# The product of finitely many cyclic groups is cyclic

How to prove that the direct product of finitely many cyclic groups $C_{n_1}\times C_{n_2}\times\cdots\times C_{n_m}$ is cyclic if the $n_i$'s are pairwise relatively prime?

Hint: Let $g_i$ be a generator of $C_{n_i}$. What is the order of $(g_1, \ldots, g_m$)? What's the order of $C_{n_1} \times \cdots \times C_{n_m}$?

Let $g_k$ be the generator of $C_{n_k}$ for $k=1,...,m$. There is the element $g:=(g_1,g_2,...,g_m)\in C_{n_1}\times...\times C_{n_m}$. If $g^l=e$, the neutral element of the product, then $(g_1^l,...,g_m^l)=(e_1,e_2,...,e_m)$, so $l$ is a multiple of $n_1,...,n_m$. If $l$ is the order of $g$, then it is the least common multiple.

What do you know about the lcm of coprime numbers?

Once you know what the order of $g$ is, how can you conclude that it is a generator?

Hint: There is a proposition that says that if $n_1$ and $n_2$ are coprime, then $C_{n_1} \times C_{n_2}\cong C_{n_1\times n_2}$. Then to prove what you want, use induction.

• Maybe you can explain the proposition, which is slightly what Billy did.- – Pedro Tamaroff Jul 17 '13 at 16:46
• I was hoping she may have seen it prior to this result. I'll relabel it as a hint. – BlackAdder Jul 17 '13 at 16:51
• @BlackAdder Can you please expand on how to do the induction proof. I have no problem in the 2 groups case, but for some reason, I can't handle the case with n groups (the proof for the 2 groups case uses the fact that gcd(a,b) divides a and b, but when we get to 3 numbers say: a, b and c I need the fact that gcd(a,b,c) divides the product of any two of a,b,c, but that is not correct). Or if you have a link for the proof that would be great also. – I0_0I Mar 3 at 15:38

Suppose $m$ and $n$ are coprime. How large is the cyclic subgroup generated by $(1,1) \in C_m \times C_n$? In other words, what's the order of $(1,1)$? (I'm writing my groups additively, so $0$ is the identity and $1$ is a generator in each cyclic group.) Of course, it's the smallest non-zero $k$ such that $(k,k) = (0,0)$ in $C_m \times C_n$ - or, in other words, such that $k = 0 \mod m$ and $k = 0 \mod n$. So (by the Chinese remainder theorem, if you like, but you can also check it directly) $k = mn$. So $(1,1)$ generates a cyclic subgroup of order $mn$ inside $C_m\times C_n$ - but $C_m\times C_n$ has order $mn$, so this cyclic subgroup must be the whole group.

Now use induction.

First, remember every cyclic group of order $n$ is isomorphic to the integers modulo $n$, that is $\Bbb Z/n\Bbb Z$, so to make things simpler, let's work with that. Recall that if $(m,n)=1$ then the system $$\begin{cases}x\equiv a\mod m\\x\equiv b\mod n\end{cases}$$

has a unique solution $\mod mn$. The proof is not complicated. From the first equation, we know that must be of the form $x=mk+a\; ;\;k\in\Bbb Z$. Then \begin{align}mk+a&\equiv b&\mod n\\mk&\equiv b-a&\mod n\\k&\equiv m'(b-a)&\mod n\end{align}

where $m'm=1\mod n$ is possible because $(m,n)=1$. Thus $k$ is of the form $nj+m'(b-a)=nj+\sigma\; ;\;j\in\Bbb Z$ and \begin{align}x&=m(\sigma+jn)+n\\&=m\sigma+n+j(mn)\\&=\omega+j(mn)\end{align}

is unique modulo $mn$. Then, assuming $n_1,\dots,n_k$ are pairwise coprime, we can extend the claim above by induction. Reason: pairwise coprimality ensures that $(n_1n_2\cdots n_{k-1},n_k)=1$. We now say that the system $$\begin{cases}x\equiv a_1\mod n_1\\x\equiv a_2\mod n_2\\\cdots\\x\equiv a_k\mod n_k\end{cases}$$

has a unique solution modulo $n_1n_2n_2\cdots n_k$. Now, we're ready to show that, assuming $n_1,\cdots,n_k$ are pairwise coprime, $$\Bbb Z/n_1\Bbb Z\times \Bbb Z/n_2\Bbb Z\times \cdots\times \Bbb Z/n_k\Bbb Z\simeq \Bbb Z/m\Bbb Z$$ where $m=n_1n_2\cdots n_k$. The proof is not complicated either. Take an element $(a_1,\dots,a_k)$ in the product. By our claim, there is a unique $x$ modulo $m=n_1\cdots n_k$ such that

$$\begin{cases}x\equiv a_1\mod n_1\\x\equiv a_2\mod n_2\\\cdots\\x\equiv a_k\mod n_k\end{cases}$$

holds. Then the map $(x_1,\ldots,x_k)\to x\in \Bbb Z/m\Bbb Z$ is an isomorphism.

The other answers are great because they provide direct proofs. I think my answer provides a different and also conceptual way of looking at the problem.

If $C_k$ and $C_l$ are cyclic groups of orders $k$ and $l$, respectively, $l$ divides $k$, then there is a surjection $C_k\to C_l$.

Exercise 1: Prove this claim.

Let $C_{k}$ be the cyclic group of order $k=n_1\times\cdots\times n_m$, the product of the orders of the cyclic groups $C_{n_1},\dots,C_{n_m}$ in question. Surjections $f_i:C_{k}\to C_{n_i}$ for $1\leq i\leq m$ define a homomorphism $f:C_{k}\to C_{n_1}\times\cdots\times C_{n_m}$ (by the universal property of the product, if you're familiar with category theory; explicitly, by the rule $f(x)=(f_1(x),\dots,f_m(x))$).

Exercise 2: The kernel of $f:C_{k}\to C_{n_1}\times\cdots\times C_{n_m}$ is equal to the intersection of the kernels of the $f_i:C_k\to C_{n_i}$ for $1\leq i\leq m$.

Exercise 3: Prove that the kernel of $f:C_k\to C_{n_1}\times\cdots\times C_{n_m}$ is trivial. (Hint: Use Lagrange's theorem, the hypothesis that the $n_i$'s are relatively prime for $1\leq i\leq m$, and Exercise 2.)

Therefore, $f:C_k\to C_{n_1}\times\cdots\times C_{n_m}$ is injective. The pigeon-hole principle implies that $f$ is also surjective, hence an isomorphism. Therefore, $C_{n_1}\times\cdots\times C_{n_m}$ is a cyclic group!

Exercise 4: What goes wrong in the proof above if the orders of the cyclic groups in question aren't relatively prime? (Hint: the conclusion of one of the exercises above isn't valid.)

I hope this helps!