Exercise 3.1.5 (Order of $gN$ in $G/N$)

Question

Use the preceding exercise to prove that the order of the element $gN$ in $G/N$ is $n$, where $n$ is the smallest positive integer such that $g^n \in N$ (and $gN$ has infinite order if no such positive integer exists). Give an example to show that the order of $gN$ in $G/N$ may be strictly smaller that the order of $g$ in $G$.

Richard Ganaye · Accepted Answer

\begin{proof}

Consider the set 
$$A = \{\alpha \in \Z^+ \mid g^\alpha \in N\}.$$
If $A = \varnothing$, then for all $\alpha \in Z^+$, $g^\alpha 
ot \in N$, therefore, using Exercise 3.1.4,
$$(g N)^\alpha = g^\alpha N 
e N = 1_{G/N}.$$
This shows $gN$ has infinite order in $G/N$.

We suppose now that $A 
e \varnothing$. Then we may define
$$n = \min (A)= \min\{\alpha \in \Z^+ \mid g^\alpha \in N\},$$
so that $g^n \in N$ and $g^\alpha 
ot \in N$ if $0<\alpha < n$. Then
\begin{align*}
(g N)^n &= g^n N = N = 1_{G/N},\
(gN)^\alpha& = g^{\alpha} N 
e N = 1_{G/N} 	ext{ if }0<\alpha<n.
\end{align*} 
This shows that the order of $gN$ is $n$.

In conclusion, the order of the element $gN$ in $G/N$ is $n$, where $n$ is the smallest positive integer such that $g^n \in N$.

\bigskip
Consider the group $G = Z_4 = \langle x angle$ and $N = \langle x^2 angle$. Then
$$G/N = \{\overline{1} , \overline{x}\} = \{N, x N \},$$
where $(xN)^2 = x^2 N = N$ and $xN 
e N$, so the order of $xN$ is $2$,  strictly smaller that the order of $x$ in $G$, which is $4$.
\end{proof}

Exercise 3.1.5 (Order of $gN$ in $G/N$)

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Exercise 3.1.5 (Order of $gN$ in $G/N$)

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