It seems to me that the proof of the first structure theorem for finitely generated torsion modules (over principal ideal domains) assumes that if there exists some decomposition then it must be an invariant factor one. It doesn´t seem so obvious so im trying to prove the following.
Let $M$ be a finitely generated module over a principal ideal domain. If $M=M_1\oplus\dots\oplus M_k$ is the direct sum of cyclic modules with respective orders/exponents $r_1,\dots,r_k$ (non of which are units) then there is some permutation $\sigma$ of $\{1,\dots,k\}$ such that $r_{\sigma(k)}|r_{\sigma(k-1)}|\dots|r_{\sigma(2)}|r_{\sigma(1)}$ and $r_{\sigma(1)}=\exp(M)$.
Im trying this:
Let $r:=\exp(M)$ and $M_i:=R\langle a_i\rangle$ with $|a_i|=r_i$. There exists $0\ne a\in M$ st $|a|=r$ and also $r_i|r$ for $i=1,\dots,k$. Now if $r$ is not an associate of any $r_i$ and $a\in M$ is such that $|a|=r$ then ... contradiction ... then we can take $r=r_j$, $\sigma(1)=j$ and continue in a similar manner with the remaining $r_i$
¿How can a contradiction be derived? Ive tried some things with the pid thing, or that $a$ cannot be generated by ${a_1,\dots,a_k}$ but doesn't work. Thanks
This is false: not every decomposition as a direct sum of cyclic modules is an invariant factor decomposition. For instance, for $R=\mathbb{Z}$, $M=\mathbb{Z}/(6)$ has a decomposition $M\cong\mathbb{Z}/(2)\oplus\mathbb{Z}/(3)$ but $2$ and $3$ are not the invariant factors of $M$.