Trying to find answers for this year french intern agregation (Algebra exam), i found this question about a proof that a particular group representation of $\mathcal{S}_{3}$ on vector space $\mathbb{C}^{3}$ is indeed correct. We define the following arrow :
$\theta_{4} \left.: \begin{cases} \mathcal{S}_{3} & \rightarrow GL(\mathbb{C}^{3}) \\ \sigma & \mapsto g_{\sigma} \end{cases} \right. $
Where every $g_{\sigma}$ is the endomorphism defined by : $$ \forall ( x_{1}, x_{2}, x_{3}) \in \mathbb{C}^{3}, g_{\sigma} ( x_{1} , x_{2} , x_{3}) = (x_{\sigma^{-1}(1)}, x_{\sigma^{-1}(2)} , x_{\sigma^{-1}(3)}) $$
The first time I tried to prove it was indeed a linear map, I noticed that this calcul was false :
\begin{align} \forall (x_{1},x_{2},x_{3}) \in \mathbb{C}^{3} , \theta_{4} (\sigma \circ \mu) (x_{1},x_{2},x_{3}) &= g_{\sigma \circ \mu} (x_{1},x_{2},x_{3}) \\ &= (x_{(\sigma \circ \mu)^{-1}(1)},x_{(\sigma \circ \mu)^{-1}(2)},x_{(\sigma \circ \mu)^{-1}(3)}) \\ &= (x_{\mu^{-1} \circ \sigma^{-1} (1)},x_{\mu^{-1} \circ \sigma^{-1}(2)},x_{\mu^{-1} \circ \sigma^{-1}(3)}) \\ &= g_{\mu} (x_{\sigma^{-1} (1)},x_{\sigma^{-1}(2)},x_{\sigma^{-1}(3)}) \\ &= g_{\mu} \circ g_{\sigma} (x_{1},x_{2},x_{3}) \\ &= \theta_{4}(\mu) \circ \theta_{4}(\sigma) (x_{1},x_{2},x_{3}) \end{align}
It seems we can avoid this permutation between $\sigma$ and $\mu$ by introducing a $y_{i} = x_{\sigma^{-1} } (i) $ (are we allowed to do that ?), but I don't understand : where is the mistake in my calcul ? Is $\theta_{4}$ correctly defined ?
Your mistake is in the equality $$ (x_{\mu^{-1}\circ \sigma^{-1}(1)},x_{\mu^{-1}\circ \sigma^{-1}(2)},x_{\mu^{-1}\circ \sigma^{-1}(3)}) = g_\mu(x_{\sigma^{-1}(1)},x_{\sigma^{-1}(2)},x_{\sigma^{-1}(3)}). $$ You are misunderstanding the definition of $g_\mu$. Let me elaborate on the definition of $\theta_4$. A permutation $\sigma$ does not act on the indices of the components of the vector $x=(x_1,x_2,x_3)$, it acts permuting the components! Thus you have to see the position of the component, no its index. The equation $$ g_\mu(x_1,x_2,x_3) = (x_{\mu^{-1}(1)},x_{\mu^{-1}(2)},x_{\mu^{-1}(3)}) $$ means that the component $x_{\mu^{-1}(1)}$ is now the first component, that $x_{\mu^{-1}(2)}$ is now the second component and that $x_{\mu^{-1}(3)}$ is now the third component. Thus if we write $y=(y_1,y_2,y_3)=(x_{\mu^{-1}(1)},x_{\mu^{-1}(2)},x_{\mu^{-1}(3)})$ you get the correct formula: $$ g_\sigma(g_\mu(x_1,x_2,x_3))=g_\sigma(y)=(y_{\sigma^{-1}(1)},y_{\sigma^{-1}(2)},y_{\sigma^{-1}(3)}) $$ and since $y_i=x_{\mu^{-1}(i)}$ this implies $y_{\sigma^{-1}(i)}=x_{\mu^{-1}\circ \sigma^{-1}(i)}$, that is $$ g_\sigma g_\mu(x)=g_{\sigma\mu}(x). $$
Another way to look at this is to see how a permutation $\sigma$ acts on a basis vector. Take the standard basis of $\mathbb{C}^3$: $$ e_1=(1,0,0), \quad e_2=(0,1,0), \quad e_3=(0,0,1). $$ They can be described by $e_j=(\delta_{1j},\delta_{2j},\delta_{3j})$, where $\delta_{ij}$ is the Kronecker symbol. Then for $\sigma\in S_3$ we have $$ g_\sigma e_j = (\delta_{\sigma^{-1}(1)j},\delta_{\sigma^{-1}(2)j},\delta_{\sigma^{-1}(3)j}) = (\delta_{1\sigma(j)},\delta_{2\sigma(j)},\delta_{3\sigma(j)}) = e_{\sigma(j)}, $$ and consequently, by linearity you have that $$ g_\sigma(x_1,x_2,x_3) = s_\sigma(x_1e_1+x_2e_2+x_3e_3) = x_1e_{\sigma(1)}+x_2e_{\sigma(2)}+x_3e_{\sigma(3)}. $$ Thus you obtain another proof of the above formula: $$ g_\sigma g_\mu(x_1,x_2,x_3) = g_\sigma(x_1e_{\mu(1)}+x_2e_{\mu(2)}+x_3e_{\mu(3)}) = x_1 e_{\sigma\mu(1)} + x_2 e_{\sigma\mu(2)} + x_3 e_{\sigma\mu(3)} = g_{\sigma\mu}(x_1,x_2,x_3). $$
Hope it helps!