# Definition of an irreducible subgroup, and how to tell if any subgroup of $\mathrm{GL}_{n}(p)$ is irreducible [closed]

I'm not entirely sure what the proper definition of an "irreducible subgroup". I want an intuitive definition what an irreducible subgroup is in the simplest, most pedagogical terms as reasonably possible.

As a secondary question, I also want to know how to tell if any given subgroup of $$\mathrm{GL}_{n}(p)$$ (the complete group of linear automorphisms of an $$n$$-dimensional vector space over a finite Galois field of prime order $$p$$) is irreducible.

I don't know how to read any of the academic papers on the subject. They either have a payment lock, or have confusing notation that look like a foreign language to an average ungraduate student.

• Regarding the issue of paywalls: many university libraries have subscriptions to a lot of papers and books, that faculty and students can access by logging into the library website. Check with your department if they may have that. Dec 6, 2021 at 17:58
• In the same spirit as @R.vanDobbendeBruyn, most more recent papers can be found on the arxiv.org Dec 6, 2021 at 18:30

The notion exists both for (ordinary) groups and for algebraic groups, which are groups that are simultaneously algebraic varieties in a compatible way. Over fields that are not algebraically closed, the modern way to deal with these uses scheme-theory, which is probably beyond undergraduate mathematics. So let me focus on subgroups of the abstract group $$\operatorname{GL}_n(k)$$ instead of the algebraic group $$\operatorname{GL}_{n,k}$$.

Definition. Let $$k$$ be a field, and $$G \subseteq \operatorname{GL}_n(k)$$ a subgroup. Then $$G$$ is irreducible if the representation $$G \to \operatorname{GL}_n(k)$$ is irreducible. In other words, there is no subspace $$V \subseteq k^n$$ such that $$g V \subseteq V$$ for all $$g \in G$$, except $$V = 0$$ or $$V = k^n$$.

The other definition you often find is in terms of parabolic subgroups:

Definition. A subgroup $$P \subseteq \operatorname{GL}_n(k)$$ is parabolic if there is a flag $$0 = V_0 \subsetneq \ldots \subsetneq V_r = k^n$$ of subspaces for which $$P$$ is the stabiliser, i.e. $$gV_i = V_i$$ for all $$i$$ if and only if $$g \in P$$.

Then these notions are closely related:

Lemma. Let $$G \subseteq \operatorname{GL}_n(k)$$ be a subgroup. Then $$G$$ is irreducible if and only if the only parabolic subgroup $$P \subseteq \operatorname{GL}_n(k)$$ containing $$G$$ is $$P = \operatorname{GL}_n(k)$$.

Proof. If $$G \subseteq P$$ for a parabolic subgroup $$P \subsetneq \operatorname{GL}_n(k)$$ that is the stabiliser of some flag $$0 = V_0 \subsetneq \ldots \subsetneq V_r = k^n$$, then we must have $$r \geq 2$$ as $$P \neq \operatorname{GL}_n(k)$$. Then $$V_1$$ is a nontrivial $$G$$-subrepresentation of $$k^n$$, hence $$G$$ is not irreducible.

Conversely, if $$G$$ is not irreducible, there exists some subspace $$V \subseteq k^n$$ different from $$0$$ and $$k^n$$ such that $$gV = V$$ for all $$g \in G$$. Then the stabiliser of the flag $$0 \subsetneq V \subsetneq k^n$$ is a parabolic subgroup $$P \subsetneq \operatorname{GL}_n(k)$$ containing $$G$$. $$\square$$

Example. For concreteness' sake, you should probably compute some stabilisers. For example, the stabiliser in $$\operatorname{GL}_2(k)$$ of $$V = \operatorname{span} {1 \choose 0} \subseteq k^2$$ is the group of upper triangular matrices. In general, parabolic groups are conjugate to the block upper triangular form: $$\left(\begin{array}{c|c|c} * & * & * \\\hline 0 & * & * \\\hline 0 & 0 & *\end{array}\right),$$ where the blocks on the diagonal are square (but not necessarily the others). Hint: choose a basis $$v_1,\ldots,v_n$$ of $$k^n$$ such that $$V_i = \operatorname{span}(v_1,\ldots, v_{d_i})$$ for $$d_i = \dim V_i$$.

Remark. As for deciding irreducibility, you have to do some work. Usually you show that a representation $$V$$ is irreducible by checking that for any nonzero vector $$v \in V$$, the conjugates $$\{gv\ |\ g \in G\}$$ span $$V$$. For example, the permutation matrices $$S_n \subseteq \operatorname{GL}_n(k)$$ are never irreducible, as $$V = \operatorname{span}\Big(\begin{smallmatrix}1 \\ \vdots \\ 1\end{smallmatrix}\Big)$$ is fixed by $$S_n$$. But $$\{(x_1,\ldots,x_n)\ |\ \sum x_i = 0\} \subseteq k^n$$ is irreducible as $$S_n$$-representation, at least when $$n$$ is invertible in $$k$$ (i.e. $$\operatorname{char} k = 0$$ or $$\operatorname{char} k = p > 0$$ and $$p \nmid n$$). (Exercise!)

• The word "usual" in the first sentence is a bit loaded. I think the definition you give for subgroups of $\mathrm{GL}_n(k)$ is bog-standard. Dec 6, 2021 at 18:35
• @SeanEberhard I should have known that Serre himself (who seems to have coined the term) would be interested in finite groups, not algebraic groups. Dec 6, 2021 at 21:01
• It may be worth remarking that 'irreducible' in the context of algebraic groups qua schemes has an entirely different meaning. Dec 6, 2021 at 21:35
• @LSpice although I've never seen it used like that, as for groups this is equivalent to connected. Dec 6, 2021 at 22:34
• @LSpice I meant between representation and subgroup. Sorry, two conversations at once Dec 13, 2021 at 9:15