# counting non-isomorphic groups of a given cardinality

Given an infinite cardinal $\kappa,$ is there some nice way to construct $2^\kappa$ non-isomorphic groups of that cardinality? In the answer to this stackexchange question, there is a fairly high-powered argument to show that that many (abelian) groups do exist, but it seems shocking that there is not a direct construction, like there is for $\aleph_0.$ (by the way, I doubt that allowing arbitrary, instead of just abelian, groups helps that much, but one never knows.) This question came from a conversation with our own @Joel David Hamkins.

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Perhaps it would be worth saying "infinite cardinality": the question for finite cardinalities seems to me to be rather different. – Qiaochu Yuan Apr 24 '12 at 23:19
In ZFC there cannot be more than $2^\kappa$ many groups on a fixed set of size $\kappa$. If there are already $2^\kappa$ many non-isomorphic Abelian groups then we have hit the maximal number. Of course there are many many more which are non-Abelian, but the cardinality of this collection cannot grow much further. – Asaf Karagila Apr 24 '12 at 23:25
Asaf, the question is about the lower bound, not the upper bound. The answer to the math.SE question uses stability theory, but one expects that if there really are $2^\kappa$ many nonisomorphic groups of size $\kappa$, then we might hope to describe them quite explicitly. For example, for countable groups, take any set of primes and consider the direct product of cycle groups of those orders. This gives continuum many non-isomorphic countable groups. Can one similarly give such an easy account of $2^\kappa$ many non-isomorphic groups of size $\kappa$ for larger $\kappa$? – Joel David Hamkins Apr 24 '12 at 23:41
@Qiaochu: True, fixed. – Igor Rivin Apr 24 '12 at 23:48
Sorry, I meant direct sum of cyclic groups, not direct product, since we want a countable group. – Joel David Hamkins Apr 24 '12 at 23:52

If you just want a direct construction which avoids nontrivial set theory such as stationary sets etc., how about this?

Step One: For each subset $S \subseteq \kappa$, let $M(S)$ be the structure $\langle \kappa; < , S \rangle$, where $S$ is regarded as a unary relation. Obviously, if $S \neq T$, then $M(S)$ and $M(T)$ are non-isomorphic.

Step Two: For each subset $S \subseteq \kappa$, encode $M(S)$ into a corresponding graph $\Gamma(S)$ so that if $S \neq T$, then $\Gamma(S)$ and $\Gamma(T)$ are non-isomorphic. (This is an easy exercise.)

Step Three: For each subset $S \subseteq \kappa$, encode the graph $\Gamma(S)$ into a suitable group $G(S)$ with generators $\Gamma(S)$ and relations $R(S)$ which encode the adjacency relation. (This can be done using small cancellation theory.)

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I am sure I am being very dense, but why do non isomorphic graphs lead to non-isomorphic groups? – Igor Rivin Apr 25 '12 at 1:55
I was being too terse. The relations $R(S)$ are supposed to be chosen so that this is true. For example, for every vertex $v$, include the relation $v^{p}=1$, where $p$ is a large prime. If $u$, $v$ is a pair of vertices, then include the relation $(uw)^{q}$ or $(uw)^{r}$, depending on whether or not $u$, $w$ are adjacent, where $q$ and $r$ are much bigger primes. (This trick is borrowed from Jay William's recent PhD thesis.) – Simon Thomas Apr 25 '12 at 2:07
Simon, could we also see your direct construction using stationary sets? That would be nice to see... – Joel David Hamkins Apr 25 '12 at 2:16
I was thinking of Saharon's proof via stability theory which involves coding stationary sets. For example, consider the usual construction of $2^{\aleph_{1}}$ dense linear orders of cardinality $\aleph_{1}$ as unions of countable dense linear orders, where the embeddings are end extensions. – Simon Thomas Apr 25 '12 at 2:22
@Simon: After you got graphs, there is no need to use small cancelation. Use Right Angled Artin Groups (see arxiv.org/pdf/1007.1118.pdf, theorem 1.10). – Mark Sapir Apr 25 '12 at 4:52