Usually, for a family $\{G_\alpha\}_{\alpha \in A}$ of abelian groups, one defines
$$ \bigoplus_{\alpha \in A} G_\alpha \:= \{ a \in \prod_{\alpha \in A} G_\alpha : \sharp \{ \alpha \in A : \pi_\alpha(a) \ne 0 \} < \infty \} < \prod_{\alpha \in A} G_\alpha $$
where $\pi_\beta : \prod_{\alpha \in A} G_\alpha \to G_\beta$ is the projection. One has the inclusion maps $i_\beta : G_\beta \to \bigoplus_{\alpha \in A} G_\alpha$ such that $\pi_\alpha \circ i_\alpha = \mbox{id}$ and $\pi_\alpha \circ i_\beta = 0$ for $\alpha \ne \beta$.
It can then be shown that $\bigoplus_{\alpha \in A} G_\alpha$ has the following universal property:
(a) For any abelian group $H$ and any family of homomorphisms $\phi_\beta : G_\beta \to H$ there is a unique homomorphism $\phi : \bigoplus_{\alpha \in A} G_\alpha \to H$ such that $\phi \circ i_\alpha = \phi_\alpha$.
From the definition of $\bigoplus_{\alpha \in A} G_\alpha$, it follows that
(b) Any element $g \in \bigoplus_{\alpha \in A} G_\alpha$ can be written as $i_{i_1}(g_{\alpha_1})+…+i_n(g_{\alpha_n})$, where $\{\alpha_1,…,\alpha_n\} \subset A$ is finite and $g_{\alpha_1} \in G_{\alpha_1}$,…,$g_{\alpha_n} \in G_{\alpha_n}$.
If we define a direct sum of the $G_\alpha$s as a couple $(G,\{j_\alpha\}_{\alpha \in A})$ with $G$ and $j_\alpha : G_\alpha \to G$ with the property (a), can we show that (b) is true for $G$ and the $j_\alpha$s, without using an explicit construction of the group $G$, and without appealing to the existence of an isomorphism between $G$ and $\bigoplus_{\alpha \in A} G_\alpha$?
Best Answer
Assume $G$ has property (a). Let $H$ be the subgroup of $G$ generated by $j_{\alpha}(G_{\alpha})$ for $\alpha\in A$. I claim that $H=G$.
Since $G$ is abelian and $H$ is a subgroup, $H$ is normal. Let $K=G/H$ and let $\pi$ be the canonical homomorphism. If we let $\phi_{\beta}\colon G_{\beta}\to K$ be the zero map for all $\beta$, then $\pi\circ i_{\alpha}=\phi_{\alpha}$ for all $\alpha$; but if $\zeta\colon G\to K$ is the zero map, then we also have $\zeta\circ i_{\alpha}=\phi_{\alpha}$ for all $\alpha$. By the uniqueness clause of (a), $\pi=\zeta$, which means that $K$ is the trivial group, so $G=H$, as claimed.
Since the subgroup generated by a family $j_{\alpha}(G_{\alpha})$ of subgroups consists precisely of all finite sums of elements of the union of the subgroups, and each element in the union can be written as $j_{k}(g_{k})$ for some $k\in A$ and $g_k\in G_{j}$, property (b) follows.
Of course, property (a) is already enough to define direct sum up to unique isomorphism (respecting the maps $j_{\alpha}$). By property (a) for $G$, we have a unique map $\phi\colon G\to \oplus G_{\alpha}$ such that $i_{\alpha}=\phi\circ j_{\alpha}$; and by prperty (a) for $\oplus G_{\alpha}$, we have a unique map $\psi\colon \oplus G_{\alpha}\to G$ such that $j_{\alpha}=\psi\circ i_{\alpha}$. Then $\phi\circ\psi\colon \oplus G_{\alpha}\to\oplus G_{\alpha}$ satisfies $i_{\alpha}=(\phi\circ\psi)\circ i_{\alpha}$, so by the uniqueness clause $\phi\circ\psi=\mathrm{id}_{\oplus G_{\alpha}}$. And $\psi\circ\phi\colon G\to G$ satisfies $j_{\alpha}=(\phi\circ\psi)\circ j_{\alpha}$, so by the uniqueness clause $\phi\circ\psi=\mathrm{id}_G$. Thus, $G\cong \oplus G_{\alpha}$, which is why we talk about the direct sum and no merely a direct sum.