` = e} is a subgroup of G.

Suppose that G is an abelian group and H is a subset of G then, prove that the subset H = {a∈ G: `a^3` = e} is a subgroup of G.    

Solution:

To prove that the subset H = {a ∈ G: a³ = e} is a subgroup of an abelian group G, we need to show three conditions hold: closure, identity element, and inverse element.

Closure: Let's take two elements, a and b, from H. We need to show that their product, ab, is also in H. Since a and b are in H, we know that a³ = e and b³ = e. Now, consider the product (ab)³:

(ab)³ = ababab.

Since G is abelian, we can rearrange the product:

(ab)³ = aaabb = a³b³.

Since a³ = e and b³ = e, we have:

(ab)³ = e * e = e.

Therefore, (ab)³ = e, which means that ab is in H. The closure property is satisfied.

Identity Element: We need to show that the identity element of G is in H. Since G is an abelian group, it has an identity element, denoted as e. We know that e³ = e since any element cubed in G gives us the identity element. Therefore, the identity element e is in H.

Inverse Element: Finally, we need to show that for every element a in H, its inverse, denoted as a⁻¹, is also in H. Let's consider an element a in H, which means a³ = e. We want to find its inverse a⁻¹ such that a⁻¹³ = e. By raising both sides of a³ = e to the power of 2, we get:

(a³)² = e².

Simplifying this, we have:

a⁶ = e.

Therefore, a⁻¹ = a⁵. Since a is in H, we have a⁵ = (a³)a² = ea² = a². This means a⁻¹ = a², which is in H.

Since H satisfies all three conditions, closure, identity element, and inverse element, we can conclude that H is a subgroup of the abelian group G.