Suppose | G | = n and m ∈ N is such that gcd(m, n) = 1. If g ∈ G and g^m = e,prove that g = e

To prove that g = e, we need to use the fact that the order of an element divides the order of the group.

Step 1: Let's denote the order of g as k. This means that g^k = e and k is the smallest such positive integer.

Step 2: Since g^m = e, the order of g, k, must divide m (because m is the power to which we raise g to get the identity, and the order is the smallest such power).

Step 3: However, we also know that k divides n (since the order of an element divides the order of the group).

Step 4: Since k divides both m and n, and gcd(m, n) = 1, it must be the case that k divides 1.

Step 5: The only positive integer that divides 1 is 1 itself. Therefore, k must be 1.

Step 6: If the order of g is 1, this means that g^1 = e. Therefore, g = e.

So, we have proved that if g^m = e and gcd(m, n) = 1, then g = e.