Do both versions (invariant and primary) of the Fundamental Theorem for Finitely Generated Abelian Groups hold at the same time?

For your example, $\mathbb{Z}_{20}$,

• The first decomposition gives you $\mathbb{Z}_{20}\simeq\mathbb{Z}_{2^2}\times\mathbb{Z}_5$.

• The second decomposition gives you $\mathbb{Z}_{20}\simeq\mathbb{Z}_{20}$. In other words, this one doesn't break up the abelian group at all.

• You can note that $\mathbb{Z}_{20}\not\simeq\mathbb{Z}_2\times\mathbb{Z}_{10}$ since $\mathbb{Z}_{20}$ has an element of order $20$ while the maximum order of an element of $\mathbb{Z}_2\times\mathbb{Z}_{10}$ is $10$.

The idea of the statements is that it is possible to write it in this form, not that all products of the given form are isomorphic to the given group.

A more interesting example might be given by $$ \mathbb{Z}_4\times\mathbb{Z}_6\times\mathbb{Z}_5. $$

• The first decomposition gives you $$ \mathbb{Z}_4\times\mathbb{Z}_6\times\mathbb{Z}_5\simeq \left(\mathbb{Z}_{2^2}\right)\times\left(\mathbb{Z}_2\times\mathbb{Z}_3\right)\times\mathbb{Z}_5\simeq \mathbb{Z}_2\times\mathbb{Z}_{2^2}\times\mathbb{Z}_3\times\mathbb{Z}_5. $$

• On the other hand, the second decomposition gives you $$ \mathbb{Z}_4\times\mathbb{Z}_6\times\mathbb{Z}_5\simeq\mathbb{Z}_2\times\mathbb{Z}_{2^2\cdot 3\cdot 5}=\mathbb{Z}_2\times\mathbb{Z}_{60}. $$ The group on the right combines the highest powers of each prime that appear in the fully expanded product of the first decomposition, then you work your way down by induction.

You are correct about the Chinese remainder theorem indeed $\mathbb{Z}_{20} \not\cong \mathbb{Z}_2 \times \mathbb{Z}_{10}$. The problem with applying the 2nd version in this manner is the rank uniquely determines the group. So for example $\mathbb{Z}_{20}$ is rank 1, that is it already is written in that form (trivially)