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I already heared a lot about: irreducible things are the building blocks everywhere: In groups: the building block are the simple groups. In representation theory: the building blocks are the irreducible representations. In module theory: the building blocks are the simple modules. And so on and so forth.

But what does that mean? The main problem is the classification up to isomorphism problem. If I classify all simple groups/irreducible representations/simple modules, do I automatically classify all groups/representations/modules?? It makes no sense in my eyes. In the example of representation theory of finite groups with char(K) does not divide |G|, we have Maschke's theorem. So every representation is decomposable into irreducible representations. This makes a little bit sense for me, but what about the other cases where this theorem does not hold?

Thank you for your time!

  • In the case of group representations over $\mathbb{C}$, every representation is a direct sum of irreducible representations, so classifying all irreducibles does immediately classify all representations of that group. For simple groups, the relationship is more complicated. See e.g., this question. – Jair Taylor Jun 02 '20 at 19:38
  • Representations of finite groups in characteristic zero form a semisimple category. That's the keyword. This means you can decompose arbitrary representations in terms of irreducible ones. Unfortunately not all categories you might encounter are semisimple. – Abdelmalek Abdesselam Jun 02 '20 at 20:46
  • Both of your examples are similar to the case where Maschke's theorem holds. Then I can understand, that irreducible representations are the building blocks. But why should that also be true in the other cases? –  Jun 02 '20 at 21:49
  • The nicest module are semisimple, so it's the same situation for those. Even for other modules, and for groups, you have Jordan-Holder. And that's all the examples you mention. For one you didn't mention, numbers have prime factorizations (but in view of cyclic groups the fundamental theorem of arithmetic is arguably just a special case of Jordan-Holdern.) – anon Jun 02 '20 at 22:21

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Let $V$ be a finite-dimensional representation of a finite group $G$. A composition series of $V$ is a filtration $0 = V_0 \subseteq V_1 \subseteq \cdots \subseteq V_n = V$ of subspaces such that each $V_i$ is a submodule, and each quotient $V_{i} / V_{i-1}$ (called a composition factor) is a simple module. The Jordan-Hölder theorem for representations applies, which states that although there may be many composition series for $V$, the list of composition factors and their multiplicities is always the same. Therefore we get a useful invariant of $V$: which simples appear in a composition series, with what multiplicity. This might be what people mean when they say that simple modules are the "building blocks" of modules.

In general, this list does not classify $V$ up to isomorphism: there could be many non-isomorphic modules with the same composition factors. But if $V$ is semisimple (for example, when the characteristic of the field does not divide $|G|$), then that list does classify $V$ up to isomorphism. (It is easy to check that if $V \cong S_1 \oplus \cdots \oplus S_n$ where each $S_i$ is simple, then any composition series of $V$ has composition factors given by the $S_i$ in some order).

For an example where a module is not semisimple, consider the group $G = \{1, g\}$ of order $2$ acting on the vector space $V = \mathbb{F}_2\{e_1, e_2\}$ of $2$-element vectors over the finite field $\mathbb{F}_2$, where $g$ acts by switching $e_1$ and $e_2$. We can write out all of the $G$-submodules explicitly, just by checking the orbits of the four vectors $0, e_1, e_2, e_1 + e_2$: $$ \{0\}, \quad V_1 = \{0, e_1 + e_2\}, \quad V,$$ so there is a unique 1-dimensional submodule, isomorphic to the trivial module, which I've called $V_1$. Therefore a composition series for $V$ is $0 \subseteq V_1 \subseteq V$, with composition factors $V_1$ and $V / V_1$ both isomorphic to trivial modules. However $V$ is not isomorphic to a direct sum of two trivial modules since there is a unique $1$-dimensional submodule of $V$. If we had another two-dimensional module $U$ which had trivial modules as its composition factors, we would still need more information to tell whether it were isomorphic to $V$ or not.

Joppy
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  • Great answer thanks! So the phrase ''irreducible representations are the building blocks'' is just in some cases true and not in all cases, right? Like the classification of semi-simple representations is equivalent to the classification of irreducible representations. –  Jun 03 '20 at 08:24
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    @ZPlaya I think the notion of 'building blocks' always has merit, it just turns out that for non-semisimple representations there is some nontrivial 'glue' holding those blocks together. So there is more structure than just the building blocks, we need to determine how they are assembled. – Joppy Jun 03 '20 at 22:49