Final answer:
To prove that a*b=b*a for all a in a group G where a is unique and a*a=e, we use the property that 'a' acts as its own inverse and the associative property of the group operation, confirming the commutative multiplication of 'a' with any element b in the group.
Step-by-step explanation:
The question pertains to the properties of elements within a group structure in abstract algebra, a branch of mathematics. Let's consider group (G,*), where the operation '*' is the group operation. The problem statement asserts that if an element 'a' in G satisfies the condition a*a=e, where 'e' is the identity element, and 'a' is unique, then a*b=b*a for all 'b' in G. To prove this assertion, we can proceed as follows:
Since 'a' is an element of the group such that a*a=e, and every element in a group has a unique inverse, 'a' functions as its own inverse because combining 'a' with itself yields the identity element 'e'. In a group, the inverse of an element, when multiplied by the element itself, will always result in the identity, regardless of the order of multiplication. Hence, a * b = a-1 * b, which we can rearrange to b = a * a * b. By associativity, a group property, we can write this as b = (a * a) * b = e * b = b. Furthermore, because 'a' is its own inverse, we substitute and get a * b = b * a for all 'b' in G, proving the assertion that a*b=b*a.