If all subgroups of direct product are direct products, then orders are coprime

Question

I am wondering if the following assertion is true:

Let $K$ and $G$ be finite groups such that all subgroups of their direct product $K\times G$ are of the form $L\times H$ for some subgroups $L$ of $K$ and $H$ of $G$. Then the orders of $K$ and $G$ are coprime.

I know that the converse is true but I do not know how the above assumption gives information about the orders of $K$ and $G$. I tried to find integers $a$ and $b$ such that $|K|\cdot a + |G|\cdot b = 1$, which is equivalent to saying that the greatest common divisor of $|K|$ and $|G|$ is $1$ (Bézout's identity). However, I have not yet got something useful.

Many thanks for any help!

If $p$ is a prime divisor of both $|K|$ and $|G|$ then we have a subgroup $C_p\times C_p$ which in turn has subgroups not of the required form. — ancient mathematician, Jul 29 '22 at 12:57
In fact, the result holds even if you do not assume $G$ and $K$ are finite. See duplicate. — Arturo Magidin, Jul 29 '22 at 15:52

score 4 · Accepted Answer · edited Jul 30 '22 at 12:05

4

Suppose the orders of $K$ and $G$ are not coprime. Let $p$ be a common prime factor. By Cauchy's theorem there are elements $k\in K,g\in G$ such that $\lvert k\rvert=\lvert g\rvert=p$. Consider the "diagonal" subgroup $\Delta=\{(k^a,g^a)\mid a=1,2,\dots p\}$. Then $\Delta$ is not a direct product of subgroups.

Thus your conjecture is true.

edited Jul 30 '22 at 12:05

Simba

445

answered Jul 29 '22 at 13:05

calc ll

8,427

If all subgroups of direct product are direct products, then orders are coprime

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