Let $\Phi$ be an irreducible complex representation of the group $S_n$ and $\Phi'(\sigma)=\Phi(\sigma) \operatorname{sgn}(\sigma).$ $(\sigma \in S_n).$
Prove that
1) $\Phi'$ is representation of the group $S_n$;
2) the following three statements are equivalent
a. $\Phi'\cong \Phi$
b. The restriction of $\Phi$ to $A$ is a reducible representation
c. $\chi_{\Phi} (\sigma)=0$ as $\sigma$ is odd permutation
I can prove 1). What about 2). This is my attempt so far.
$$\Phi'\cong \Phi \Leftrightarrow $$ $$\chi_{\Phi'}=\chi_{\Phi} \Leftrightarrow $$ $$\chi_{\Phi(\sigma)}=-\chi_{\Phi(\sigma)} \text { as $\sigma$ is odd permutation} \Leftrightarrow $$ $\chi_{\Phi} (\sigma)=0$ as $\sigma$ is odd permutation. I can not prove that the statement b is equivalent to the statement a (or c).
Help me please. Sorry for my english
If $\Phi$ restricted to $A_n$ is reducible, then the associated ${\mathbb C}G$-module is a direct sum of two irreducible submodules $V_1 \oplus V_2$ interchanged by any odd permutation $\sigma$. (That follows from Clifford's Theorem, but it is not hard to see it directly, because if it split into more than two submodules then $\Phi$ itself could not be irreducible.) So, if you write the matrices of the representation using the union of bases of $V_1$ and $V_2$, you see that the matrix of $\sigma$ has blocks in the top right and bottom left, so its diagonal entries are all zero, and hence $\chi(\sigma) = 0$. That gives b implies c.
Conversely, if $\Phi$ restricted to $A_n$ is irreducible, then by Schur's Lemma, the only endomorphisms of its associated module are scalars. Since a module isomorphism from $\Phi$ to $\Phi'$ would restrict to a module endomorphism of the restriction of $\Phi$, there can be no such isomorphism, so $\Phi$ and $\Phi'$ are not equivalent, giving a implies b.