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If we let $G$ be a finite group and take $N\lhd G$. Then take $\chi$ to be an irreducible character on $G$ such that we have $[\chi\downarrow_N,\mathbb{I}]_N\neq 0$. I am trying to show that $N\leq \textrm{ker}(\chi)$

Now if $N\leq \textrm{ker} (\chi)$ that means that $\chi(n)=\chi(1)$ $\forall n\in N$ which with the above condition is just trying to show that $\chi_N=e\mathbb{I}_N$

Is this just a simple application of Clifford's Theorem?

Clifford's Theorem gives us that $\chi_N=\tau_i^{e_i}$ for some irreducible $\tau_i$ that are all in the same orbit under the $G$-action.

But as the $G$-action on the trivial character is trivial; that is $^g\mathbb{I}(h)=\mathbb(g^{-1}hg)=1=\mathbb{I}(h)$ and we know by assumption that one of the $\tau_i$ must be $\mathbb{I}$ then we know that they all are and we are done?

Is what I have done above correct?

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  • $\begingroup$ Seems fine. You can also use Frobenius reciprocity: $[\chi,1_N^G] \neq 0$, but $\ker(1_N^G) = N$, and the kernel of a reducible character is the intersection of the kernels of its irreducible constituents. $\endgroup$ – Jack Schmidt May 13 '14 at 13:58

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