5

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 ?

Community
  • 1
KON3
  • 4,111
  • Compare with this question. – Dietrich Burde Dec 02 '15 at 09:02
  • 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 02 '15 at 11:20
  • 1
    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 02 '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 02 '15 at 12:15
  • @DerekHolt If $m$ and $n$ are powers of the same prime, say $p$, doesn't follow that the order of $(a,b)$ is a power of $p$ and then one can set $c=1$ and $d$ a suitable power of $p$? – user26857 Dec 02 '15 at 12:27
  • Ori it could be that you need to set $c$ to be a power of $p$ and $d=1$. For example $Z_4 \oplus Z_2$ has quotients $Z_4 \oplus Z_1$ and $Z_2 \oplus Z_2$. That's why I said that not allpossible choices of $c,d$ in $i$ will work in (ii). – Derek Holt Dec 02 '15 at 13:06
  • 1
    @Anjan3 Since $\langle (a,b)\rangle$ is a cyclic group you should have in (i) that $\gcd(c,d)=1$. – user26857 Dec 02 '15 at 15:39
  • @DerekHolt Dear Sir. The isomorphism is group isomorphism. Not ring isomorphism. But I am curious about that also. Secondly, I apologize for using symbol $\mathbb{Z}_m$ instead of $\mathbb{Z}/(m\mathbb{Z})$. In fact, as I learnt Abstract Algebra from Contemporary Abstract Algebra by Gallian, so I am habituated with the symbol that has been used in the book. Please blame him why did prof. Gallian require to mislead people like us by using the same notation $\mathbb{Z}_m$ for group and ring. – KON3 Dec 04 '15 at 12:57
  • Compare to you, my knowledge is definitely less and so I am sorry that I made mistake. Thank you for your correction. There are indeed several choice for $c, d$ but my intention was for just a suitable choice. Although I failed to make the appropriate question format. Extremely Sorry for that. between did you mean $\mathbb{Z}_4\oplus \mathbb{Z}_1$ as external direct product ? Cause I learnt from MSE that people use $\times $ to denote the same. For finite case of direct product, one can use $\oplus $ symbol. I hope you meant that. – KON3 Dec 04 '15 at 13:02
  • @user26857 Many many thanks to you for your reply. I understood my fault. I knew $c, d$ will be relatively prime which I showed in my attempt but forgot to mention in the question. Sorry for that. – KON3 Dec 04 '15 at 13:07

1 Answers1

2

(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 $d_2\mid n$.

user26857
  • 52,094
  • 1
    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 02 '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 02 '15 at 14:49
  • This is an example where (in your notation) $d_2=mn$ does not divide $n$. – Derek Holt Dec 02 '15 at 16:26
  • 1
    @DerekHolt From what you said in the first comment I've understood that there is something wrong in my decomposition using SNF. Now I think you wanted to say that the OP's claim in (ii) is wrong, which I agree. – user26857 Dec 02 '15 at 16:44
  • Thank you so much for your beautiful answer. It was really nice explanation. Would you please reply what is SNF ? The line "By using the SNF for the matrix whose rows are (a,b), (m,0), and (0,n) one find...." was not clear to me. :-( – KON3 Dec 04 '15 at 13:11