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 $mn$. 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 :)
 A: 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: 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 $
A: 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.
A: 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) = $mn$.
