# $G$ cyclic of order $n \implies G/mG$ cyclic of order $d = (m,n)$

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?

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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

## 2 Answers

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.

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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.

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