**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.5**

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.

Solution: 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 $|gN| = n$.

The group $G$ 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.