Show that for a $∈ G, Na ∈ G/N$, order(Na) is a factor of order(a). Let $(G, ·$) be a group and $N$ be a normal subgroup and $G/N$ be
the corresponding quotient. Show that for a $∈ G, Na ∈ G/N$ 
order(Na) is a factor of order(a).
I am really confused about how to proceed with this. I understand that it is asking me to prove that the order of an element in the quotient group divides the order of an element in a group that it is the quotient group of, but I am unsure of how to proceed.
Here is what I have so far:
Assume that $ord(a) = m $
$(Na)^m = N(a^m) = N(e) = Ne = N$
But i am not really sure where I can take this with the work above. Any help would be appreciated!
 A: We denote the identity element of $G$ by $e$.
Let the order of $a$ in $G$ be $n$, and the order  of $(Na)$ in $G/N$ be $s$.  We need to show that $s \mid n$.  
Since $N$ is normal in $G$, $G/N$ inherits a group structure from $G$, whence
$Na^s = (Na)^s = N, \tag 1$
so
$a^s \in N; \tag 2$
We note we cannot have
$0 \le n < s, \tag 3$
for if (3) binds, since
$Na^n = (Na)^n = Ne = N \Longrightarrow a^n \in N, \tag 4$
the order of $Na$ in $G/N$ is then less than $s$, contrary to our hypothesis.  So 
$n \ge s. \tag 5$
We now exploit the euclidean algorithm to write
$n = qs + r, \tag 6$
where $0 \le r < s$; then 
$r = n - qs, \tag 7$
and
$a^r = a^{n - qs} = a^n a^{-qs} = e a^{-qs} = a^{-qs}; \tag 8$
now since $a^s \in N$, and $N$ is a subgroup of $G$, 
$a^{-s} = (a^s)^{-1} \in N, \tag 9$
whence
$a^{-qs} = (a^{-s})^q \in N; \tag{10}$
but by (8), (10) forces
$a^r \in N; \tag{11}$
but now $0 < r < s$ contadicts the fact that $s$ is the order of $Na$ in $G/N$ is $s$, since (11) yields
$(Na)^r = Na^r = N. \tag{12}$
Thus $r = 0$ and hence $s \mid n$.
A: The proof is essentially the same as the proof that if $\psi:G\rightarrow H$ is a homomorphism, then for all $g\in G$, $o(\psi(g))$ divides $o(g)$ (provided $o(g)$ is finite).
Suppose that $o(g)=n$ and $o(\psi(g))=m$.  Then, observe that $\psi(g)^n=\psi(g^n)=\psi(e)=e$.  Since $o(\psi(g))=m$, it follows that $\psi(g)^m=e$.  Let $d=\gcd(m,n)$, then there are integers $a$ and $b$ so that $d=am+nb$.  Then $\psi(g)^d=\left(\psi(g)^m\right)^a\left(\psi(g)^n\right)^b=e^ae^b=e.$  Therefore, $o(\psi(g))\leq d$, but since $o(\psi(g))=m$, $m\leq d$.  However, since $d=\gcd(m,n)$, $d\leq m$, so $d=m$.  Since $\gcd(m,n)=m$, it follows that $m$ divides $n$.
Now, use the fact that $q:G\rightarrow G/N$ is a homomorphism and you're done.  Alternatively, take the $\gcd$ part out of the argument and apply it directly in this case.
