$\gcd(|H_{1}|,|H_{2}|,\cdots,|H_{n}|) > 1$ implies $H_{1} \times H_{2} \times \cdots \times H_{n} = G$ not cyclic

Question

I need to prove a version of the following: "Suppose $G = H_{1} \times H_{2}$ and $\gcd(|H_{1}|,|H_{2}|)>1 $, then $G$ is not cyclic" for $G$ expressed as the direct product of $n$ subgroups.

I.e., I need to prove the following: "Suppose $\mathbf{G=H_{1} \times H_{2} \times \cdots \times H_{n}}$ and $\mathbf{\gcd(|H_{1}|,|H_{2}|,\cdots |H_{n}|) > 1}$, then $\mathbf{G = H_{1} \times H_{2} \times \cdots \times H_{n}}$ is not cyclic."

This is related to a question I posted earlier here. However, I am having difficulty extending it to $n$ subgroups, primarily because the relationship between the $\gcd$ of $n$ integers and the $lcm$ of $n$ integers is very different than it is between only $2$ integers.

For example, for $3$ integers $a_{1}, a_{2}, a_{3}$, we have that

$\displaystyle lcm(a_{1}, a_{2}, a_{3}) = \frac{abc}{\gcd((a_{2}\cdot a_{3}), (a_{3} \cdot a_{1}), (a_{1} \cdot a_{2}))}\,\,\,\,\,\,\,(*)$

Also, for these $3$ integers, we have that $lcm$ and $\gcd$ are associative:

$lcm(a_{1}, a_{2}, a_{3}) = lcm(lcm(a_{1},a_{2}),a_{3})$

and

$\gcd(a_{1}, a_{2}, a_{3}) = \gcd(\gcd(a_{1},a_{2}),a_{3})$

But, using these properties in order to get somewhere with an inductive proof that $\gcd(a_{1}, a_{2}, \cdots a_{n})>1$ (where $|H_{1}| = a_{1}$, $|H_{2}|=a_{2}$, $\cdots$ $|H_{n}|=a_{n}$) implies that the order of any element $(h_{1},h_{2},\cdots, h_{n}) \in H_{1} \times H_{2} \times \cdots \times H_{n} = G$ is $\leq a_{1}\cdot a_{2}\cdot \, \cdots\, \cdot a_{n}$ gets one quickly into trouble.

For example, when trying to make some headway for showing this for the $n=3$ case:

$\gcd(a_{1},a_{2},a_{3}) = \gcd(\gcd(a_{1},a_{2}),a_{3})$. Then, trying to reduce it to the $n=2$ case, we have that $\gcd(a_{1},a_{2},a_{3})\cdot lcm(a_{1},a_{2},a_{3}) = \gcd(\gcd(a_{1},a_{2}),a_{3})\cdot lcm(lcm(a_{1},a_{2}), a_{3})$, and I'm stuck, because the first argument in the $\gcd$ is not equal to the first argument in the $lcm$, and so I cannot say that this is equal to $a_{1} \cdot a_{2} \cdot a_{3}$.

The reason why I'm stuck is, of course, because of what I said before regarding what the $lcm$ of three integers is in equation $(*)$.

The proof of this result is needed in order for me to be able to prove what I wanted to in this question.

I'm really stuck. Could somebody provide me with a proof of this result? (it really is secondary, but essential, to what I need to show here) Or at least give me the tools I need to prove it myself (I just hope it isn't too long and frustrating; I've already spent a whole day on it)?

Thanks.

Answer 1

You're overthinking this.

$H_{1} \times H_{2} \times \cdots \times H_{n}$ is not cyclic because $H_{1} \times H_{2}$ is a subgroup that is not cyclic, by the case $n=2$.

Alternatively, $H_{1} \times H_{2} \times \cdots \times H_{n}$ is not cyclic because all elements have order less than $a_1 a_2 \cdots a_n$ since $g^m=1$ for all $g \in G$, where $m=\operatorname{lcm}(a_1, a_2, \dots, a_n) < a_1 a_2 \cdots a_n$. Again, this follows from the case $n=2$ because $\operatorname{lcm}(a_1, a_2) < a_1 a_2$.

Answer 2

Let $p$ be a prime divisor of $\gcd( |H_1|, \dots , |H_n|)$. By Cauchy's theorem every group $H_i$ has a subgroup $P_i$ of order $p$. Hence $$P_1 \times \dots \times P_n \cong C_p^n$$ is a subgroup of $G$. If $G$ were cyclic, then $C_p^n$ would be cyclic (subgroups of cyclic groups are cyclic): but this is not the case, so necessarily $G$ is not cyclic.