I'm currently reading Serre's Linear Representations of Finite Groups, and I'm kind of confused regarding the concepts of irreducibility and indecomposability.

If I'm understanding it correctly, an irreducible representation is a representation $(\rho, V)$ of a group $G$ which does not has a non-trivial subrepresentation (i.e. a representation $(\rho|_W, W)$ where $W\subseteq V$ is a $G$-stable subspace). While an indecomposable representation is a representation that is not isomorphic to any direct sum of other representations.

Wikipedia and other sources say that irreducibility implies indecomposability, which seems logical (not 100 % sure why though), while Serre says the following:

Let $\rho:G \rightarrow GL(V) $ be a linear representation of $G$. We say that it is irreducible or simple if $V$ is not $0$ and if no vector subspace of $V$ is stable under $G$, except of course $0$ and $V$. By theorem I [which says that there exists a $G$-stable complement $W^0$ of a $G$-stable subspace $W\subseteq V$], this second condition is equivalent to saying $V$ is not the direct sum of two representations.

Does this not mean that irreducibility is equivalent indecomposability, and thereby going against what Wikipedia says?


P.S. I'm getting kind of annoyed at this book for being "to concise" and skipping a lot of details. Does anyone have any other recommendations on introductory books on representation theory?

  • $\begingroup$ in the case Serre is dealing with (finite dimensional reps of finite groups over the complex numbers iirc) irreducibility and indecomposability are equivalent. In general they're not, so both sources are correct. $\endgroup$ Feb 1, 2019 at 11:25

2 Answers 2


Let $G=(\mathbb{R},+)$ and let $V=\mathbb{R}^2$. Consider $\rho\colon G\longrightarrow GL(V)$ defined by$$\rho(\lambda)(x,y)=(x+\lambda y,y).$$Then $\rho$ is a representation of $G$ which is indecomposable (the only non-trivial subspace is $\mathbb{R}\times\{0\}$) but not irreducible.

However, Serre is dealing with finite-dimensional complex representations of finite groups, and in that case, yes, every indecomposable representation is irreducible.

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    $\begingroup$ Also, Serre is in characteristic $0$. $\endgroup$ Feb 1, 2019 at 11:34
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    $\begingroup$ @Trettman You will find an answer here. $\endgroup$ Feb 1, 2019 at 11:39
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    $\begingroup$ @Trettman It is mentioned in the quote you gave, as theorem I. $\endgroup$ Feb 1, 2019 at 11:41
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    $\begingroup$ @Trettman It generalizes to continuous representations of compact groups. $\endgroup$ Feb 1, 2019 at 11:58
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    $\begingroup$ @Trettman Look at the proof. At some point, it sums over the elements in the group and divides by the order of the group. Neither of these make sense for an infinite group (though interestingly, the same idea applies using integrals for for example Lie groups). $\endgroup$ Feb 1, 2019 at 11:58

An easier setting to understand is the following example from Linear Algebra and Jordan Canonical Forms.

Consider a $2$-dimensional complex vector space and let $T:V\rightarrow V$ to be a linear map given by the following matrix in some basis $B$: \begin{bmatrix} 2 & 0 \\ 1 & 2 \\ \end{bmatrix} Then, one can use the operator $T$ to produce a representation of the polynomial algebra $\mathbb C[x]$ in the vector space $V$ by the following formula: $p\cdot v = p(T)(v)$. Since $\mathbb C[x]$ is a commutative algebra, every irreducible submodule of $V$ is $1$-dimensional, i.e., irreducible submodules correspond to susbpaces generated by eigenvectors.

Now, if we write $B = \{\epsilon_1, \epsilon_2\}$, the above matrix shows that $\mathbb C\epsilon_2$ is a proper submodule of $V$. In particular, $V$ can not be an irreducible representation. However, the representation $V$ is indecomposable, because it is impossible to find $W\subseteq V$ such that $V= \mathbb C \epsilon_2 \oplus W$, as you can directly check.

This examples shows you a representation which is indecomposable, but is not irreducible.


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