$\mathbb{Z}_4\times\mathbb{Z}_4 \simeq\mathbb{Z}_2\times\mathbb{Z}_8$ true or false? True or false exercise

Is $\mathbb{Z}_4\times\mathbb{Z}_4$isomorph to $\mathbb{Z}_2\times\mathbb{Z}_8$ ?

Acording to the Theorem of finitely generated abelian groups:
$\mathbb{Z}_4\times\mathbb{Z}_4\simeq\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$
$\mathbb{Z}_2\times\mathbb{Z}_8\simeq\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$
Therefore  $\mathbb{Z}_4\times\mathbb{Z}_4\simeq\mathbb{Z}_2\times\mathbb{Z}_8$
However the solution claims the statement to be false.
According to John Hughes comment:

Theorem of finitely generated abelian groups:If $G$ is an finitely generated abelian so that $G\simeq\mathbb{Z}_{{p_1}^{n_1}}\times...\mathbb{Z}_{{p_k}^{n_k}}\times \mathbb{Z}^n$ or $G\simeq\mathbb{Z}^n$
where $p_1,...,p_k$ are prime and not necessarily distinct, $n_1,...,n_k\in\mathbb{N}$ and $n>0$.

Question:
What am I doing wrong? Is the solution wrong?
Thanks in advance!
 A: Every element of $\mathbb{Z}_2\times{}\mathbb{Z}_2\times{}\mathbb{Z}_2\times{}\mathbb{Z}_2$ has order 2, but this is not true for $\mathbb{Z}_4\times{}\mathbb{Z}_4$. So the two groups aren't isomorphic.
A: I think you already make an implicit mistake when replacing $\mathbb Z_4$ with $\mathbb Z_2\times\mathbb Z_2$. Those two groups are not isomorphic, they just have the same number of elements.
It is true that $\mathbb Z_p\times\mathbb Z_q $ is isomorphic with $\mathbb Z_{pq} $ if $p $ and $q $ are coprime (Chinese remainder theorem), but obviously 2 isn't coprime with 2.
A: I think you make confusion between 
1) Every abelian group of order $16$ is isomorphic to one of the following group : $$\mathbb Z_{2}\times \mathbb Z_2\times \mathbb Z_2\times \mathbb Z_2,\quad \mathbb Z_2\times\mathbb Z_8 ,\quad \mathbb Z_4\times \mathbb Z_4,\quad \mathbb Z_{16},$$
and
2) All abelian group of order 16 are isomorphic.
The last proposition is of course wrong, and it's the first one that is true. As you can see, none of groups of 1) can be isomorphic since for example $\mathbb Z_{16}$ has an element of order $16$, whereas $\mathbb Z_2\times \mathbb Z_8$ has none element of order $16$. 
A: The theorem you state is simply wrong - $\mathbb Z_4$ is a counterexample. In the correct version of the theorem $p_j$ is not required to be prime, it is just a power of a prime.
