# Relation between irreducible and completely reducible representations

While studying representations of finite groups I got confused by the the statement that any irreducible representation is at the same time a completely reducible representation. This doesn't seem to make any sense to me, since an irrep has per definition no (non-trivial) invariant subspace and therefore the carrier space can't be a direct sum of the invariant subspaces.

Furthermore I am puzzled by the statement that any representation of a finite group is equivalent to a completely reducable representation.
Let's consider for instance the symmetric group $S_3$ and it's 1-dim. representation $D_1: S_3 \rightarrow \mathbb{R}$ , $g_i \mapsto 1$. According to the just mentioned statement $D_1$ has to be equivalent to a completely reducable representation and therefore the direct sum of it's invariant subspaces (which is the set $\{1\}$) should make up the carrier space. Which is obviously not the case.

Any ideas?

I might add in view of the latest discussion about the question if not any arbitrary rep is completely reducible and hence the concept being useless:
A representation $(D,V)$ is completely reducible if:
$V$ is the direct sum of invariant subspace (true for any representation, if one considers $V$ and $0$ as invariant subspaces) and
the projection of $D$ on the invariant subspaces $D|_{V_i}$ is irreducible (which is not true in general if one just considers $V$ and $0$ as the only invariant subspaces).
Therefore the statement, that any representation is completely reducible is not true due to the second criterion.

Thanks Philipp

• It has another invariant subspace: The entire space. – Tobias Kildetoft Dec 30 '14 at 18:30
• I know this sounds strange, but it is really just a matter of definition. If $V$ is the underlying space of an irreducible representation, then the only irreducible subspace of $V$ is $V$ itself, but $V$ is the direct sum of its subspaces $\{ V \}$, so it fulfils the definition of completely reducible. – Derek Holt Dec 30 '14 at 18:30
• Thank you all for your answers. It seems that my notion of reducibility doesn't match up with yours. As far as I am concerned an unreducible representation can be defined as such a representation that has no non trivial invariant subspace. With saying "non trivial" I want to point out that $0$, $\emptyset$ and $V$ - the carrier space itself - is excluded in the definition of irreducibility. – Philipp Dec 31 '14 at 11:42
• $0$ and $V$ are excluded from the consideration in the definition of irreducible. But $V$ (and $0$) is ok, when considering complete reducibility. – Jyrki Lahtonen Dec 31 '14 at 12:07
• Pardon me if this sounds patronizing. You can think of it this way. An irreducible representation is an atom. A completely reducible element is something built from the atoms (in the most obvious way. Here "most obvious"="direct sum" - there are subtleties that I skip for now). By this definition the trivial rep corresponds to an empty set of atoms, and is thus completey reducible. An irreducible rep corresponds to a set of a single atom, an is thus completily reducible et cetera. The set of irreducible summands can have zero, $1$ or more parts. – Jyrki Lahtonen Dec 31 '14 at 12:25

So an irreducible representation (like your $1$-dimensional $S_3$ rep) is completely reducible because the carrier space itself is irreducible and so the carrier space is indeed the sum of it's irreducible subspaces because it is the sum of a single subspace: itself.