Abelian group, invariant and p-group

Question

For an abelian group $G$ I have shown that $$ G=Z(p^{l_1})\times...\times Z(p^{l_r})\Rightarrow G^p\cong Z(p^{l_1-1})\times...\times Z(p^{l_r-1}) \, \text{and ord} (G/G^p)=p^r. $$

A $p$-group $G$ is defined by the fact that for $p$ prime, for all $a\in G$ there exists exponents $n(a)\in \mathbb{N}$ with $\text{ord}(a)=p^{n(a)}$.

Why does it follow from the above that the number $r$ of cyclical factors is an invariant of the $p$-group $G$?

//edit: Why does it follow that the number of non-trivial factors in $G^{p^l}$ is invariant?

Answer 1

The order of $G/G^p$ is independent of any way of expressing $G$ as a product of subgroups and so $p^r$ and hence $r$ is an invariant.

Answer 2

If it was not, you would have that $G = C_{p^{a_1}}\times C_{p^{a_2}}\times ...\times C_{p^{a_k}}$, and also $G = C_{p^{b_1}}\times ... \times C_{p^{b_s}}$, with $k \ne s$. Then, you would have that the elements of order $p$ in the first presentation are exactly $p^k-1$, and in the second presentation there are $p^s-1$ such elements. Since $k \ne s$, $p^k-1 \ne p^s-1$; however, the two products of cyclic groups are two presentations of the same abelian group, and are thus isomorphic. This leads to contradiction, and so the number of cyclic factors is an invariant for presentation.