Vtyjkyuk

Given a pair of positive integers $n,q$ and a subgroup $H leq mathbbZ_q^n$ such that $H cong mathbbZ_q^m$ for a positive
integer $m < n$ then show that
$$ mathbbZ_q^n / H cong mathbbZ_q^n-m.$$

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 $mathbbZ_q$- 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).

Let $e_i$ be the component vectors of $G=mathbbZ_q^n$, 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
$$ h_1bigcup e_jmid jneq i$$
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 $G/Hcong mathbbZ_q^n-m$.

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
$$ G = P_1times P_2timescdotstimes P_r $$
where each Sylow subgroup is the unique one in $G$. Order considerations show
$$ H = (Hcap P_1)(Hcap P_2)cdots(Hcap P_r) $$
and this product is direct since the factors have trivial intersection.

Thus
$$ G/Hcong P_1/(Hcap P_1)times P_2/(Hcap P_2)cdots $$
and these groups have the desired form by the first part of this answer.

$requireAMScd$
Here is a slightly different viewpoint, which might clear up what's happening.

Let $A=mathbbZ^n$ with basis $e_1,ldots,e_n$. Also let $K=qA$, which is a characteristic subgroup of $A$, since for $phiintextrmAut(A)$, $phi(qa)=qphi(a)in K$.

So $Gcong A/K$; 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 $B=langle b_1,ldots,b_mrangle$ maps to $H$.

Now there exists an automorphism $phiintextrmAut(A)$ 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

beginCD
A @>phi>> A\
@VVV @VVV \
G @>>widetildephi> G
endCD

with $widetildephiintextrmAut(G)$. Note that we must have $gcd(n_i,q)=1$ in order for $widetildephi(h_i)$ to have order $q$. Thus we have
beginalign*
G &=langle widetildef_1,ldots,widetildef_nrangle\
widetildephi(H) & =langle n_1widetildef_1,ldots,n_mwidetildef_mrangle\
&= langlewidetildef_1,ldots,widetildef_mrangle
endalign*
The problem is then trivial.