I am restricting to "finite" abelian groups, and their description in terms of invariant factors.

The problem I am considering is the basic one, but in some different way if possible.

Let $C_n$ denotes cyclic group of order $n$. Suppose we have decomposition of finite abelian group as $$ C_{d_1} \oplus C_{d_2} \oplus \cdots \oplus C_{d_k} \cong C_{e_1} \oplus C_{e_2} \oplus \cdots \oplus C_{e_l} \hskip5mm (d_i|d_{i+1} \mbox{ and } e_i|e_{i+1}) $$ Is there a direct way to proceed from this decomposition (i.e. without further factoring $d_i$'s or $e_i$'s to show that $$k=l \hskip5mm \mbox{ and } \hskip5mm d_i=e_i \hskip5mm \mbox{ for all } i?$$

The books or notes which I saw, mostly pass to prime-power factorization from the above decomposition and use the uniqueness of elementary divisors.

My question is simply, can we prove uniqueness of invariant factors for finite abelian groups without moving to prime power factorization? If possible, how? and if not, why?

  • $\begingroup$ Hint: The largest invariant factor corresponds to the exponent of the group. Cancel it (prove that you can!) and use induction. $\endgroup$
    – lhf
    Commented Oct 6, 2021 at 11:59
  • $\begingroup$ How cancellation is possible directly? I don't think it works directly even in abelian $p$-groups. $\endgroup$ Commented Oct 6, 2021 at 12:01

1 Answer 1


I am not sure what sort of proofs you want to reject. But here is how it can be proved.

Suppose $k\ne l$. Then without loss assume $k>l$; let $p$ is a prime divisor of $d_1$. Then $\{x\in G\mid x^p=1\}$ has order $p^k$ (left hand side) and at most $p^l$ right hand side. Hence $k=l$. Moreover $p$ divides $e_1$, or we still have a contradiction.

Now argue by induction on $|G|$. Consider $H=\{x^p\mid x\in G\}$ and see that $d_i/p=e_i/p$ for all $i$.


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