While looking at the french wikipedia page for the finite abelian groups structure theorem (that is, every finite abelian group $G$ is isomorphic to a group of the form $\mathbb{Z}/a_1\mathbb{Z} \times \dotsb \times \mathbb{Z}/a_n\mathbb{Z}$, with $a_1 \mid \dotsb \mid a_n$), which can be found here for french-speaking people, I found written that there are short proofs of the theorem relying on representation theory.

But I have not found a proof of this kind anywhere; all the classical books prove it with the usual generalization to finitely generated $A$-modules with $A$ a principal ring.

Would someone know anything about such a proof ? If so, could someone give me a reference where it would be done, or write the proof here ?

Thanks in advance.

  • 1
    $\begingroup$ I was under the impression that the structure theorem for modules over a principal ideal domain is the way of proving this with representation theory (representation theory of the principal ideal domain $\mathbb{Z}$). $\endgroup$ – Joppy Nov 2 '17 at 20:12

What confuses me is that this wikipedia page distinguishes between a proof using representation theory of finite groups, and another(?) one using characters, but here's what I can think of:

From character theory, I'll use in particular the following:

Fact. Let $G$ be a finite abelian group and let $g\in G$ have order $n$. Then there is a linear character $\lambda\colon G\to \mathbb{C}^*$ which sends $g$ to a primitive $n$th root of unity.
Proof. Sending $g$ to $e^{2\pi i/n}$ defines a character of $\langle g \rangle$, and there must be some irreducible (i. e., linear) character of $G$ lying over this character. (When $g$ has prime power order, one can also argue that there must be some character which is nontrivial on an element of prime order contained in $\langle g \rangle$.)

Now among all linear characters $\lambda\colon G\to \mathbb{C}^*$, choose one such that the image $\lambda(G)$ has maximal order. Since $\lambda(G)$ is a finite subgroup of $\mathbb{C}^*$, it is cyclic, say $\lambda(G)= \langle \lambda(g_1)\rangle$ for some $g_1\in G$. I claim that $G= \langle g_1 \rangle \times \ker \lambda$ and that the order of $g_1$ is the exponent of $G$. Then one can continue by induction.

For any $g\in G$ we have $\lambda(g) = \lambda(g_1)^k = \lambda(g_1^k)$ for some $k$ and thus $gg_1^{-k}\in \ker\lambda$. As $g\in G$ was arbitrary, this shows $G=\langle g_1 \rangle \ker \lambda$. By the above Fact, we see that $\lambda(g_1)$ and $g_1$ must have the same order. Thus $\langle g_1 \rangle \cap \ker \lambda = \{1\}$ and we have shown $G=\langle g_1 \rangle \times \ker \lambda$.

Finally, let $\mu$ be a linear character of $\ker \lambda$. Then $(\lambda\times \mu)$ is a character of $G$ and $(\lambda \times \mu)(G)$ contains $\lambda(g_1)$, and thus by maximality we have $(\lambda\times \mu)(G) = \langle \lambda(g_1) \rangle$. This yields $\mu(\ker\lambda) \subseteq \langle \lambda(g_1)\rangle$. So again from the Fact, the exponent of $\ker \lambda$ divides the order of $g_1$.


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