While I was reading "Abelian Groups" by Fuchs $(2015)$, I encountered Example $1.2$ in the chapter on Mixed Groups, which stated the following:
Let $p_1,p_2,\dots,p_n,\dots$ denote different primes, and define $T_1=\bigoplus_{n=1}^\infty\langle a_n\rangle$ with $o(a_n)=p_n$. Thus $T_1$ is the torsion subgroup of the direct product $B=\prod_{n=1}^\infty \langle a_n\rangle$, a mixed abelien group.
Let $b_0=(a_1,\dots,a_n,\dots)\in B$. For $i\neq n$, the equation $p_nx=a_i$ is uniquely solvable in $<a_i>$, thus $B$ contains a unique element $b_n$ such that : $$p_nb_n=(a_1,\dots,a_{n-1},0,a_{n+1},\dots)=b_0-a_n$$
$T_1$ is the torsion part of the group $A_1=\langle T_1,b_1,\dots,b_n,\dots\rangle$ such that $A_1/T_1$ is torsion free of rank $1$. We claim that $A_1$ is not splitting. Otherwise, we had $A_1=T_1\oplus G$ for some torsion-free $G<A_1$, so $b_n=t_n+g_n$ with $t_n\in T_1 ,g_n\in G$ for each $n$. Thus : $$p_nt_n+p_ng_n=p_nb_n=b_0-a_n=(t_0-a_n)+g_0$$
Equating the $T_1$-coordinates we get $p_nt_n=t_0-a_n$ for $n=1,2,\dots$.
There is a prime $p_j$ for wich the equation $p_jx=t_0$ is solvable in $T_1$, and if $x=t\in T_1$ is solution, then we get $p_j(t-t_j)=a_j$, an abvious contradiction.
How did we get from $[p_nt_n+p_ng_n=p_nb_n=b_0-a_n=(t_0-a_n)+g_0]$ to $p_nt_n=t_0-a_n$ for $n\geq 1$ ? And where is the contradiction?