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Let $G$ be a finite group acting on a finite set, 2-transitively, and $(\rho,V)$ be the corresponding permutation representation of $G$ over $\mathbb{C}$. Then it is known that $V$ is direct sum of a trivial representation and an irreducible representation. The known proofs of this fact use character theory.

I tried to prove by "character free method" as follows:

Let $\{ e_1,\cdots, e_n\}$ be a basis of $V$ on which $G$ is acting 2-transitively. Then $V$ is direct sum of two $G$-invariant subspaces:

$V_0=\langle e_1+e_2+\cdots + e_n\rangle$, and $V_1=\langle e_1-e_2,e_2-e_3,\cdots, e_{n-1}-e_n\rangle$, and $G$ acts trivially on $V_0$.

To show that $V_1$ is irreducible, I proceed as follows:

Let $W\subseteq V_1$ be $G$-invariant subspace.

Case $1$: If $e_i-e_{i+1}\in W$ for some $i$, then by 2-transitivity of $G$, $\exists g\in G$ such that $g.e_i=e_j$ and $g.e_{i+1}=e_{j+1}$, hence $g.(e_i-e_{i+1})=e_j-e_{j+1}$, hence $W$ contains all basis vectors of $V_1$, hence $W=V_1$.

Question How can we proceed in Case $2$ for the proof?

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To prove that $V_1$ is irreducible, it is sufficient to prove that $\mathrm{End}_{\mathbb{C}G}(V_1)$ is one-dimensional. (Subproof: if $V_1$ has a non-zero proper subrepresentation $W$ then $V_1 = W \oplus C$ for some complementary representation $C$. The projection maps onto $W$ and onto $C$ corresponding to this decomposition are linearly independent.)

Let $\theta \in \mathrm{End}_{\mathbb{C}G}(V_1)$. Extend $\theta$ to an endomorphism of $V = V_0 \oplus V_1$ by setting $\theta(V_0) = 0$.

By definition $G$ acts on $\{1,2,\ldots, n\}$. Let $H = \mathrm{Stab}_{G}(1)$. Suppose that

$$\theta(e_1) = a e_1 + \sum_{i=2}^n b_i e_i$$

where $a \in \mathbb{C}$ and $b_i \in \mathbb{C}$ for each $i \in \{2,\ldots, n\}$. Since $h \cdot 1 = 1$ for each $h \in H$, and $\theta$ is a $\mathbb{C}G$-homomorphism, the vector

$$\sum_{i=2}^n b_i e_i \in V_1 $$

is $H$-invariant. But $H$ has a single orbit on $\{2,\ldots,n\}$ because $G$ is $2$-transitive. Therefore $b_i$ is constant for $i \in \{2,\ldots,n\}$. Let $b$ be the common value.

To complete the proof, pick $g_1, \ldots, g_n \in G$ such that $g_i \cdot 1 = i$ for each $i \in \{1,2,\ldots, n\}$. We have

$$ 0 = \theta\bigl( \sum_{i=1}^n e_i \bigr) = \theta\bigl( \sum_{i=1}^n g_i \cdot e_1 \bigr) = \sum_{i=1}^n g_i \cdot \theta(e_1). $$

Comparing the coefficient of $e_1$ on both sides gives $a+(n-1)b = 0$. So $\theta$ is determined by $a$ and, as claimed, $\mathrm{dim}\ \mathrm{End}_{\mathbb{C}G} (V_1) = 1$.

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There is a way of reconciliating the point of view of Beginner and that of Mark Wildon. Let $v=\sum x_i e_i$ a non-zero vector in $W$. Then $x_i \not= 0$ for some $i$ and translating $v$ by an element of $G$, one may assume that $i=1$. Let $H$ be the stabilizer of $e_1$ in $G$. By assumption it acts transitively on $\{ 2,3,...,n\}$. So $w_1=\sum_{h\in H} h.v$ has the form $\sum y_i e_i$, with $y_1\not=1$ and $y_2 =y_3 =\cdots = y_n$. By scaling $w_1$, we may assume that $y_1 =1$, so that $y_i = -1/(n-1)$ for $i\geqslant 2$. Similarly, one constructs a vector $w_2 =\sum z_i e_i$ in $W$ such that $z_2 =1$ and $z_1 =z_3 =\cdots =z_n =-1/(n-1)$. Then $e_1 -e_2 =\lambda (w_1 -w_2 )\in W$, where $\lambda (1+1/(n-1))=1$, that is $\lambda = (n-1)/n$. Using the $2$-transitive action of $G$, we have that $e_i -e_j\in W$ for all $i\not= j$ and we are done since the $e_i -e_j$ generate $W$ as a $\mathbb C$-vector space.

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  • $\begingroup$ Perhaps you can expand on why $e_1-e_2= w_1-w_2$, especially why $y_i=z_i$ for $i\neq 1,2$ $\endgroup$ – Max May 31 '18 at 12:36
  • $\begingroup$ @max I modified my post according to your remark. $\endgroup$ – Paul Broussous Jun 1 '18 at 10:54

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