Compute the order of a quotient group element

Let G be a group and N a normal subgroup of G. Prove that the order of the element gN in G/N is n, where n is the least positive integer such that g^n \in N, and is infinite if no such n exists. Give an example to show that the order of gN in G/N may be strictly smaller than the order of g in G.

Suppose g^n \notin N for all positive integers n. Then we have (gN)^n = (g^n)N \neq N = 1_{G/N} for all positive integers n, hence |gN| = \infty.

Suppose g^n \in N for some positive integer n; let n be the least such integer. Then (gN)^n = g^nN = 1, so that |gN| \leq n. If k is some integer strictly less than n such that (gN)^k = g^kN = 1, then we have g^k \in N, a contradiction. Thus |gN| = n.

G is normal in itself for all groups G, and G/G is the trivial group. So every element in G/G has order 1, but elements of G may have arbitrarily large order.

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