# Is there an idea of “primeless isomorphism” studied somewhere in finite group theory?

What I mean by "primeless isomorphism" is essentially a relation on finite groups by identifying groups whose structure differs only in which primes divide the groups' orders. The groups aren't exactly isomorphic, but they are close to it; I'm having trouble formulating the idea rigorously, so I think it might be best to explain what I mean through examples.

In the simplest case, take cyclic groups $C_{p}$ and $C_q$ for distinct primes $p$ and $q$. Obviously these aren't isomorphic, but they would be "primeless isomorphic" as the only difference in the group structure is what the primes actually are. By contrast, $C_{p}$ and $C_n$ would not be considered primeless isomorphic for composite $n$, because $C_n$ has proper nontrivial subgroups and $C_p$ doesn't - a structural difference independent of which primes divide $n$.

For another example, we could look at the class of Frobenius groups $C_pC_q$ where $C_q$ acts fixed point freely on $C_p$ (again with distinct primes $p,q$). There are constraints on what these primes can be in that $q$ has to divide $p-1$ for the group to exist, but among those groups that do, it doesn't seem like they are qualitatively different. Groups of order $p^3$ have been classified in exactly the way that I mean; the same thing goes for Dihedral groups $D_{2n}$ of squarefree order, which split into different "primeless isomorphism classes" depending on the number of prime divisors.

The set of all groups with order $p^3$ for some prime $p$ would be divided into seven equivalency classes: $[C_{p^3}],[C_{p^2}\times C_p],[(C_p)^3],[Q_8],[D_8],[\text{Heis}\,Z_p],$ and $[G_p]$ (where for the last two classes $p$ is odd).

It's hard to say precisely what I mean, but hopefully you get my drift.

• Has it been studied? Is there a name for it?

• If not, is that because it is somehow logically difficult (or impossible) to define?

If nobody's heard of this,

• Can anyone think of a good way to formulate a definition for two groups to be "primeless isomorphic?" The definition should give rise to an equivalence relation on any given set of groups that partitions it into classes which are only "quantitatively" different, but not "qualitatively" in the way I've been getting at.
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Grouping finite groups together according to the multiset of exponents in the prime factorization of order seems to throw away quite a lot of structural information; for all but small-exponent / few-primes situations I would expect the groups in these equivalence classes to manifest potentially vast qualitative differences. – anon Oct 8 '12 at 4:21
For the groups of order $p^3$, the classification for $p=2$ is different from the classification for odd $p$, if I recall correctly (even though the number of groups is the same). – Ted Oct 8 '12 at 4:38
Right - that's why I'm saying I want equivalency classes within those sets, according to the "primeless isomorphism" relation. For example $\Sigma_{(1)}$ would be one equivalency class - all cyclic groups of prime order would be identified as the same thing under this relation. $\Sigma_{(2)}$ would have two equivalency classes, $[C_{p^2}]$ and $[C_p\times C_p]$. $\Sigma_{(3)}$ would have classes $[C_{p^3}],[C_{p^2}\times C_p],[(C_p)^3],[Q_8],[D_8],[\text{Heis}\,\mathbb{Z}_p],[G_p]$, where for the last two classes p is odd. Does this make sense? – Alexander Gruber Oct 8 '12 at 4:44
Isn't the number of these equivalency classes going to explode pretty quickly? For instance there are over $10^{15}$ groups of order $2^{11}=2048$ so wouldn't you expect there to be in excess of $10^{15}$ classes in $\Sigma_{(11)}$? – JSchlather Oct 8 '12 at 6:18
The lie ring approach to $p$-groups is mostly independent of $p$. This is similar to Qiaochu Yuan's answer, as it views them as defined by equations over algebras. Many qualitative properties only depend on the equations, not on the coefficient ring. However, isomorphism is still too strong an idea to try and capture, since most of the p-groups are qualitatively the same. @user1729: there are at least as many groups of order $p^n$ as there groups of order $2^n$, and usually more. – Jack Schmidt Oct 8 '12 at 13:26

Here's a special case which seems to admit a reasonable answer. To start with two of your examples, the cyclic groups $C_p$ are the specializations of the additive group scheme $\mathbb{G}_a$ to the finite fields $\mathbb{F}_p$, whereas the Heisenberg groups are the specialization of the Heisenberg group scheme $\text{Heis}$ to the finite fields $\mathbb{F}_p$. For the purposes of this answer, by "group scheme" I mean a functor

$$G : \text{CRing} \to \text{Grp}$$

from the category of commutative rings to the category of groups. (The actual definition has some additional technical assumptions which are not relevant to this answer.) The additive group scheme is the functor which assigns a commutative rings its underlying abelian group, while the Heisenberg group scheme is the functor which assigns to a commutative ring $R$ the group of matrices of the form

$$\left[ \begin{array}{ccc} 1 & R & R \\\ 0 & 1 & R \\\ 0 & 0 & 1 \end{array} \right].$$

So one way to say that a family of finite groups is related even though they are not isomorphic is to say that they are specializations of the same group scheme to finite fields. Several families of finite groups of Lie type have this property, such as the groups $\text{GL}_n(\mathbb{F}_p)$.

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In 1940 Philip Hall introduced isoclinism, an equivalence relation on groups than isomorphism (being isomorphic implies being isoclinic, but not vice versa). The concept of isoclinism was introduced to classify p-groups, although the concept is applicable to all groups. Isoclinism can be extended to isologism, which is similar to isoclinism, but then w.r.t. a variety of groups. There exists a vast literature on isoclinism and isologism.

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