# Primitive action vs. irreducible representation

Let $\rho:G\rightarrow \rm{GL}_n(K)$ be a representation of group $G$.

One knows well what do we mean by $\rho$ is irreducible.

An alternate way to say that $\rho$ is irreducible could be that $\rho(G)$ can not be (conjugate to) a subgroup of $\rm{GL}_{n_1}(K)\times \cdots \times \rm{GL}_{n_r}(K)$. Is this correct?

My question is then regarding primitive permutation action of $G$.

We say that $G$ acts primitively on $\Omega$ is there are no blocks except singleton or $\Omega$.

I was thinking in the following language:

Let $|\Omega|=n$, so that $\rho:G\rightarrow S_n$ is a homomorphism corresponding to the action.

Then $\rho$ (or action of $G$ on $\Omega$) is primitive means $\rho(G)$ can not be (conjugate to) a subgroup $S_{n_1}\times \cdots \times S_{n_r}$.

My question is that is the above interpretation of primitive action correct?

• Your characterization of irreducible is not correct. What you write corresponds to being indecomposable. – Tobias Kildetoft Dec 7 '16 at 10:25
• you may be write; I would have to take "$G$ is finite and $K=\mathbb{C}$"; then this may be ..... right? – p Groups Dec 7 '16 at 10:27

If $V$ is a representation which is decomposable as $V=\bigoplus V_i$ then $G\to\mathrm{GL}(V)$ can be tightened to a homomorphism $G\to\prod_i\mathrm{GL}(V_i)$. This is true. However, in general, "indecomposable" and "irreducible" do not coincide. They do coincide if $G$ is a finite group, the field contains all $m$th roots of unity where $m$ is the exponent of $G$, and $|G|$ is invertible.

If $X$ is a $G$-set and $\Gamma=\{X_i:i\in I\}$ is some nontrivial $G$-stable partition, it does not follow that the image of $G\to\mathrm{Perm}(X)$ lands inside $\prod \mathrm{Perm}(X_i)$. The reason is that elements $g\in G$ can move elements (in $X$) from one block of $\Gamma$ to another. Here is the minimal counterexample: suppose we have a partition $\Gamma=\{\{1,2\},\{3,4\}\}$ of $X=\{1,2,3,4\}$. Let $G$ be the subgroup of $S_4$ which stabilizes this partition. This will include $\{(),(12),(34),(12)(34)\}$ (which is the direct product of $\mathrm{Perm}(\{1,2\})$ and $\mathrm{Perm}(\{3,4\})$), but it will also include four other elements,

$$(14)(23), \quad (13)(24), \quad (1423), \quad (4132).$$

These permutations permute the blocks. For example,

$$(14)(23)\cdot \{1,2\}=\{3,4\}, \quad (14)(23)\cdot\{3,4\}=\{1,2\}.$$

In general, the stabilizer of a partition $\Gamma$ (inside $\mathrm{Perm}(X)$) will be a copy of

$$\prod_{n\ge1} \left(S_n\wr S_{m(n)} \right)$$

where $m(n)$ denotes the number of blocks of size $n$ and $\wr$ denotes the wreath product.

It's understandable what you're trying to do though: make irreducible representations and primitive group actions analogous. And the are analogous. Explaining how requires thinking with categories.

There is a category $\mathsf{Rep}(G)$ whose objects are representations of $G$ and there is also a category $\mathrm{Set}(G)$ of $G$-sets whose morphisms of $G$-equivariant functions. Both categories have terminal objects (in $\mathsf{Rep}(G)$, the trivial/zero representation, and in $\mathsf{Set}(G)$ a singleton set). In the category $\mathsf{Rep}(G)$, a representation $V$ is irreducible if and only if it has no homomorphic images other than itself and the terminal object (up to isomorphism), and similarly in $\mathsf{Set}(G)$ a $G$-set has a primitive action if and only if it has no homomorphic images other than it and the singleton set, up to isomorphism.

We can expand our analogy with a couple other examples. In $\mathsf{Grp}$, the analogous objects are simple groups, and in $\mathbb{N}$ (natural numbers as objects, with unique morphisms $n\to d$ precisely when $d$ is a divisor of $n$) the analogous objects are prime numbers.

Why is an action being primitive equivalent to no nontrivial homomorphic images? It's helpful to intimately understand the relationship between three essentially equivalent ideas: equivalence relations, partitions, and surjections. Surjections induce partitions of the domain into fibers, and partitions induce equivalence relations, and vice versa on both counts. The same essential equivalence exists between $G$-congruence relations, $G$-stable partitions, and $G$-equivariant surjections.