# Does $G\oplus G$=$H\oplus H$ imply $G$=$H$ for divisible abelian groups?

Suppose $G$ and $H$ are two divisible abelian groups. Futhermore we have $G\oplus G$=$H\oplus H$. Is $G$ isomorphic to $H$?

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If this is homework, what have you tried? –  Tyler May 10 '11 at 16:50
Do you know the structure theorem for divisible abelian groups? –  Arturo Magidin May 10 '11 at 16:55
@Tyler: @Arturo:What I have tried is to write them as a direct sum of variouus Q and some prufer groups. So I think, I can delete the same ones from two sides. –  Basil R May 10 '11 at 17:02

If $A$ is a divisible abelian group, and we let $\delta_{\infty}(A) = \dim_{\mathbb{Q}}(A/A_{\mathrm tor})$, and $\delta_p=\dim_{\mathbf{F}_p}(A[p])$, where $p$ is a prime and $A[p] = \{a\in A\mid pa = 0\}$, then two divisible groups $A$ and $B$ are isomorphic if and only if $\delta_{\infty}(A)=\delta_{\infty}(B)$ and $\delta_p(A)=\delta_p(B)$ for all primes $p$. Apply it to $G$ and $H$, using the fact that $G\oplus G \cong H\oplus H$.
As you advised, I cosidered $G$/$t$($G$) and $H$/$t$($G$) and the fact that if G⊕G≅H⊕H then we have, $G$/$t$($G$)⊕$G$/$t$($G$)=$H$/$t$($G$)⊕$H$/$t$($G$). Since each term in previous equality is a direct sum of some $Q$, so we have for example a+a=b+b where "a" indicates the number of $Q$ in direct sum(for $G$/$t$($G$)) and b is for $H$/$t$($H$). The question for ending this problem is, does a+a=b+b imply a=b here? Thanks. –  Basil R May 19 '11 at 12:35
@Basil: Yes: either both $a$ and $b$ are finite, in which case you have $2a=2b$, hence $a=b$; or else both are infinite, in which case you have $a = a+a = b+b= b$ (since for infinite cardinals, $\kappa+\kappa=\kappa$). –  Arturo Magidin May 19 '11 at 16:11