$H \leq \mathbb{Z}_q^n$ and $H \cong \mathbb{Z}_q^m$ implies that $\mathbb{Z}_q^n / H \cong \mathbb{Z}_q^{n-m}$ 
Given a pair of positive integers $n,q$ and a subgroup $H \leq \mathbb{Z}_q^n$ such that $H \cong \mathbb{Z}_q^m$ for a positive
  integer $m < n$ then show that 
  $$  \mathbb{Z}_q^n / H \cong \mathbb{Z}_q^{n-m}.$$

This is a problem I came up with trying to generalize the trivial case when $q$ is a prime number (It is trivial because in this case a $\mathbb{Z}_q$- module is a vector space).
I actually found a proof but I believe it's way too long for such a statement so I'm not asking for hints or suggestions: my question is if somebody is able to find a better proof (even a sketch would be fine).
 A: Let $e_i$ be the component vectors of $G=\mathbb{Z}_q^n$, which have $1$ in the $i$th place, and zero elsewhere.  Consider the generators of $H$; let's write one as
$$ h_1 = (a_1, \ldots, a_n) $$
This element has order $q$, so one of the components has order $q$: say $a_i$. Then
$$ \{h_1\}\bigcup \{e_j\mid j\neq i\}$$
is a generating set for $G$ of $n$ order-$q$ elements; there is thus an automorphism of $G$ sending $h_1$ to $e_i$. We can thus quotient out $e_1$, and induction on $m$ then shows $G/H\cong \mathbb{Z}_q^{n-m}$.
Edit: The above only works for $q$ a prime power, but that ends up being enough. Because $G$ can be written as a direct product of its Sylow subgroups
$$ G = P_1\times P_2\times\cdots\times P_r $$
where each Sylow subgroup is the unique one in $G$.  Order considerations show
$$ H = (H\cap P_1)(H\cap P_2)\cdots(H\cap P_r) $$
and this product is direct since the factors have trivial intersection.
Thus
$$ G/H\cong P_1/(H\cap P_1)\times P_2/(H\cap P_2)\cdots $$
and these groups have the desired form by the first part of this answer.
A: $\require{AMScd}$
Here is a slightly different viewpoint, which might clear up what's happening.
Let $A=\mathbb{Z}^n$ with basis $e_1,\ldots,e_n$. Also let $K=qA$, which is a characteristic subgroup of $A$, since for $\phi\in\textrm{Aut}(A)$, $\phi(qa)=q\phi(a)\in K$.
So $G\cong A/K$; let $H$ be generated by $h_1,\ldots,h_m$, all of order $q$. Let $b_i$ be a preimage of $h_i$ in $A$ [it can just have the same coordinates of $h_i$]. Then $B=\langle b_1,\ldots,b_m\rangle$ maps to $H$.
Now there exists an automorphism $\phi\in\textrm{Aut}(A)$ sending $e_i$ to $f_i$, such that $b_i$ gets sent to $n_if_i$ for some integers $n_i$. We thus have the commutative diagram
\begin{CD}
A @>{\phi}>> A\\
@VVV @VVV \\
G @>>{\widetilde{\phi}}> G
\end{CD}
with $\widetilde{\phi}\in\textrm{Aut}(G)$. Note that we must have $\gcd(n_i,q)=1$ in order for $\widetilde{\phi}(h_i)$ to have order $q$. Thus we have 
\begin{align*}
G &=\langle \widetilde{f_1},\ldots,\widetilde{f_n}\rangle\\
\widetilde{\phi}(H) & =\langle n_1\widetilde{f_1},\ldots,n_m\widetilde{f_m}\rangle\\
&= \langle\widetilde{f_1},\ldots,\widetilde{f_m}\rangle
\end{align*}
 The problem is then trivial.
