Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The permutation representation of $S_n$ is $\mathbb C^n$ with elements of $S_n$ permuting the basis vectors $\{e_1, e_2, \ldots, e_n\}$. It has a trivial subrepresentation spanned by the vector $v = \sum_i e_i$. By Maschke's theorem there is a complement subrepresentation (given by the condition $\sum_i x_i = 0$, where $x_i$ is the i'th coordinate). This last representation (the standard representation of $S_n$) is irreducible. Why is this?

If we try characters, we can get the following: denote the standard representation by $V$ (dimension n-1), and the full permutation representation by $Perm$. Then $$ Perm = V\oplus Triv, $$ where the latter is the trivial representation. This means that $\chi_V(g) = |X^g| - 1$, and $$ \langle\chi_V, \chi_V\rangle = \frac{1}{|G|}\sum(|X^g|-1)^2|C(g)| = 1+\frac{1}{|G|}\sum (|X^g|^2 - 2|X^g|)|C(g)|, $$ where $X = \{1,2,\ldots, n\}$ and $X^g$ is the set of fixed points of permutation $g\in S_n$ and $C(g)$ is the conjugacy class of $g$. The sum is taken over all different conjugacy classes of $S_n$.

Now, since I know that $V$ is irreducible, the last sum must be zero. I don't see, however, why that should be.

Perhaps there is a direct proof or irreducibility?

Thank you.

share|cite|improve this question
There is a direct proof (it is worth trying to figure out yourself; take a vector and explicitly show the subrepresentation it generates is the whole thing). The character-theoretic proof is explained here:… – Qiaochu Yuan Apr 26 '12 at 16:44
@Qiaochu: What do you mean by the subrepresentation a vector generates? – joriki Apr 26 '12 at 17:27
@joriki: I mean the intersection of all subrepresentations containing a vector. Concretely, I mean $\text{span}\{ gv : g \in G \}$. – Qiaochu Yuan Apr 26 '12 at 18:14
@Qiaochu: At first I thought that's what you meant, but if so, I don't understand your argument that "the subrepresentation it generates is the whole thing". In the above sense, each $e_i$ generates the entire permutation representation, but that doesn't show that the permutation representation is irreducible. I don't see what's different in the case of the complement of the trivial subrepresentation. – joriki Apr 26 '12 at 18:35
possible duplicate of Permutation module of $S\_n$ (and I needed to upvote some of the answers there to be able to mark this as duplicate! please consider doing some more upvoting) – Marc van Leeuwen Apr 17 '13 at 7:47

The question seems to be answered in comments, so I am posting this CW-answer, so that it does not remain unanswered.

Qiaochu posted link to this MO thread where a proof using characters can be found. He also mentioned that there is a simple solution not using characters, which is a nice exercise; so you might want to try to do the proof yourself. I have tried to write down this simple solution in an answer to another question.

share|cite|improve this answer

Another nice proof comes the following fact:

if a finite group $G$ acts on a set $X$ two-transitively, then the obvious representation of $G$ on the vector space with $X$ has a basis is irreducible,

whose proof, in turn, I will leave as an exercise.

share|cite|improve this answer

Coming late to this, but it's worth pointing out that there is an interesting proof by induction that avoids characters and whatnot, "from scratch", in Hernández-Lamoneda, L.; Juárez, R.; Sánchez-Sánchez, F. Dissection of solutions in cooperative game theory using representation techniques. Internat. J. Game Theory 35 (2007), no. 3, 395–426.

Let $\mathbb{R}^n$ be generated by $\{e_i\}_{1\leq i\leq n}$. Basically, when $n=2$ it's clear that $V$ is irreducible, so let $n>2$ and use induction. Note that the subgroup that fixes $e_n$ is (isomorphic to) $S_{n-1}$, so the permutation representation of of that subgroup (by induction) decomposes as desired. Then you look back at the standard representation under $S_n$ and note if it were reducible, its invariant subspaces would also be invariant under $S_{n-1}$ and hence decompose into the specific subspaces given by induction. But those subspaces aren't $S_n$-modules, so it's not reducible.

I have no idea whether this is original to this paper, nor whether there is something a little incomplete about that last step, which somehow seems fishy (but I prefer using characters so I don't often make such arguments). I do like that it doesn't use a lot of heavy tools, and that it uses induction.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.