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.
(b) Let a,b G. Prove that
aba-¹b-¹ = e.

Answer 1

Every element of a finite group G where each element squared equals the identity is its own inverse. Furthermore, the product aba^{-1}b^{-1} is equal to the identity element because a and b each being their own inverse leads to the expression simplifying to abab, which equals e.

Step-by-step explanation:

Finite Group Properties

Given a finite group G with identity element e, and the property that for all elements a ∈ G, we have a² = e.

(a) To prove that every element a of G is its own inverse, we consider the given property a² = e. By definition, an inverse of an element a in a group is such that aa^{-1} = e. Comparing this with the given property a² = e (which is equivalent to aa = e), it is clear that a is its own inverse because the element multiplied by itself yields the identity element. Therefore, a = a^{-1} for all a ∈ G.

(b) Let a and b be elements in G. We need to show that aba^{-1}b^{-1} = e. Since a and b are their own inverses from part (a), we have a^{-1} = a and b^{-1} = b. Substituting into the original expression we get abab = e because ab is also an element of G and must adhere to the property that squaring it gives the identity. Hence, the expression simplifies to e.