# How to show $\frac{\mathbb{Z}_m\times \mathbb{Z}_n}{\langle (a,b)\rangle}\simeq \mathbb{Z}_{\frac mc}\times \mathbb{Z}_{\frac nd}$?

Suppose we have the group $\mathbb{Z}_m\times\mathbb{Z}_n$, and $(a,b)\in\mathbb{Z}_m\times\mathbb{Z}_n$. We need to justify that

(i) There exist $c, d$ such that $\langle (a,b)\rangle$ is isomorphic to the group $\mathbb{Z}_c\times \mathbb{Z}_d$ with $c\mid m, d\mid n$.

(ii) $\dfrac{\mathbb{Z}_m\times \mathbb{Z}_n}{\langle (a,b)\rangle}\simeq \mathbb{Z}_{\frac mc}\times \mathbb{Z}_{\frac nd}$.

How to show these ?

The first one I tried as: Since $\langle (a,b)\rangle$ is cyclic there is $\alpha$ such that $\langle (a,b)\rangle\simeq \mathbb{Z}_\alpha$. And then $\alpha\mid mn$ which means we can find two relatively prime $c,d$ such that $cd=\alpha, c\mid m, d\mid n$ and $\mathbb{Z}_\alpha\simeq \mathbb{Z}_c\times \mathbb{Z}_d$. Then ?

• Compare with this question. – Dietrich Burde Dec 2 '15 at 9:02
• My first reaction to the title is that it is unclear whether the isomorphism is intended to be an isomorphism of rings or of additive groups. (I have never been able to understand why so many people use ${\mathbb Z}_m$ to denote an additive group.) – Derek Holt Dec 2 '15 at 11:20
• The question is misleading in the sense that there may be several choices for $c$ and $d$ that satisfy (i), but they will not necessarily all work in (ii). – Derek Holt Dec 2 '15 at 12:05
• Note also that, since a cyclic group is a direct products of cyclic groups of prime power order, the problem reduces to the case when $m$ and $n$ are both powers of the same prime $p$. But it is still nontrivial! – Derek Holt Dec 2 '15 at 12:15
• @Anjan3 Since $\langle (a,b)\rangle$ is a cyclic group you should have in (i) that $\gcd(c,d)=1$. – user26857 Dec 2 '15 at 15:39

(i) In fact, $\alpha$ is the order of $(a,b)$ in $\mathbb Z_m\times\mathbb Z_n$, so $\alpha=\operatorname{lcm}(m',n')$, where $m'=\operatorname{ord}(a)\mid m$ and $n'=\operatorname{ord}(b)\mid n$. Set $d'=\gcd(m',n')$ and write $m'=d'm_1'$, $n'=d'n_1'$ with $\gcd(m_1',n_1')=1$. Then $\alpha=d'm_1'n_1'$.
Now your goal is to write $\alpha=\alpha_1\alpha_2$ with $\gcd(\alpha_1,\alpha_2)=1$ and $\alpha_1\mid m'$, $\alpha_2\mid n'$. This can be done as follows. If $\gcd(d'm_1',n_1')=1$ or $\gcd(m_1',d'n_1')=1$ then set $\alpha_1=d'm_1'$ and $\alpha_2=n_1'$, or $\alpha_1=m_1'$ and $\alpha_2=d'n_1'$, respectively. Otherwise, there are some primes in common between $d'$ and $n_1'$, respectively between $d'$ and $m_1'$ (but there are no primes in common between these two sets of primes since $\gcd(m_1',n_1')=1$). Now write $d'=ed_1'd_2'$ where $d_1'$ is the product of all common primes between $d'$ and $m_1'$, $d_2'$ is the product of all common primes between $d'$ and $n_1'$, and $e$ is the product of primes in $d'$ which don't show up neither in $m_1'$ nor in $n_1'$. Then $\alpha=(ed_1'm_1')(d_2'n_1')$ and notice that $\gcd(ed_1'm_1',d_2'n_1')=1$.
(ii) By using the SNF for the matrix whose rows are $(a,b)$, $(m,0)$, and $(0,n)$ one find $$\frac{\mathbb{Z}_m\times \mathbb{Z}_n}{\langle (a,b)\rangle}\simeq \mathbb{Z}_{d_1}\times \mathbb{Z}_{d_2},$$ where $d_1=\gcd(a,b,m,n)$ and $d_2=\gcd(bm,an,mn)/d_1$. We have $d_1\mid m$, but I can't see why $d_2\mid n$.
• I don't think that the argument in (ii) works in general. If $m$ and $n$ are coprime and $(a,b)=(0,0)$ then the SNF gives $Z_1 \times Z_{mn}$. – Derek Holt Dec 2 '15 at 13:39
• @DerekHolt And since $\mathbb Z_1=0$ we get $\mathbb Z_{mn}$. I don't see where is the mistake. – user26857 Dec 2 '15 at 14:49
• This is an example where (in your notation) $d_2=mn$ does not divide $n$. – Derek Holt Dec 2 '15 at 16:26