2

Let $G$ be a finite, non-trivial group and let $K$ be a compact Lie group such that each subgroup of $G\times K$ is of the form $H\times L$ for some subgroups $H\subset G$ and $L\subset K$. Is it true that then $K$ is also finite?

I tried in vain to find a counterexample, and I guess that the claim is true, but I do not know how to prove it. The answers to the question When must the subgroup of a product be the product of subgroups? come quite close to my problem, but I am not sure how to use them to prove the above statement.

Shaun
  • 44,997
Simba
  • 445
  • 2
    In what way do the linked answers fall short? It seems that it would suffice to show that every compact torsion Lie group is finite. – Greg Martin Aug 20 '22 at 17:43
  • You are right. But why is every compact tosrion Lie group finite? Does this follow from the fact that any infinite compact Lie group contains a maximal torus? – Simba Aug 20 '22 at 18:07
  • 2
    It certainly seems so, since tori always contain non-torsion points. – Greg Martin Aug 20 '22 at 19:56

1 Answers1

1

Yes, in this case $K$ is finite.

Indeed, if $K_0$ is a connected component of $1$, then by the definition of Lie group $K_0$ is an open normal subgroup of $K$. Since the factor group $K/K_0$ is discrete and compact, it is finite.

If $K_0\neq\{1\}$, then $K_0$ contains a connected abelian compact subgroup $A$ (e.g. the image of a one-dimensional subalgebra from the Lie algebra of group $K_0$ under exponential mapping). Since $A\cong\mathbb{R}/\mathbb{Z}$ there is an element of order $p$ in $A$ for any prime $p$. If $a\in G$ and $b\in A$ are elements of order $p$ for a prime $p$, then the diagonal of the subgroup $\operatorname{gr}(a)\times\operatorname{gr}(b)$ has order $p$ and hence cannot be represented as a direct product of non-trivial subgroups.

It follows that $K_0=\{1\}$ and the group $K$ is finite.

kabenyuk
  • 10,712