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Say you have two groups $G = \langle g \rangle$ with order $n$ and $H = \langle h \rangle$ with order $m$. Then the product $G \times H$ is a cyclic group if and only if $\gcd(n,m)=1$.

I can't seem to figure out how to start proving this. I have tried with some examples, where I pick $(g,h)$ as a candidate generator of $G \times H$. I see that what we want is for the cycles of $g$ and $h$, as we take powers of $(g,h)$, to interleave such that we do not get $(1,1)$ until the $(mn)$-th power. However, I am having a hard time formalizing this and relating it to the greatest common divisor.

Any hints are much appreciated!

  • 1
    As a supplement to the various detailed answers below, the following short remark might help: gcd$(m,n) = 1$ if and only lcm$(m,n) = mn$. It is the latter fact (about the lcm) which is more directly relevant to your problem than the gcd fact (although they are equivalent). Try combining your intuition about "interleaving" with the formal notion of lcm$(m,n)$.2010-10-03

4 Answers 4

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Use these facts:

1) Cyclic groups of the same order are isomorphic.

2) http://mathforum.org/library/drmath/view/73205.html

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HINT \rm\quad\ \ gcd(m,n) > 1

$\rm\quad\iff\ lcm(m,n) < mn$

$\rm\quad\iff\ \mathbb Z_m \times \mathbb Z_n\ $ has all elts of order $\rm < mn$

$\rm\quad\iff\ \mathbb Z_m \times \mathbb Z_n\ $ is noncyclic

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    Very clear and simple argument.+12017-04-18
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Note that $|G\times H|=|G||H|=nm$; so $G\times H$ is cyclic if and only if there is an element of order $nm$ in $G\times H$.

In any group $A$, if $a,b\in A$ commute with one another, $a$ has order $k$, and $b$ has order $\ell$ then the order of $ab$ will divide lcm$(k,\ell)$ (prove it).

Now take an element of $G\times H$, written as $(g^a,h^b)$, where $G=\langle g\rangle$, $H=\langle h\rangle$, $0\leq a\lt n$, $0\leq b\lt m$. Then $(g^a,h^b)=(g^a,1)(1,h^b)$. In this case, what is the order? Under what conditions can you get an element of order exactly $nm$, which is what you need?

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    You want to prove an "if and only if". So you want to show that *if* $G$ is cyclic, *then* gcd$(m,n)=1$ (equivalently, lcm$(m,n)=mn$). **And** you want to prove that *if* gcd$(m,n)=1$ (equivalently, lcm$(m,n)=mn$), **then** $G$ is cyclic. So showing that if $G$ is cyclic then lcm$(m,n)=mn$ is only *half* of what you are trying to do.2010-10-03
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The order of $G\times H$ is $n.m$. Thus, $G\times H$ is cyclic iff it has an element with order $n.m$. Suppose $gcd(n.m)=1$. This implies that $g^m$ has order $n$, and analogously $h^n$ has order $m$. That is, $g\times h$ has order $n.m$, and therefore $G\times H$ is cyclic.

Suppose now that gcd(n.m) >1. Let $g^k$ be an element of $G$ and $h^j$ be an element of $H$. Since the lowest common multiple of $n$ and $m$ is lower than the product $n.m$, that is, $lcm(n,m) < n.m$, and since $(g^k)^{lcm(n,m)} = e_{G}$, $(h^j)^{lcm(n,m)} = e_{H}$, we have $(g^k\times h^j)^{lcm(n,m)} = e_{G\times H}$. It follows that every element of $G\times H$ has order lower than $n.m$, and therefore $G\times H$ is not cyclic.

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    in the $G$, $H$ cyclic, $\gcd(|G|,|H|) = 1$ $\implies$ $G \times H$ cyclic direction, don't you mean to say that $|g^{m} \times h^{n}| = n\cdot m$ and not that $|g \times h| = n \cdot m$?2016-10-28