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Let $n,m \in \mathbb{N}$, prove that $\mathbb{Z_m} \times \mathbb{Z}_n \cong \mathbb{Z}_d \times \mathbb{Z}_l $ where, $d=gcd(m,n)$ and $l=lcm(m,n)$.

At first I tried to define a homomorphism $\varphi: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}_d \times \mathbb{Z}_l $ in the natural way, and prove that $ker(\varphi)=(m)\times(n)$, but I did not see the way.

So I try using elemental divisors as follows:

Since $gcd(m,n)\cdot lcm(m,n)=mn$, we know that

$$|\mathbb{Z_m} \times \mathbb{Z}_n| =mn=dl=| \mathbb{Z}_d \times \mathbb{Z}_l| $$

Lets write $n$ and $m$ in their "complete prime factorization" (Using all the primes "in" $n$ and $m$ with powers zero if the prime do not appear in the factorization)

$n=p_1^{\alpha_1}\cdots p_k ^{\alpha_k}$ and $m=p_1^{\beta_1}\cdots p_k ^{\beta_k}$, with $\alpha_i,\beta_i\geq0$, $p_i$ prime, then $mn=p_1^{\alpha_1+\beta_1}\cdots p_k ^{\alpha_k+\beta_k}$.

If this is the case then, $d=p_1^{\delta_1}\cdots p_k^{\delta_k}$ and $l=p_1^{\sigma_1}\cdots p_k^{\sigma_k}$, where $\delta_i=min\{\alpha_i,\beta_i\}$ and $\sigma_i=max\{\alpha_i,\beta_i\}$.

And as $dl=mn$ then, $\alpha_i+\beta_i=\delta_i+\sigma_i$, for all $i=1,\cdots k$.

Therefore,

$$ \mathbb{Z_n} \cong \mathbb{Z}_{p_1^{\alpha_1}} \times \cdots \times \mathbb{Z}_{p_k^{\alpha_k}}\;, \mathbb{Z_m} \cong \mathbb{Z}_{p_1^{\beta_1}} \times \cdots \times \mathbb{Z}_{p_k^{\beta_k}}$$ and

$$\mathbb{Z_d} \cong \mathbb{Z}_{p_1^{\delta_1}} \times \cdots \times \mathbb{Z}_{p_k^{\delta_k}}\; , \mathbb{Z_l} \cong \mathbb{Z}_{p_1^{\sigma_1}} \times \cdots \times \mathbb{Z}_{p_k^{\sigma_k}}$$

Finally, without loss of generality, if $\delta_i=\alpha_i$ for some $i \in \{1,\cdots,k\}$, then $\sigma_i=\beta_i$, so $\mathbb{Z_m} \times \mathbb{Z}_n$ and $\mathbb{Z}_d \times \mathbb{Z}_l $ has the same elemental divisors, then they are isomorphic.

The proof is ok? and if it is, how can I improve it if is the case

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    $\begingroup$ If you are asking people to review your proof, please use the [solution-verification] or [proof-verification] tags; if you want feedback on improving the writing, etc., please use the [proof-writing] tag. $\endgroup$ Jun 30, 2021 at 22:49
  • $\begingroup$ I will take it into account for future posts, thanks $\endgroup$
    – George
    Jun 30, 2021 at 22:51
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    $\begingroup$ You can divide any $x \in \mathbb{Z_n}$ by $d$ and get $x = qd+r$, with $r \in \mathbb{Z}_d$. How about sending $(x, y)$ to $(r, qm+y)$? $\endgroup$ Jun 30, 2021 at 23:12
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    $\begingroup$ Using a calculation along the lines of Smith normal form, I think what I get is: suppose $d = am + bn$. Then $\begin{bmatrix} a & b \\ -n/d & m/d \end{bmatrix}$ gives an automorphism of $\mathbb{Z}^2$ which should send $m\mathbb{Z} \times n\mathbb{Z}$ to $d\mathbb{Z} \times \ell\mathbb{Z}$; so it induces the desired isomorphism between quotients. (It's possible the calculations are off, but the basic idea of using Smith normal form manipulations to find an explicit isomorphism should be solid.) $\endgroup$ Jun 30, 2021 at 23:38
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    $\begingroup$ See math.stackexchange.com/questions/2205618/… $\endgroup$
    – lhf
    Jul 1, 2021 at 0:33

1 Answer 1

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It may help to know the following classification for direct sums: A group $G$ is isomorphic to the direct sum $H\times K$ iff there exists $H\triangleleft G$ and $K\triangleleft G$ such that $H\cap K=1$ and $HK=G$.

To apply this classification to your problem observe that $1\times\mathbb{Z}_d\triangleleft \mathbb{Z}_m\times\mathbb{Z}_n$ and $\langle 1\times 1\rangle\triangleleft \mathbb{Z}_m\times\mathbb{Z}_n$ and that $(1\times\mathbb{Z}_d) \cap \langle 1\times 1\rangle=1$ and $(1\times\mathbb{Z}_d) \langle 1\times 1\rangle=\mathbb{Z}_m\times\mathbb{Z}_n$. Finally note that $1\times\mathbb{Z}_d\simeq\mathbb{Z}_d$ and $\langle 1\times 1\rangle=\mathbb{Z}_l$.

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