Does there exist a group $G$, such that $G\otimes H \cong G$ for all finite groups, $H$?

Question

A friend and I, whom only have elementary knowledge of group theory, were playing around some of the natural transformations between group theory and other categories. We probably shouldn't be playing around with as high-level questions as these without properly establishing our understanding beforehand, but it is in the sake of recreation. An obvious fact is that the trivial group behaves as a zero-object, in that $$\{e\}\otimes G\cong G$$ for all groups, $G$ (where $\otimes$ is the direct product). The next idea was if there were something akin to an additive identity. This may have a trivial answer, but it evades me at the moment,

Does there exist a group $G$ (possibly infinite), such that $G\otimes H \cong G$ for all finite groups, $H$?

Originally, I suspected that such a $G$ probably doesn't exist, on behalf of the fact that it would have to behave like the empty-set, which is not a group. However, akin to the fact that $\infty$ in the extended real numbers also behaves like an additive identity, I wonder if there exists an infinite group which behaves as an absorbing identity to finite groups (I'm not sure if the finitude of $H$ is neccesary, assuming it exists, but it seems like a intuitive restriction).

If not, I suppose this idea could be salvaged into,

Does there exist a group $G$, such that $G * H \cong H$ for all finite groups, $H$?

where $*$ is the free product, considering that it is indeed the coproduct in $\text{Grp}$. The trivial group, might trivially have the property that $\{e\} * H \cong H$, but my understanding of what a free product is, is very limited.

Any and all insight would be greatly appreciated, despite the possibility that this is simply a silly idea, and I am a little out of my pay-grade.

Answer 1

For the first part of the question, the answer is yes, though it takes a slightly 'monstrous' construction. Since the finite groups are countable, we can enumerate them as $G_1$, $G_2$, $\ldots$; now, we want to make sure that whenever we direct-sum another finite group onto this, it remains isomorphic. Well, $G_1\times G_1$ almost certainly isn't isomorphic to $G_1$, but if we take a direct product of $\omega$ copies of $G_1$, $G_1^{\omega}$, then this can 'absorb' another instance of $G_1$ easily. Maybe the easiest way to describe this direct product is as the set of functions ${\mathfrak g}():\mathbb{N}\mapsto G_1$, where $\mathfrak g()\cdot \mathfrak h()$ is defined 'pointwise' as the function $\mathfrak f()$ such that $\mathfrak f(n)=\mathfrak g(n)\mathfrak h(n)$ for all $n\in\mathbb{N}$, and the identity and inverses are likewise defined pointwise. Under this definition, we can get an isomorphism $G_1\times G_1^{\omega}\mapsto G_1^{\omega}$ that absorbs the extra copy of $G_1$ by mapping the pair $\langle g, \mathfrak{g}()\rangle$ where $g\in G_1$ and $\mathfrak{g}\in G_1^{\omega}$ to the function $\mathfrak{g'}$ defined by $\mathfrak{g'}(0)=g$ and $\mathfrak{g'}(n+1)=\mathfrak{g}(n)$ for all $n$.

This group is 'immense' — its cardinality is $\mathfrak{c}$, the cardinality of the reals, for any (nontrivial!) finite group $G_1$. For a somewhat smaller example, rather than the direct product we can take the direct sum of these groups. I won't go into details on the difference in definitions (ask your local category theorist), but for our purposes the direct sum of infinitely many copies of $G_1$ can be defined as the subgroup of $G_1^\omega$ consisting of all of the functions $\mathfrak{g}()$ that are the identity for all but finitely many elements; in other words, it's the set of finite sequences of elements of $G_1$, with the product of two elements again considered componentwise (and all components beyond the last considered the identity) and therefore finite. For instance, $\langle g_1, g_2, g_3\rangle\cdot \langle h_1, h_2\rangle$ $=\langle g_1h_1, g_2h_2, g_3\rangle$. It's a nice and very standard exercise to convince yourself that this group, which I'll write as $G_1^{\lt\omega}$, has 'only' countably many elements.

Next, to make sure that this works for all finite groups, we can do the same for all the other factors and take the group $G_1^{\omega}\times G_2^{\omega}\times\ldots$; this group will have exactly the property you want. We can also make the smaller construction here by considering the infinite direct sum $G_1^{\lt\omega}\oplus G_2^{\lt\omega}\oplus\ldots$; similarly to the above, this consists of finite-length sequences $\langle \mathfrak{g}_1, \mathfrak{g}_2, \ldots, \mathfrak{g}_n\rangle$ where each $\mathfrak{g}_i\in G_i^{\lt\omega}$ is in turn a finite sequence $\langle g_{i;j}\rangle$ of elements of $G_i$. Since all of its pieces are countable, this group is also countable, and it also 'absorbs' any finite group in exactly the same way as the infinite direct product.

Answer 2

First, it is not customary to use $\otimes$ as the symbol for direct products. We instead use $\times$.

Second, yes, there is such a $G$.

Take a set $S$ of groups such that for all finite groups $H$, there is a $W \in S$ such that $H \cong W$. We could do this by taking $S = \{W \mid W$ is a group with underlying set $\{1, \ldots, n\}$ for some $n\}$.

Then define $G = \prod\limits_{W \in S} W^\mathbb{N}$. Given a finite group $H$, take some $W \in S$ such that $H \cong W$. Then $G \times H \cong G \times W \cong G$.

Answer 3

Yes. For each $n\in{\mathbb N}$ there is at most a finite number of groups of order $n$, up to isomorphism. Thus there is a sequence $G_1, G_2, \ldots$ such that each finite group $H$ is isomorphic to precisely one of the $G_n$. Now define $G$ to be the direct product of all the $G_n$, each with countable multiplicity. This $G$ satisfies $G\times H\cong G$ for each finite group $H$. (I guess this is what the OP wanted, since $G\times H\cong H$ for all finite groups $H$ clearly is impossible.)

Answer 4

There is a way to show this large group exists without finding it.

Consider the "set" of all commutative groups (technically "a class"). We call this "set" by $S$. Suppose $G_1,G_2\in S$. We will say that $G_2$ "dominates" $G_1$, and write $G_1\leq G_2$ if and only if $G_1\oplus H = G_1$ implies that $G_2\oplus H = G_2$ for all groups $H$ in $S$. In other words, if $G_1$ has the desired "absorbing property" with respect to $H$ then so does $G_2$.

Now suppose that $G_1\leq G_2$. We claim that $G_1\oplus G_2$ dominates both $G_1$ and $G_2$. Suppose $G_1\oplus H = G_1$. Then it follows that $G_2\oplus H = G_2$ also. Therefore, $(G_1\oplus G_2)\oplus H = G_1\oplus (G_2\oplus H) = G_1\oplus G_2$. Hence, $G_1\oplus G_2$ dominates $G_1$. Furthermore, if $G_2\oplus H = G_2$ then a similar argument will show that $(G_1\oplus G_2)\oplus H = G_1\oplus G_2$.

The above argument can be generalized to an arbitrary sum, as long as the groups form a dominating chain. If $(G_i)_{i\in I}$ is a family of groups such that whenever $i_1,i_2\in I$ we have that $G_{i_1}\leq G_{i_2}$ or $G_{i_2}\leq G_{i_1}$. Then, based on the same method as above, $\bigoplus_{i\in I} G_i$ will be a group which will dominate each $G_i$. Therefore, every "chain" has an upper bound.

By Zorn's Lemma we conclude that there is a maximal element in this "set" $S$. Let us call it $M$. We argue that $M$ has the desired property, i.e. $M\oplus H = M$. Indeed, if it did not then $M\oplus H$ would dominate $M$ and violate the maximality property of $M$.