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.

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    $\begingroup$ 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. $\endgroup$ Dec 6, 2021 at 17:58
  • $\begingroup$ In the same spirit as @R.vanDobbendeBruyn, most more recent papers can be found on the arxiv.org $\endgroup$ Dec 6, 2021 at 18:30

1 Answer 1


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!)

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    $\begingroup$ 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. $\endgroup$ Dec 6, 2021 at 18:35
  • $\begingroup$ @SeanEberhard I should have known that Serre himself (who seems to have coined the term) would be interested in finite groups, not algebraic groups. $\endgroup$ Dec 6, 2021 at 21:01
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    $\begingroup$ It may be worth remarking that 'irreducible' in the context of algebraic groups qua schemes has an entirely different meaning. $\endgroup$
    – LSpice
    Dec 6, 2021 at 21:35
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    $\begingroup$ @LSpice although I've never seen it used like that, as for groups this is equivalent to connected. $\endgroup$ Dec 6, 2021 at 22:34
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    $\begingroup$ @LSpice I meant between representation and subgroup. Sorry, two conversations at once $\endgroup$ Dec 13, 2021 at 9:15

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