Edit: Actually, I just mixed up indices in my thinking. This was not a deep question/problem at all.
I have started reading Tate's paper about $p$-divisible group. The definition given is the following.
I am actually confused about the last statement concerning the case of ordinary abelian groups. Say, if I place myself in such case, I must have $G_0 = \{0\}$ the group of order $1$, and $G_1$ an abelian group of order $p^h$. That is in all generality a product of groups $\mathbb Z / p^{k_i}\mathbb Z$ for $1\leq i \leq r$ with $k_1+\ldots +k_r=h$. In this context, the condition $(ii)$ seems to state that the multiplication-by-$p$ homomorphism in $G_1$ (that is iteration $p$-times of the group law that is seen as addition) must be injective. This can't be true.
I assume that my interpretation of condition $(ii)$ is wrong. Could somebody please help me see how $(ii)$ is supposed to translate in the language of ordinary abelian groups ?
The axiom says $G_0$ is the kernel of the multiplication by $p^0$ map, not the multiplication by $p$ map. The multiplication by $p^0$ map is indeed injective, and $G_0$ is trivial, so there's no contradiction.