6
$\begingroup$

I'm trying to form a ring (or ring-like structure) out of the set of all finite groups.

Has anyone created/encountered an operation "+" before with the follow properties.

Let $G_1, G_2$ be finite groups and |G| denote the size of G, then

$$ |G_1| + |G_2| = |G_1 + G_2|$$

Where the left hand side is addition amongst natural numbers and the right hand side is our "abstract group addition"

And let $\times$ denote the direct product. Then:

$$ G_1 \times ( G_2 + G_3) = (G_1 \times G_2) + (G_1 \times G_3) $$

The "+" ideally would be commutative and associative, but I'm not sure if such an operator can necessarily exist.

My goal is to try to denote a notion of "rational groups" and dedekind cuts to groups to see if I can create a "fractional group theory" so to speak.

$\endgroup$
  • $\begingroup$ That would fail the distributive multiplicative identity. Since the RHS would be cyclic, but the LHS would a direct product of some group and a cyclic group $\endgroup$ – frogeyedpeas Jun 21 '17 at 5:19
  • 1
    $\begingroup$ BTW, great story about linear optimization and drug dealers! $\endgroup$ – Asvin Jun 21 '17 at 5:45
  • 2
    $\begingroup$ Technicality: "All finite groups" is a class, not a set. "All isomorphism classes of finite groups" is a set. $\endgroup$ – zibadawa timmy Jun 21 '17 at 6:00
  • 3
    $\begingroup$ Every homomorphism from this additive magma to an abelian group would factor through $\mathbf{Z}$. Indeed, denoting by $c_n$ the image of the cyclic group of order $n$, we would have $pc_1=c_p$ for all prime $p$. Then I claim that for every $n\ge 1$, if $g$ is the image of a group of order $n$, then $g=nc_1$. Proof by induction on $n$; it's clear for $n=1$. For given $n\ge 2$, there is a prime $p$ with $n\le p<2n$; the case $n$ prime being ok, we can suppose $p>n$. Then $g+c_{p-n}=c_p$, so $g+(p-n)c_1=pc_1$, thus $g=nc_1$. So the group associated with such a structure would be quite poor. $\endgroup$ – YCor Jun 21 '17 at 8:11
  • 1
    $\begingroup$ A related notion is en.wikipedia.org/wiki/Distributive_category. Note it mentions Groups do not form such a category, but without explanation. It looks like your structure would be akin to a distributive category structure on Groups that is compatible with the distributive structure on (finite) Sets via the forgetful functor. $\endgroup$ – PrimeRibeyeDeal Jun 21 '17 at 14:22
5
$\begingroup$

I believe that we can get a commutative, associative operation as follows. Let $G=A\times B$ and $H=A\times C$, where $B$ and $C$ have no common factor (when written as a direct product of indecomposable groups of order greater than one). Define $G+H=A\times D$, where $D$ is a product of cyclic groups of prime order such that $|D|=|B|+|C|$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.