If X is a completely regular topological space, then the abelian topological group F(X) is a (Markov) free abelian topological group on X if X is a subspace of F(X), X generates F(X) algebraically and for every continuous mapping \(\phi\) of X into any abelian topological group G there exists a continuous homomorphism \(\Phi\) of F(X) into G that agrees with \(\phi\) on X. In the non-abelian case the third author [J. Lond. Math. Soc., II. Ser. 12, 199-205 (1976; Zbl 0318.22002)] showed that the free topological group on any finite dimensional compact metrizable space can be embedded via a topological group isomorphism in the free topological group on the closed unit interval I. The authors conjecture that in the abelian case, \(F(I^ 2)\) cannot be so embedded in F(I), noting that although F((0,1)) can be embedded in F(I), the embedding is not an obvious one. Here it is proved that F(I) has a closed subgroup topologically isomorphic to \(F(S^ 1)\) and in general, that \(F(S^ n)\) and \(F((S^ 1)^ n)\) embed in \(F(I^ n)\).