Let $m, n$ be relatively prime and $G, G'$ be cyclic groups of orders $m, n$, respectively. Show that $G \times G'$ is cyclic of order $mn$.

Question

I easily see that the order is $mn$, but am confused on how to show that $G \times G'$ will be cyclic.

I know I need to find an element $(x,y) \in G \times G'$ that has order mn but I'm not sure how to find that element. For example, if we are considering two groups of order $5$ and $11$, I don't know what to look for.

Any guidance is much appreciated :)

If $g\in G$ and $g'\in G'$, can you figure out what the order of $(g,g')$ will be in terms of the orders of $g$ and $g'$? — rogerl, Jul 31 '20 at 22:55
Correct me if I'm wrong here but would it be that: if n = ord(g) and m = ord(g') then ord(g,g') = nm? — buffalodude, Jul 31 '20 at 22:57

score 2 · Answer 1 · answered Aug 01 '20 at 06:43

Let $g$ be the generator of $G$ and $g'$ the generator of $G'$. Then I claim $(g,g')\in G\times G'$ is a generator.

Let $(h,h')\in G\times G'$ be arbitrary. Then there exists integers $k$ and $\ell$ such that $h=g^k$ and $h'=(g')^\ell$. Since $n$ and $m$ are coprime, there exists $a,b\in\mathbb Z$ such that $an+bm=k-\ell$. Rewriting this, we have $-bm+k=an+\ell$. Denote this as $N$. Now, $(g,g')^N=(g^{-bm+k},(g')^{an+\ell})=(g^k,(g')^\ell)=(h,h')$. This proves $(g,g')$ generates $G\times G'$.

A learner · Answer 2 · 2020-08-01T10:07:16.007

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We know a simple rule that each cyclic group of order $n$ is isomorphic to $\mathbb{Z}_n $.

So, $G \cong \mathbb{Z}_m $ and $G' \cong \mathbb{Z}_n $. So, $ G × G' \cong \mathbb{Z}_m × \mathbb{Z}_n $.

As, $gcd(m,n)=1 $, $ \mathbb{Z}_m × \mathbb{Z}_n \cong \mathbb{Z}_{mn} $.

Since, $\mathbb{Z}_{mn} $ is cyclic group of order $mn$.

So, , as, $ G × G' \cong \mathbb{Z}_{mn} $, , $ G × G' $ becomes cyclic group of order $mn$.

Edit: why? , $\mathbb{Z}_{mn} \cong \mathbb{Z}_m × \mathbb{Z}_n $

As, $\mathbb{Z}_{mn} $ is cyclic, it has normal subgroup of order $m$ , which is isomorphic to $\mathbb{Z}_{m} $ and a normal subgroup of order $n$ , which is isomorphic to $\mathbb{Z}_{n} $ And , since ,$\mathbb{Z}_{m} ∩ \mathbb{Z}_{n} = \{0\} $, as $gcd(m,n)=1$.

By the rule of internal direct product, $\mathbb{Z}_{mn} \cong \mathbb{Z}_m × \mathbb{Z}_n $

edited Aug 01 '20 at 10:07

answered Aug 01 '20 at 08:03

A learner

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But why is $\mathbb Z_m\times \mathbb Z_n\cong\mathbb Z_{mn}$? – Kenta S Aug 01 '20 at 09:28
As $m,n$ coprime. In detail, take homomorphism $f : \mathbb{Z} \to \mathbb{Z}_m × \mathbb{Z}_n $ defined as, $f(a)=(a mod m, a mod n)$ – A learner Aug 01 '20 at 09:35
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You have just restated the problem in terms of $\mathbb Z_n$, no? – Kenta S Aug 01 '20 at 09:37
What do you mean?? – A learner Aug 01 '20 at 09:42
You should probably include that homomorphism into your answer. – Kenta S Aug 01 '20 at 09:43
@Kenta S there also can be easier way to show that, like by internal direct product. – A learner Aug 01 '20 at 09:57

score 0 · Answer 3 · answered Aug 01 '20 at 12:47

Take $x \in G$ of order $n$ and $y \in G’$ of order $m$. If $(x,y)$ has order $k$, then the order of $x$ divides $k$ and the order of $y$ does as well. So both $n$ and $m$ divide $k$ and as $n$ and $m$ are coprime we have $nm$ divides $k$. Since $k$ is at most $nm$, we’re done.

score 0 · Answer 4 · answered Aug 01 '20 at 13:11

Suppose $g\in G$, $g'\in G'$ with $g$ of order $r$ and $g'$ of order $s$, and that the order of $(g,g')$ is some integer $k$. Then $g^k=e$ and $(g')^k=e$, so that $r\mid k$ and $s\mid k$. But $k$ must be the smallest integer that is a multiple of both $r$ and $s$ since it is the order of $(g,g')$. It follows that $k=\mathrm{lcm}(r,s)$.

You can apply this to your case, where $r=n$ and $s=m$ are coprime, so that $\mathrm{lcm}(r,s) = $mn$.

Let $m, n$ be relatively prime and $G, G'$ be cyclic groups of orders $m, n$, respectively. Show that $G \times G'$ is cyclic of order $mn$.

4 Answers4