7

Is there a group $U$ such that for any group $G$, $G$ is isomorphic to a subgroup of $U$? 0 For any group $G$, it is isomorphic to a subgroup of a symmetric group (Cayley's theorem) so I'm wondering if we could do something like that to construct $U$. For example, every finite group is a subgroup of

$$S = \bigcup_{n \in \mathbb{N}} S_n,$$

I'm curious as to whether a construction of $U$ might be similar, if it exists.

Robin
  • 3,227
  • 8
    Presumably the answer is "no" because of some argument like "groups of all cardinalities exist, and the class of all cardinalities is too large to be a set (much less a group)". – Greg Martin Dec 28 '23 at 19:03
  • I think every set can be turned into a group. – Brauer Suzuki Dec 28 '23 at 19:03
  • 2
    See here for an argument for @BrauerSuzuki's comment (excepting the empty set), assuming the axiom of choice. Ergo, groups of all cardinalities exist (except $0$). – Brian Tung Dec 28 '23 at 19:05
  • 3
    @BrauerSuzuki: That every nonempty set can be given a group structure is equivalent to the axiom of choice. – Arturo Magidin Dec 28 '23 at 19:06
  • I beat @ArturoMagidin to linking to his own answer. :-) – Brian Tung Dec 28 '23 at 19:07
  • 1
    Have you heard of the issue with "the set containing all sets"? I believe a similar thing would happen with the group that contained all groups... This may be the wrong approach as there is a difference between containing in the set sense and being a subgroup, but could a group contain the product of every other group? – cable Dec 28 '23 at 19:07
  • 2
    @cable Yeah, I was wondering if the group structure would make it more restrictive, but didn't consider the obvious thing that Daniël Apol pointed out, namely that $U$ must have a cardinality of itself, and another group will have a larger one. – Robin Dec 28 '23 at 19:10
  • @BrianTung: By 61 seconds. – Arturo Magidin Dec 28 '23 at 19:12
  • 7
    Though set theoretic constraints are really the only constraints. If you allow a group to be defined on a proper class, and that you’re only considering universal among all “set-based” groups, then you can construct such a universal “class-based” group. – David Gao Dec 28 '23 at 19:16

2 Answers2

15

You need to bound the size of your groups. Namely, if $U$ exists, it has a certain cardinality $\kappa$, but it is easy to see that there are groups of cardinality $>\kappa$. However, if we index the collection of all isomorphism classes of groups of cardinality $<\kappa$ for some cardinal $\kappa$ by a set $A$, then the product of all (representatives of isomorphism classes of) groups contained in $A$ forms a universal group in the sense of your question: every group of cardinality $<\kappa$ is a subgroup of this object.

Daniël Apol
  • 3,964
  • You have to work a little bit to get a set of groups that contains at least one representative of each isomorphism class in $A$, especially since an isomorphism class of groups isn't a set. – ronno Dec 29 '23 at 03:43
  • 3
    @ronno Not hard work at all. Every isomorphism class of groups of cardinality $<\kappa$ contains a group whose underlying set is a cardinal $<\kappa$, and the collection of such groups is a set. – BrianO Dec 29 '23 at 11:27
12

The axiom of choice was mentioned, but we don't need that. Neither the result that every non-empty set has a group structure.

Assume that there is a group $U$ such that every group admits a monomorphism to $U$.

Let $|U|$ denote the underlying set of $U$. Consider the group $\prod_{u \in |U|} \mathbb{Z}/2\mathbb{Z}$. Its underlying set identifies with $P(|U|)$. By assumption, there will be an injective map $P(|U|) \to |U|$. This yields a surjective map $|U| \to P(|U|)$, which cannot exist as we know.

The same proof shows: Let $\mathcal{V}$ be any variety in the sense of universal algebra that has a $\mathcal{V}$-algebra with at least two elements. Then there is no "universal" $\mathcal{V}$-algebra. So there is no universal ring, etc.

Martin Brandenburg
  • 163,620