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Let $X:S_n → GL_n(\mathbb{R})$ be the defining representation of $S_n$. Prove that $\det X(\pi) = \operatorname{sgn}(\pi)$ for all $\pi \in S_n$.

attempt: I was thinking in trying to use $X(e) = I$ and then show for every $\pi \in S_n$ given $\pi = \sigma_1.....\sigma_n$ , we can have $X(\sigma_i)$ implies $detX(\sigma_i) = -1$.

For example if we are given a permutation 3214 then we can start with the identity, and swap colums or rows 1 and 3. Then this produces a negative sign from linear algebra.

I was also told I could use this $detX(\pi) = \Sigma_{\sigma \in S_n} sgn(\sigma) [a_{1\sigma(1)}.....a_{n\sigma(n)}]$.

If I prove $detX(\pi) = \Sigma_{\sigma \in S_n} sgn(\sigma) [a_{1\sigma(1)}.....a_{n\sigma(n)}]$, would I be able to conclude $detX(\pi) = sgn(\pi)$?

In addition, I don't understand. Can someone please help me? Any feedback or suggestion would really help. Thank you!

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  • $\begingroup$ I don't understand your question "If I prove...". You were already told you could use $\det X(\pi) = \Sigma_{\sigma \in S_n} \text{sgn}(\sigma) [a_{1\sigma(1)}\cdots a_{n\sigma(n)}]$. That means you do not need to prove it! All the effort should be in getting from there to $\det X(\pi) = \text{sgn}(\pi)$. $\endgroup$ – Erick Wong Feb 26 '16 at 3:56
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For any $\pi \in S_n$, we have $\operatorname{sgn}(\pi) = (-1)^k$, where $k$ is the number of transpositions that appear in a decomposition of $\pi$. Thus, since the determinant is multiplicative, it suffices to show that $\det X(\tau) = -1$ for any transposition $\tau$, which you can easily do.

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  • $\begingroup$ So use $det(\pi_1)det(\pi_2) = det(\pi_1\pi_2)$ for two permutations, this would show it's a homomorphism.? So $X(\tau) $ needs to imply $detX(\tau) = -1$? Can we use the identity matrix ? $\endgroup$ – user2943 Feb 26 '16 at 3:42
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    $\begingroup$ @user2943 By definition of a representation $X$ is a homomorphism. So for any decomposition of $\pi$ into transpositions $\pi = \tau_1 \cdots \tau_k$ we have $X(\pi) = X(\tau_1) \cdots X(\tau_k)$. Thus $\det X(\pi) = \det X(\tau_1) \cdots \det X(\tau_k)$, and hence it suffices to prove that $\det X(\tau_i) = -1$. $\endgroup$ – Alex Provost Feb 26 '16 at 3:43
  • $\begingroup$ So we would have two cases, when k is even and when k is odd? When k is even we get 1. So we would only have to work when k is odd? $\endgroup$ – user2943 Feb 26 '16 at 3:50
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    $\begingroup$ @user2943 There is no need to separate it into two cases. $\endgroup$ – Alex Provost Feb 26 '16 at 3:51

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