# Irreducible action of a group on a set

Let $p$ be a prime. I am solving a problem and I am told that I should use that the action of $\text{SL}_2(p)$ on $\mathbb{F}_p^2$ is irreducible. But I don't know what this means?

• $\mathbb{F}^2_p$ is not just a set. it is a vector space. In this case, it usually means it must not have nontrivial invariant subspace.
– user56706
Commented Jul 22, 2014 at 17:23

I have never heard of an "irreducible action of a group on a set." However, in the case of a group $G$ acting on a vector space $V$ as linear isomorphisms, this means that there are no $G$-invariant subspaces other than $\{0\}$ and $V$.
A $G$-invariant subspace is a vector subspace $W \subset V$ so that for any $g \in G$, $g(W) \subset W$.
• Thank you for that explanation! It makes sense regarding my exercise. However, how does one prove that $SL_2(p)$ is irreducible on $\mathbb{F}_p^2$ in the first place? Commented Jul 22, 2014 at 18:58
• Since $\mathbb F_P^2$ is a $2$-dimensional vector space over $\mathbb F_p$, proper subspaces are of dimension $1$. Thus, existence of an $SL_2(p)$-invariant subspace is equivalent to existence of an eigenvector shared by all elements of $SL_2(p)$. Can you show this is impossible? Commented Jul 22, 2014 at 20:45