Context and Observations:
I am currently working on the Fundamental theorem of finitely generated Abelian groups and its different uses. In particular, I started to observe that if we take any Abelian group, and we look at its order, we could quickly see to what kind of direct product of cyclic group of prime-power-order this group is isomorphic to. For example, if an Abelian group $G$ has order $8 = 2^4$, I know (by the fundamental theorem) that $G$ will be isomorphic to one of the following : $\mathbb{Z}_{2^3}$, $\mathbb{Z}_{2^2} \times \mathbb{Z}_2$, or $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$.
Now what I did in this example in order to know precisely to which of the $3$ is $G$ isomorphic to, is that I counted how many elements of each order are in $G$, and try to make this number correspond with one of the cyclic group I mentioned above.
Using this method with our example, if all the element of $G$ are exactly of order $2$ we have : $\mathbb{Z}_{2^3}$ has an element of order $8$, so they are not isomorphic. Similarly, $\mathbb{Z}_{2^4}$ has an element of order $4$, therefore $G$ is not isomorphic to $\mathbb{Z}_{2^2} \times \mathbb{Z}_2$. Hence, it has to be isomorphic to the remaining one, that is : $ G \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$.
Now, I am trying to see if there is a theorem that encapsulates this last statement: two direct product of cyclic groups of prime power orders can't have the same exactly the same number of element of any given order unless they are equal.
If anyone has any indication about this kind of result, I would love to hear about them.
Yes, this is true. In fact, it is part of the proof of the uniqueness of the decomposition into primary factors.
Explicitly, say an abelian group $G$ has order $p^n$. Then the elements of order dividing $p^i$ are precisely the elements of the kernel of the map $x\mapsto x^{p^i}$ (using multiplicative notation).
Write the group as a direct product/sum of cyclic groups of prime power order, $$G = C_{p^{a_1}}\times C_{p^{a_2}}\times\cdots\times C_{p^{a_k}}$$ where $C_m$ is the cyclic group of order $m$, and we let $1\leq a_1\leq\cdots\leq a_k$.
Then $$G^p\cong C_{p^{a_1-1}}\times C_{p^{a_2-2}}\times\cdots\times C_{p^{a_k-1}}.$$ Since $|G^p| = p^{n-k}$, and $|\mathrm{ker}\phi||G^p| = |G|$ (where $\phi$ is the map sending $x$ to $x^p$), then the kernel has order $p^k$. The number of elements of order dividing $p$ tells you the number of cyclic factors.
The number of elements of order dividing $p^2$ divided by the number of elements of dividing $p$ will tell you the number of nontrivial cycles in the decomposition of $G^2$, which is the number of cyclic factors of order at least $p^2$.
Similarly, the number of elements of order dividing $p^r$ divided by the number of elements of order dividing $p^{r-1}$ will tell you the number of cyclic factors of order at least $p^r$.
Thus, if you know the number of elements of order $1$, $p$, $p^2$, etc., it tells you exactly what the cyclic decomposition of $G$ is. Which means that the numbers completely determine the isomorphism type of $G$.