# “Canceling out” factors in quotient group.

Let $\mathbb{Z}_n^m$ be a direct product of integers modulo $n$ over itself $m$ times, or $\mathbb{Z}_n^m=\{(a_1,a_2,...,a_m)\ |\ a_i\in\mathbb{Z}_n\text{ for all } 1\leq i\leq m\}$ over coordinate-wise addition mod $n$. I need a result that explicitly states that the quotient group $\mathbb{Z}_n^m/\mathbb{Z}_n\cong\mathbb{Z}_n^{m-1},$ but I don't quite know how to prove this, if true at all. Intuitively it seems to make sense, and this question seems to have a similar result in it. Clearly $\mathbb{Z}_n\unlhd\mathbb{Z}_n^m$, as $\mathbb{Z}_n$ is isomorphic to $Z=\{(a,0,0,...,0)\ |\ a\in\mathbb{Z}_n\}\subset\mathbb{Z}_n^m,$ and $Z$ is a normal subgroup of $\mathbb{Z}_n^m$, so the quotient group exists to begin with. I've tried actually finding the isomorphism without success. I know that the order of the quotient group is the same as that of $\mathbb{Z}_n^{m-1}$, so it would seem very plusible that the two groups are indeed isomorphic. How could I proceed from here?

• Technically $\mathbb{Z}_n^m / \mathbb{Z}_n$ doesn't make sense since $\mathbb{Z}_n$ is not a subset of $\mathbb{Z}_n^m$. Do you mean $\mathbb{Z}_n^m / Z$? – Adayah Nov 3 '17 at 17:54
• To be exact, yes. I can represent all my groups I'm working in a similar manner as I did with $\mathbb{Z}_n$ and $Z$, so showing $\mathbb{Z}_n^m/Z\cong\mathbb{Z}_n^{m-1}$ will suffice for me. – cethy Nov 3 '17 at 18:03

Consider the homomorphism $\mathbb{Z}^m_n \mapsto \mathbb{Z}^{m-1}_n$ defined by $f(a_1,a_2,...,a_m) = (a_2,...,a_m)$. This is surjective, and its kernel equals $Z$. Therefore, by one of the fundamental theorems of quotient groups, the map $f$ induces an isomorphism $$F : \mathbb{Z}^m_n / Z \to \mathbb{Z}^{m-1}_n$$ given by the formula $F(A + Z)=f(A)$ for each tuple $A=(a_1,...,a_n)$.
As said in the comment of @Adayeh, this is the best you are going to get, because $\mathbb{Z}^m_n / \mathbb{Z}_n$ is not defined since $\mathbb{Z}_n$ is not a subgroup of $\mathbb{Z}^m_n$.