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Let $G_1, G_2, ... , G_n$ be finite cyclic groups. I want to show that $G_1 \times G_2 \times ... \times G_n$ is cyclic if, and only if, $\gcd(|G_i||G_j|) = 1$ whenever $i \ne j$. I do know that if $G$ and $H$ are finite cyclic groups, then $G \times H$ is cyclic if, and only if, $\gcd(|G||H|)=1$. I'm assuming that some form of induction would work when dealing with $n$ finite cyclic groups, but I have absolutely no idea how to apply it.

Any help would be much appreciated.

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You're exactly right, you need to use induction. I'm going to write $m_i:=|G_i|$; so in one direction, if $\gcd(m_i,m_j)=1$ for $i\neq j$, then by the induction hypothesis you have that $G_1\times\cdots\times G_{n-1}$ is cyclic. Then using the result for $n=2$ (which you say you already know), you just need to show that $\gcd(m_1\cdots m_{n-1},m_n)=1$ to deduce that $(G_1\times\cdots\times G_{n-1})\times G_n$ is cyclic as well.

On the other hand, if $G_1\times\cdots\times G_n$ is cyclic, recall a quotient of a cyclic group is cyclic, so $G_1\times\cdots\times G_{n-1}$ and $G_n$ are cyclic groups, and by the induction hypothesis on the former you see that $\gcd(m_i,m_j)=1$ whenever $1\le i,j\le n-1$ and $i\neq j$. But then again you can use the $n=2$ case to also deduce that $\gcd(m_1\cdots m_{n-1},m_n)=1$, and putting these two facts together find that $\gcd(m_i,m_j)=1$ whenever $1\le i,j\le n$ and $i\neq j$.

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  • $\begingroup$ I kind of get it. With the first part though, I knew that $gcd(|G_n|, |G_i|) = 1$ for all $i \ne n$. How do I prove from there that $gcd(|G_1 \times G_2 \times ... \times G_{n-1}|, |G_n|) = 1$? Once I know that $|G_1 \times G_2 \times ... \times G_{n-1}|$ and $|G_n|$ are coprime, then the result for $n=2$ tells us that $(G_1 \times G_2 \times ... \times G_{n-1}) \times G_n$ is cyclic. $\endgroup$ – Tim Aug 18 '19 at 22:21
  • $\begingroup$ If $\gcd(|G_1|\cdots|G_{n-1}|,|G_n|)\neq1$, then there is some prime (say $p$) dividing both $|G_1|\cdots|G_{n-1}|$ and $|G_n|$. Now recall it's a classic result that if a prime divides a product $ab$, then the prime divides either $a$ or $b$; then of course with induction you can conclude that if $p$ divides a product $a_1\cdots a_m$ then $p$ divides some $a_i$, so in our case we can conclude $p$ divides some $|G_i|$ with $1\le i\le n-1$. From this conclude that $\gcd(|G_i|,|G_n|)\neq1$ for some $i\in\{1,\dots,n-1\}$, giving a contradiction. $\endgroup$ – Alex Mathers Aug 19 '19 at 18:19

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