# Prove $\mathbb{Z_m} \times \mathbb{Z}_n \cong \mathbb{Z}_d \times \mathbb{Z}_l$

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

• 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. Jun 30, 2021 at 22:49
• I will take it into account for future posts, thanks Jun 30, 2021 at 22:51
• 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 $(r, qm+y)$? Jun 30, 2021 at 23:12
• 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.) Jun 30, 2021 at 23:38
• – lhf
Jul 1, 2021 at 0:33

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 $$\langle 1\times 1\rangle=\mathbb{Z}_l$$.