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?
 A: 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.
