will be [@fuchs Theorem 27.5] a subgroup $H$ of $G$ such that $$G=\left(\bigoplus_{p_k^{n_k}\in {{\mathbb P}}_0^{\uparrow}}
\bigoplus_{\gamma(p_k^{n_k})}B_{p_k,n_k}\right) \bigoplus H,$$

For each prime $p$, consider a $p$-basic subgroup $B_{p,H}=\oplus_{n} B_{p,n,H}$ of $H_p$, the $p$-part of $t(H)$, it is immediately checked that either $B_{p,H}$ itself (if $p\not \in
{{\mathbb P}}_0^\uparrow$) or $B_{p,H} \bigoplus \left(\bigoplus
_{\overset{p_k^{n_k}\in {{\mathbb P}}_0^{\uparrow}}{p_k=p}}
\bigoplus_{\gamma(p_k^{n_k})}B_{p_k,n_k}\right)$ (if $p \in
{{\mathbb P}}_0^\uparrow$) is also $p$-basic in $G$.

Since different basic subgroups are necessarily isomorphic [@fuchs Theorem 35], we have that $B_{p,H}$ or $B_{p,H}
\bigoplus \left(\bigoplus_{\overset{p_k^{n_k}\in
{{\mathbb P}}_0^{\uparrow}}{p_k=p}}
\bigoplus_{\gamma(p_k^{n_k})}B_{p_k,n_k}\right)$ is isomorphic to $B_p$. We have therefore that, for each $p$, either $\sup
|B_{p,n,H}|$ is not attained (case 1 above) or attained at infinitely many $p^n$’s (case 2).

Let now $n$ be any natural number. Then $|n
B_{p_k,n_k,H}|=|B_{p_k,n_k,H}|$ unless $p_k^{n_k}$ divides $n$. Since this will only happen for finitely $p_k^{n_k}$’s, we conclude, in both cases 1 and 2 that $|nB_{p,H}|=|B_{p,H}|$.

Using that $B_{p,H}$ is pure in $H_p$ and that $H_p/B_{p,H}$ is divisible we have that, $$\begin{aligned}
|nH_p|&=\left|\frac{nH_p}{nB_{p,H}}\right|+\left|nB_{p,H}\right| \\
&=
\left|n\left(\frac{H_p}{B_{p,H}}\right)\right|+\left|B_{p,H}\right|\\
&=
\left|\frac{H_p}{B_{p,H}}\right|+\left|B_{p,H}\right|=|H_p|.\end{aligned}$$ Since $ |H|=\sum_p H_p + r_0(H)|$ for every infinite group $H$ and $r_0(nH)=r_0(H)$ we have finally that $|H|=|nH|$, for every $n\in
{{\mathbb Z}}$.

The terminology introduced in the next definition is motivated, 