# Structure theorem for finitely generated modules over PID applied to an abelian group of order $10$.

When we apply the structure theorem for finitely generated modules over PID to an abelian group $G$, which is a $\mathbb{Z}$-module, we have that there is a unique representation for $G$ such that $$G \cong \mathbb{Z} /d_1 \oplus \dots \oplus \mathbb{Z} /d_n$$ where the integers $d_i$ have the property that $1 < d_1 | d_2 | \dots | d_n$.

Now consider any abelian group, $G$, of order $10$, then $G$ is a $\mathbb{Z}$-module and $$G \cong \mathbb{Z} /2 \oplus \mathbb{Z} /5$$ in however in this case $2 \not \mid 5$ and so it seems that the above statement would not be true.

• The representation with invariant factors (these are the names of the $d_i$s) is simply… $\mathbf Z/10\mathbf Z$. – Bernard Aug 19 '15 at 21:28
• ... aka a cyclic group ($\mathbb{Z}$-module). – hardmath Aug 19 '15 at 22:42
The whole point of the divisibility condition is to avoid the obstruction to uniqueness provided by the chinese remainder theorem essentially. There are a couple of ways to talk about the classification theorem of finitely generated modules over a PID, and when applied to $\mathbb{Z}$ these will give you some sum of $\mathbb{Z}$ modules that are not necessarily irreducible.
In the second you want the factors in your invariant factor decomposition to be the indecomposable submodules, i.e. the minimal submodules that could occur as factors in your product decomposition, this is the primary decomposition of a finitely generated $\mathbb{Z}$ module into its component $p$-groups for each prime $p$. You accidentally did the second, when you should've used the chinese remainder theorem and written it as the first, or just as $\mathbb{Z}/10$
There is then nothing wrong with it, the representation of $G$ is unique when $d_{1}|d_{2}| \cdots |d_{n}$, but not necessarly unique if they don't.