Is there a "universal group"?

Question

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.

Answer 1

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.

Answer 2

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.