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I'm supposed to show that each Complex finite-dimensional irreducible representation of an abelian group is one dimensional.

For any map $\phi: V \rightarrow V$ it holds that $\phi(\rho(g)v) = \rho(g) \phi(v)$. Also since the group $\rho(h) \rho(g) v = \rho(g) \rho(h) v$. From a previous exercise I know that $\phi = \lambda \cdot id_V$ for some $\lambda \in \mathbb{C}$. This transforms the previous equation into $\lambda \cdot id_V \cdot (\rho(g)v) = \rho(g) \lambda \cdot id_V \cdot v$ which implies that $\lambda \cdot id_V \cdot (\rho(g)v) =\lambda \cdot \rho(g) \cdot v$. Now I'm not quite sure how to bring into play that $G$ is abelian. Could someone give me a hint?

Cheers!

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    $\begingroup$ Why do we have $\phi(\rho_g(v)) = \rho_g(\phi(v))$ for an arbitrary linear map $\phi: V \to V$? $\endgroup$
    – Diglett
    Commented Sep 8, 2018 at 9:05
  • $\begingroup$ This is essentially the claim that commuting matrices have a common eigenvector. $\endgroup$
    – Kenta S
    Commented Dec 1, 2022 at 1:50

4 Answers 4

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Let us sort out things a bit.

Let $\rho: G \to \operatorname{GL}(V)$ be an irreducible representation of any group $G$.

You have seen that if $\varphi : V \to V$ commutes with all $\rho(g)$, for $g \in G$, that is $$ \varphi (\rho(g) v) = \rho(g) \varphi (v)\tag{comm} $$ for all $g \in G$ and $v \in V$, then $\varphi = \lambda \operatorname{id}_{V}$ for some $\lambda \in \mathbf{C}$.

Now if $G$ is abelian we have $\rho(x) \rho(g) = \rho(g) \rho(x)$ for all $g, x \in G$, so that $\varphi = \rho(x)$ satisfies (comm).

It follows that for any $x \in G$ there is $\lambda \in \mathbf{C}$ such that $$ \rho(x) = \lambda \operatorname{id}_{V}, $$ so all $\rho(x)$ are scalars, and then leave every subspace invariant.

Since the representation is irreducible, $V$ must have dimension $1$ then.

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    $\begingroup$ Why if $\varphi : V \to V$ commutes with all $\rho(g)$ then $\varphi = \lambda \operatorname{id}_{V}$? Can you please show it? Thanks! $\endgroup$
    – Nikita
    Commented Jan 20, 2018 at 11:17
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    $\begingroup$ @ Nikita: This is Schur's lemma. $\endgroup$
    – JSCB
    Commented Feb 5, 2019 at 10:52
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    $\begingroup$ It is better to evade Schur's lemma in this special, simple case $\endgroup$
    – John
    Commented Nov 28, 2019 at 2:21
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As an alternative approach:
Since all homomorphisms from a finite abelian group $G$ to $C^*$ are irreducible characters of $G$, and there are $|G|$ many of these, by dint of orthogonality relations, we conclude that they are all irreducible characters of $G$, and hence all irreducible representations of $G$ are of dimension $1$.
Inform me of any error. Thanks.


Edit:

I am not sure now what my argument was for the equation $\lvert\operatorname{Hom}(G,\mathbb C^*)\rvert=\lvert G\rvert$, but I think for a finite abelian group, we can prove this by writing $G$ as a direct product of cyclic groups, by means of the fundamental theorem of finitely generated abelian groups, and then showing that this equation holds for cyclic groups, and both sides of the equation are multiplicative with respect to the direct product.

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  • $\begingroup$ How do you know there are $|G|$ of them? $\endgroup$ Commented Jan 30, 2023 at 1:19
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    $\begingroup$ I do not recall exactly my thoughts at that time, but I try to provide a possible argument in an edit. $\endgroup$
    – awllower
    Commented Jan 31, 2023 at 2:56
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For some variety, here is a ring-theoretic argument.

For a finite abelian group and an algebraically closed field $k$, $R:=kG\cong k^{|G|}$ as rings by Artin-Wedderburn.

[$R\cong\prod_iM_{n_i}(D_i)$ is a finite product of matrix rings over division rings, $D_i\cong\operatorname{End}_R(V_i)$ a division ring (finite dimensional over $k$), $V_i$ runs over the isomorphism classes of the simple left $R$-modules, $n_i$ their multiplicity in $R$ as a left module over itself.]

Hence the dimensions of the simple left $R$-modules must all be one [since $R$ is commutative, $n_i=1$, and $D_i$ is a field because $k$ is algebraically closed, $D_i=k$].

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  • $\begingroup$ I'm not very familiar with commutative algebra, I just want to ask, why does R being commutative implies that $n_i =1$ in the decomposition? Thank you very much. $\endgroup$
    – Khoa ta
    Commented Sep 12, 2020 at 18:30
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    $\begingroup$ @Khoata The ring of $n\times n$ matrices (say with entries from ring with unit) is always noncommutative. $\endgroup$
    – yoyo
    Commented Sep 12, 2020 at 18:46
  • $\begingroup$ thank you so much. Silly me!! I don’t know why I missed that, I kept thinking there is some deeper reasoning from commutative algebra about this. $\endgroup$
    – Khoa ta
    Commented Sep 12, 2020 at 21:03
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We could go a bit heavier on linear algebra and lighter in representation theory.

Let $\rho: G\rightarrow GL(V)$ be a representation. Clearly $\rho(g)^{|G|}=I$ and so $\rho(g)$ is diagonalizable.

Furthermore, $\{\rho(g)| g\in G\}$ commute, so they are simultaneoulsy diagonalizable and there is a common eigenvector $v$.

Clearly $\text{span}(v)$ is $\rho(G)$ invariant, but $\rho$ is irreducible, so we must have $ \{v\}$ a basis for $V$.

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