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I have couple questions about terms I mentioned in the title.

  1. Why we don't define direct sum of non-abelian groups (subset of direct products which consists of elements with almost every component equal to identity element)?
  2. Why is direct sum used in definition of coproduct in category of Abelian groups but it is not "good" for category of groups in general (I know free groups are used)? Where is essential difference?
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    $\begingroup$ The direct sum of nonabelian groups is certainly well defined, but it doesn't satisfy the universal property of a "coproduct". The free product of two groups is a different construction which does satisfy the universal property of a coproduct. So there's nothing wrong with direct sums, they're just different than coproducts. $\endgroup$ – Jim Apr 2 '15 at 18:02
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Let's take your proposed definition of coproduct, call it $G+H$. Then given two groups $G$ and $H$, there are homomorphisms $G\rightarrow G*H$, and $H\to G*H$. From the universal property, we should get a homomorphism $G+H\to G*H$ with appropriate commutativity properties. Now, if $g\in G$ and $h\in H$ are not identity elements we know that $gh\in G+H$ has to map to $gh\in G*H$. And $hg\in G+H$ has to map to $hg\in G*H$. But, $gh=hg$ in $G+H$, and $gh\neq hg$ in $G*H$.

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  • $\begingroup$ I'm lost in meaning of symbols. Is G*H free product group? Apropriate commitative property- what do you mean by that? If I've understood corectly you used universal property of coproduct to show my definition is not isomorphic to free product. But what if we dont know about free product? Why would direct or weak direct product fail definition of coproduct? $\endgroup$ – Meow Apr 1 '15 at 12:13
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    $\begingroup$ @Meow Yes, $G*H$ is the free product. What I've shown is that your definition of direct sum does not work. A direct sum is a special type of coproduct. Since your definition isn't the appropriate coproduct, it doesn't work. The same can be said about any other definition you might think up, except the free product. $\endgroup$ – Joe Johnson 126 Apr 1 '15 at 18:09
  • $\begingroup$ In what part are you using that G and H are non-abelian groups? @JoeJohnson126 $\endgroup$ – user392559 Dec 4 '18 at 5:05
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Here is a down to earth example that the direct sum of groups is not the coproduct in the category of all groups.

Let $D_3 = \langle\rho, \tau \ | \ \rho\tau = \tau\rho^2, \rho^3 = \tau^2 = 1\rangle$ be the dihedral group of order $6$ and let $G = \mathbb Z/2\mathbb Z \oplus \mathbb Z/3\mathbb Z$ be the direct sum of cyclic groups of order $2$ and $3$. Note that these groups are abelian, so the problem with direct sum not equalling coproduct is not because of the groups whose direct sum you're taking, but because of the category in which you're taking the coproduct.

Now I have homomorphisms $\mathbb Z/2\mathbb Z \to D_3$ and $\mathbb Z/3\mathbb Z \to D_3$ defined by $a \mapsto \tau^a$ and $a \mapsto \rho^a$ respectively. If $G$ were indeed a coproduct then that means there should be a homomorphism $\phi\colon G \to D_3$ such that $(a, 0) \mapsto \tau^a$ and $(0, a) \mapsto \rho^a$ respectively. Note that the image of an abelian group under a homomorphism is again abelian (because $\phi(x)\phi(y) = \phi(xy) = \phi(yx) = \phi(y)\phi(x)$) and $\rho$ and $\tau$ are in the image of $\phi$, but $\rho$ and $\tau$ don't commute. This is a contradiction.

That means $\phi$ can't exist, so $G$ is not the coproduct of $\mathbb Z/2\mathbb Z$ and $\mathbb Z/3\mathbb Z$.

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Here is an example where you might see "what goes wrong". Recall that our co-product (which I will call $C$, for now) has a pair of monomorphisms:

$i_1: G \to C. i_2:H \to C$, and that, by construction, if $K$ is ANY other group with a pair of homomorphisms:

$f_1:G \to K, f_2: H \to K$, we must have a UNIQUE group homomorphism:

$\phi: C \to K$ with $\phi\circ i_n = f_n$ for $n = 1,2$.

So let's see what happens if we take $C = \{(g,h): g \in G, h \in H\}$ for some specific groups $G, H$ (here we are, by assumption, taking $i_1(g) = (g,e_H)$ and $i_2(h) = (e_G,h)$).

So consider $G \leq D_4 = \{e,r,r^2,r^3\} \cong C_4$, and $H \leq D_4 = \{e,r^2,s,sr^2\} \cong C_2 \times C_2$.

Now the direct sum is (isomorphic to) $C_4 \times C_2 \times C_2$. Note that in this group: $(r,s) = (r,e_H)(e_G,s) = (e_G,s)(r,e_H)$.

So now let $K = D_4$ with $f_1: G \to D_4,f_2:H \to D_4$ being the actual inclusion monomorphisms.

Then $r = f_1(r) = (\phi \circ i_1)(r) = \phi((r,e_H))$ and $s = f_2(s) = (\phi\circ i_2)(s) = \phi((e_G,s))$.

Since $\phi$ is a homomorphism, we have:

$rs = \phi((r,e_H))\phi((e_G,s)) = \phi((r,e_H)(e_G,s)) = \phi((r,s))$.

But also: $\phi((r,s)) = \phi((e_G,s)(r,e_H)) = \phi((e_G,s))\phi((r,e_H)) = sr$.

But this is absurd: in $K$, we certainly DON'T have $rs = sr$. And the reason things go wrong is that the copies of $G$ and $H$ "commute" in $G \times H$, but they do not in $K$.

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