Question

Let G be a finite group with identity element e. Assume that for all a∈G, we have a² = e

(a) Prove that every element a of G is its own inverse.

Answer 1

Every element of a finite group is its own inverse if the square of every element is equal to the identity element. By taking an arbitrary element 'a' and showing that 'a^2 = e' implies 'a * a = e', which means 'a' is its own inverse since it satisfies the inverse property 'ab = ba = e'.

Step-by-step explanation:

To prove that every element a of a finite group G is its own inverse, given that a² = e for all a ∈ G, where e is the identity element, we can use the definition of an inverse in a group context. In the group G, an element b is said to be the inverse of a if and only if ab = ba = e. Given the condition a² = e, this means that aa = e, which satisfies the definition of an inverse, making a its own inverse.

To see this more formally, let's consider an arbitrary element a from the group. By the given condition:

• a² = e

By the definition of the identity element, we have the property:

• ae = a

Now, since the group operation is associative, we have:

• (aa)a = a(aa) = a

Following this, we can deduce:

• aa = e
• a(aa) = aa
• a(e) = aa = e
• a = a⁻¹

This demonstration shows that a is its own inverse, as required for the proof.