Standard representation of $S_n$ is irreducible

Let $$S_n$$ act on an $$n$$-dimensional $$\mathbb{Q}$$-vector $$V$$ space with basis $$\{v_1,\cdots,v_n\}$$ by $$\sigma(v_i)=v_{\sigma(i)}$$. Consider subspaces $$W_1=\langle v_1+v_2+\cdots + v_n\rangle$$ and $$W_2=\langle v_i-v_j : i\neq j\rangle$$. Clearly $$W_1$$ is subspace with trivial action of $$S_n$$. My aim is to show that $$W_2$$ is irreducible subspace of $$V$$.

For this, first note that $$W_1\cap W_2=0$$ and $$W_1+W_2=V$$.

Next, suppose $$U$$ is a non-zero $$S_n$$-invariant subspace of $$W_2$$, and so take a vector $$0\neq v\in U$$. Then $$v=\lambda_1v_1 + \cdots + \lambda_n v_n$$ where not all $$\lambda_i$$ are equal. Without loss of generality, we can assume that $$\lambda_1\neq \lambda_2$$.

Then for $$g=(1,2)\in S_n$$, $$g.v\in U$$ and so $$g.v=\lambda_2v_1 + \lambda_1v_2+ \lambda_3v_3+ \cdots + \lambda_nv_n\in U$$. Subtracting $$g.v$$ from $$v$$ we get that $$v-g.v=(\lambda_1-\lambda_2)(v_1-v_2)\in U \mbox{ and } \lambda_1\neq \lambda_2.$$ This implies that $$v_1-v_2\in U$$. From this, it is easy to deduce that $$v_i-v_j\in U$$ for all $$i\neq j$$. This forces that $$U=W_2$$. q.e.d.

Is this proof correct?

I know that there are some proofs using 2-transitive action and/or character theory, but I am going by some different way, and want to see if it is correct.

• From the answers it seems you need to elaborate on the second to last sentence. – David Hill Oct 15 '18 at 15:51

I see one problem. How do you know that, for other indices $$i\neq j$$ apart from $$(i,j)=(1,2)$$, there is a nonzero $$u=\sum_{k=1}^n\lambda_k v_k$$ in $$U$$ such that $$\lambda_i\neq \lambda_j$$? You can assume without loss of generality that, for one fixed pair $$(i,j)$$ (which you took to be $$(1,2)$$), such an element $$u\in U$$ exists. But you start to lose "generality" when you assume further that other pairs $$(i,j)$$ can have this property too. To clarify, what prevents $$U$$ from being spanned by $$v_1-v_2$$? I would proceed in a different way.
Since $$v_1-v_2\in U$$, as you correctly deduced, it follows that $$(2\;j)\cdot(v_1-v_2)=v_1-v_j\in U$$ for $$j=3,4,5,\ldots,n$$. That is, $$v_1-v_2$$, $$v_1-v_3$$, $$\ldots$$, $$v_1-v_n$$ are in $$U$$, but then the span of these elements are precisely $$W_2$$.
• The vectors in $W_1$ are precisely the vectors of the form $\lambda v_1+\lambda v_2 + \cdots + \lambda v_n$; so a non-zero vector in the complement of $W_1$ must have different coefficients of some $v_i$ and $v_j$. We can assume it to be $v_1$ and $v_2$, with no loss of generality. – Beginner Oct 15 '18 at 6:24
• @Beginner That is not the point. You can assume without loss of generality that $v_1$ and $v_2$ have different coefficients in an element $u\in U$. But if you say further that $v_3$ and $v_4$ must have different coefficients in some $u\in U$ as well, you will lose generality. – user593746 Oct 15 '18 at 8:57
There is only one minor problem. It is not correct to assert that “From this, it is easy to deduce that $$v_i-v_j\in U$$ for all $$i\neq j$$.” What you should say is that by the same argument $$v_i-v_j\in U$$ for all $$i\neq j$$. Other than that, it's fine.
• From $v_1-v_2\in U$, apply $\sigma=(2,3)$ to this vector to get that $v_2-v_3\in U$ (hence $v_1-v_3=(v_1-v_2) + (v_2-v_3)\in U$ and so on. – Beginner Oct 15 '18 at 6:26