Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to show that $(\mathbb{Z} / m \mathbb{Z}) \otimes (\mathbb{Z} / n \mathbb{Z}) \simeq \mathbb{Z} / d \mathbb{Z}$ when $(m,n)=d$

I have shown that

$$\mathbb{Z} / m \mathbb{Z} \otimes G \simeq G/mG$$ where $G$ is any abelian group.

So what it boils down to is that $G$ cyclic of order $n \implies G/mG$ cyclic of order $d = (m,n)$.

Any hints?

share|cite|improve this question
If $\sigma$ is a generator of $G$, then $mG$ is generated by $\sigma^m$. If the order of $\sigma$ is $n$, then what powers of $\sigma$ can you get in $\langle\sigma^m\rangle$? – Alex B. Apr 11 '11 at 8:04
up vote 6 down vote accepted

What it boils down to is the following:

If $G$ is a group, and $x\in G$ is of order $n$, then $x^m$ is of order $x^{n/(m,n)}$.

This will prove that if $G$ is cyclic of order $m$, then $mG$ is cyclic of order $n/(m,n)$, hence $G/mG$ is of order $(m,n)$.

First, note that the order of $x^m$ divides $n/(m,n)$: because $$m\left(\frac{n}{(m,n)}\right) = \frac{mn}{(m,n)} = [m,n] = \mathrm{lcm}(m,n)$$ is a multiple of $n$, so $(x^m)^{n/(m,n)} = 1$.

Conversely, to show that $n/(m,n)$ divides the order of $x^m$, let $d$ be the order of $x^m$. Then $1 = (x^m)^d = x^{md}$, so $n|md$. Dividing by the gcd of $m$ and $n$, we have that $n/(m,n)$ divides $md/(n,m)$. Since $n/(m,n)$ is relatively prime to $m/(n,m)$, it follows that $n/(m,n)$ divides $d$, as claimed.

Therefore, the order of $x^m$ is $n/(m,n)$, which proves the desired result.

share|cite|improve this answer

HINT $\rm\ \ mod\ n\ \mathbb Z\::\ \ m\ \mathbb Z\ =\ m\ \mathbb Z\ + \ n\ \mathbb Z\ =\ (m,n)\ \mathbb Z$

See also this prior thread and this one too.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.