A finite group is abelian iff all its irreducible characters have dimension one (hence are linear). A common proof uses that the number of irreducible representations equals the number of conjugacy classes (hence $|G|$ in an abelian group) and that $\sum_{i=1}^k \chi_i(1)^2 = |G|$, which for $k = |G|$ is equivalent with $\chi_i(i) = 1$ (as $\chi_i(1) \ge 1$) for all $i = 1,\ldots, k$.

Is there a more computational proof? Going along the lines, let $\pi : G \to V$ be a representation of an abelian finite group with $\mbox{dim}(V) > 1$ and suppose $1 < W < V$, then $\pi(g)W \le W$...

Is there any way to continue this meaningful? I want to know if the fact that elements of $G$ commute could be used directly in a proof of the above stated property, without using such "higher level" theorems?

  • $\begingroup$ Proving that the irreps of a finite abelian group are all linear is easy: Consider the action of $G$ on some an eigenspace of some fixed $g_0\in G$. If you want something more computational, you can exhibit the irreps of $\mathbb{Z}_n$ explicitly, but that's kind of missing the point. In the opposite direction, note that the regular representation is faithful and decomposes as the direct sum of irreps. $\endgroup$ – anomaly Jan 17 '16 at 21:21
  • $\begingroup$ @anomaly: I see, you use the structure theorem of finite abelian groups. But what you mean in the converse case is not fully clear to me? Maybe you want to consider writting a full answer... $\endgroup$ – StefanH Jan 17 '16 at 21:25

$G$ abelian implies irreducibles have dimension 1: Let $V$ be an irreducible representation of $G$ (necessarily finite-dimensional). By Schur's lemma, every element $g \in G$ acts by multiplication by a scalar, so every subspace of $V$ is $G$-invariant. Hence we must have $\dim V = 1$.

Irreducibles have dimension $1$ implies $G$ abelian: This is the hard direction, and one way or another you need a result that tells you that $G$ has "enough irreducibles." I'll use the fact that a finite group $G$ acts jointly faithfully on its irreducible representations. Since these all have dimension $1$, every $g \in G$ acts by a scalar on all of its irreducibles, hence every $g, h \in G$ commute in all of the irreducibles. Hence they commute.

  • $\begingroup$ Is jointly faithfully a commen term? $\endgroup$ – StefanH Jan 18 '16 at 9:49
  • 1
    $\begingroup$ Not in this context, but there's a related usage in category theory. If it's unclear what I mean, I just mean that for any nontrivial element $g \in G$ there's an irreducible representation in which it doesn't act by the identity; in other words, irreducible representations separate points. (Without a result approximately this strong we wouldn't be able to rule out the possibility that e.g. $G$ has no nontrivial irreducible representations.) $\endgroup$ – Qiaochu Yuan Jan 18 '16 at 15:22

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