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We’re looking at the following definition in my class.

Definition. The coproduct $X_{1} \amalg X_{2}$ of $X_{1}$ and $X_{2}$, together with the morphisms $i_{j}: X_{j} \rightarrow X_{1}\amalg X_{2}$, is characterized by the following universal property: Given any object $Y$ with morphisms $f_{j}: X_{j} \rightarrow Y,$ there exists a unique $f: X_{1} \amalg X_{2} \rightarrow Y$ such that $f_{j}=f \circ i_{j}$.

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In $G_1 \times G_2$, there exist two subgroups isomorphic to $G_1$ and $G_2$, namely the sets of elements $\{(g_1,e_2): g_1 \in G_1\}$ and $\{(e_1,g_2): g_2 \in G_2\}$, respectively. These two subgroups have the property that their members commute with each other: $(g_1,e_2) \cdot (e_1,g_2) = (g_1,g_2) = (e_1,g_2) \cdot (g_1,e_2)$.

The reason given for the direct product not being the coproduct is that, in a group $Y$ with maps $f_1: G_1 \to Y$ and $f_2: G_2 \to Y$, elements of $Y$ don’t commute with each other. I’m not yet able to derive this.

My understanding of the argument is that, given the property of the two subgroups above, I ought to be able to derive $f_1(g_1) \cdot_Y f_2(g_2) = f_2(g_2) \cdot_Y f_1(g_1)$, which would not be true for a general group $Y$ (unless $Y$ is abelian).

Now if I also assume $G_1 \times G_2$ to be the coproduct, I’d have: $$f_1(g_1) \cdot_Y f_2(g_2) = (f \circ i_1)(g_1) \cdot_Y (f \circ i_2)(g_2) = f(g_1,e_2) \cdot_Y f(e_1,g_2)$$

It’s here that I don’t know how to keep going to get the relation above.

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    $\begingroup$ Your understanding is correct: the elements of $G_1\times G_2$ corresponding to elements of $G_1$ commute with all elements corresponding to those of $G_2$, which is not necessarily true for any $Y$ and diagram above, so we don't get $G_1\times G_2\to Y$. $\endgroup$ – Berci Feb 4 at 10:35
  • $\begingroup$ I think I seem to forget that in the category of groups, the morphisms are homomorphism, whose property is needed to advance the calculation above? $\endgroup$ – ensbana Feb 4 at 10:55
  • $\begingroup$ Perhaps you are trying to think about this too abstractly. The point is that, for any two nontrivial groups $G_1$ and $G_2$, there exist groups $G$ that contain isomorphic copies of $G_1$ and $G_2$ as subgroups, and in which the elements of the subgroup $G_1$ do not all commute with all of the elements of $G_2$. This implies that the coproduct is not equal to the direct product $G_1 \times G_2$. $\endgroup$ – Derek Holt Feb 4 at 11:29
  • $\begingroup$ @DerekHolt The proof that such "weird" groups exist, is of course by constructing them, i.e., ultimately by constructing the coproduct in terms of words in $G_1\cup G_2$ with th eobvious rules for compositon and modulo the obvious identifications ... $\endgroup$ – Hagen von Eitzen Feb 4 at 16:22
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    $\begingroup$ @HagenvonEitzen But there are other ways of constructing groups with the required properties. For example, you could use Cayley's Theorem and embed them both into some symmetric group, giving them overlapping support to ensure that they don't permute with each other. Then, if $G_1$ and $G_2$ are both finite, so is the constructed group,. $\endgroup$ – Derek Holt Feb 4 at 20:16
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Let $X_1, X_2$ be cyclic of order $2$ and let $x_i$ denote the non-trivial element of $X_i$. Let $Y$ be the set of permutations of $\Bbb Z$. In particular, we have two elements $y_1,y_2\in Y$, given by $y_1(n)=-n$ and $y_2(n)=1-n$. Then $y_1\circ y_1=y_2\circ y_2=\operatorname{id}_{\Bbb Z}$, i.e., these are both of order $2$. This allows us to define homomorphisms $f_i\colon X_i\to Y$ given by mapping $x_i\mapsto y_i$.

But $y_1( y_2(n))=n-1$ whereas $y_2(y_1(n))=n+1$, i.e., $y_1\circ y_2\ne y_2\circ y_1$ -- the elements do not commute in $Y$. Hence neither may the elements $i_1(x_1)$ and $i_2(x_2)$ commute in $X_1\amalg X_2$. This shows that $X_1\amalg X_2$ cannot be just the direct product.


By the way, the subgroup of $Y$ that is generated by $y_1$ and $y_2$ (together with the above homomorphisms) does have the desired universal property. Can you show this?

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As you observed, the main issue is that elements of $G_1$ always commute with elements of $G_2$ in $G_1\times G_2$, so it is not a free construction.
Notably, the direct product does coincide with the coproduct (satisfies its universal property) in the category of Abelian groups.

Choose specifically $X_1=X_2=\Bbb Z$, and consider the cocone $X_1\to X_1\times X_2\leftarrow X_2$ with $n\mapsto (n,0)$ and $n\mapsto (0,n)$.

Then a homomorphism $f_i:X_i\to Y$ basically just selects an arbitrary element $y_i$ of $Y$ (taking $y_i:=f_i(1)$), so in order for the diagram to commute, as $(1,0)$ and $(0,1)$ commute in $G_1\times G_2$, we would need that $y_1y_2=y_2y_1$ in $Y$:
If an $f:X_1\times X_2\to Y$ makes the diagram commutative, we have $$f(1,0)=f(i_1(1))=f_1(1)=y_1\\ f(0,1)=f(i_2(1))=f_2(1)=y_2\\ f(1,1)=f((0,1)+(1,0))=f(1,0)f(0,1)=y_1y_2\\ f(1,1)=f((1,0)+(0,1))=f(0, 1)f(1,0)=y_2y_1\,.$$

So, there is no arrow $X_1\times X_2\to Y$ making the diagram of the universal property commutative, whenever $y_1,y_2$ are noncommutative in $Y$.

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