There are pairs of nontrivial groups having no nontrivial homomorphisms between them, for instance $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$. (I can think of at least one proof, using Lagrange's theorem.)

With that example, it is easy to form at least one countable collection of groups such that no two groups exhibit nontrivial homomorphisms, namely $\{ \mathbb{Z}/p\mathbb{Z}\}_{p \text{ prime}}$.

What is an example of an uncountable collection of distinct groups containing no two groups with nontrivial homomorphisms between each other?


1 Answer 1


For brevity I'll use the term "non-homomorphic" to refer to any family of groups no two of which admit a nontrivial homomorphism.

For $A$ a set of primes, consider the subgroup $$G_A:=\{{z\over p_1\cdot ...\cdot p_n}: p_i\in A, z\in\mathbb{Z}\}$$ of $(\mathbb{Q};+)$ (yes, this can be written more simply, but there's a method to my madness). The key point is that if $A\not\subseteq B$ then there is no homomorphism from $G_A$ to $G_B$, since if $p\in A\setminus B$ then the only element of $G_B$ which is "divisible by $p$" is $0$ but every element of $G_A$ is "divisible by $p$."

Now to get an uncountable non-homomorphic family of groups we just need a family $\{A_i:i\in I\}$ of uncountably many sets of primes, no two distinct elements of which are $\subseteq$-comparable. This is a fun exercise, and in fact we can get such a set of size $2^{\aleph_0}$:

Fixing a bijection $$b: \mathbb{Q}\times\{0,1\}\rightarrow\mathsf{Primes},$$ for each real $r$ let $$A(r)=\{b(q, 0): q\in\mathbb{Q}, q<r\}\cup\{b(q,1): q\in\mathbb{Q}, q\ge r\}.$$ It's easy to check that if $r\not=s$ then $A(r)\not\subseteq A(s)$ and conversely: think about a rational $q$ between $r$ and $s$, and consider $b(q,0)$ vs. $b(q,1)$.

It's a cute further exercise (albeit one unrelated to the OP) to construct a size-continuum family $\mathcal{F}$ of natural numbers which are pairwise almost disjoint, that is, such that for any distinct $X,Y\in\mathcal{F}$ the intersection $X\cap Y$ is finite.

Note that the above idea is completely useless for getting even larger non-homomorphic sets of groups: it won't, for example, help us find such a set of size $2^{2^{\aleph_0}}$.

Interestingly, there's an odd "argument from authority" that any argument which builds arbitrary-cardinality non-homomorphic sets must be pretty complicated. It's due to the as-yet-seeming consistency of the following set-theoretic hypothesis:

(Vopenka's principle) Given any proper class of structures of the same type (e.g. groups), there is an elementary embedding between a pair of distinct structures in the class.

The precise definition of "elementary embedding" is a bit complicated, but all we need here is this: that all elementary embeddings are injective homomorphisms (in fact they're a lot more than that). So a fortiori, any non-homomorphic proper class of groups would be a counterexample to $\mathsf{VP}$. It's separately worth noting that $\mathsf{VP}$ is incredibly robust to modifications of the notion of elementary embedding; see e.g. the discussion here.

This roughly means that, unless $\mathsf{VP}$ is in fact inconsistent with $\mathsf{ZFC}$ (which would be a huge result), there is no single way to construct a family of groups avoiding homomorphisms between them which can "go on forever" (= produce a proper class). For example, if I have a way of constructing a non-homomorphic family of groups of size $2^{2^{\aleph_0}}$, either I have to secretly be using some additional set-theoretic assumptions (which contradict $\mathsf{VP}$) or there has to be some "obstacle" to further enlarging that family to arbitrary sizes.

Amusingly, Vopenka originally introduced his principle as a joke!

  • $\begingroup$ What is $D(r)$? $\endgroup$
    – Lee Mosher
    Jan 28 at 20:28
  • $\begingroup$ From the following line it looks like $D(r)$ is a Dedekind cut corresponding to the real number $r$. You could also use unit disks centered at $r$ (and the $\{0, 1\}$ factor would be unnecessary in this case). $\endgroup$ Jan 28 at 20:45
  • $\begingroup$ @LeeMosher Whoops, multiple-edit typo; fixed! $\endgroup$ Jan 28 at 21:23
  • $\begingroup$ An overdue comment: this is a great answer that gave me much more than I was bargaining for when I asked this question. I was mainly hoping to see an example of an uncountable collection of distinct, non-isomorphic groups (I could not even think of one.) I had never heard (nor expected to hear of something like) Vopenka’s principle. $\endgroup$ Feb 20 at 16:36

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