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Let $G$ be a group and $H\leq G$ be a subgroup of $G$ with index $n$ and $\mathbb{k}$ be a field with characteristic 0.

Let $V$ be a $m$-dimensional $\mathbb{k}$-vector space.

For a representation $\alpha\colon H\to \operatorname{GL}(V)$, we have the induced representation of $G$ $\beta\colon G\to \operatorname{GL}(V\otimes_\mathbb{k}\mathbb{k}^n)$.

Is it true that $\operatorname{Ker} \alpha=\operatorname{Ker} \beta$?

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up vote 1 down vote accepted

No, the kernels need not be equal. Take for example the field to be $\mathbb{C}$. In this case, the kernel of the induced representation is $$\bigcap_{g\in G}(\rm{ker}(\alpha))^g$$ which will not in general be all of the kernel of $\alpha$ (since the kernel of $\alpha$ need not be normal in $G$). Here, a superscript of $g$ is used to mean conjugation by $g$. (The proof of this is fairly easy, see for example Isaacs' Character Theory of Finite Groups lemma 5.11).

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What is $\operatorname{Ker}(\alpha)^g$? – user60572 Feb 1 '13 at 16:49
@user60572 Edited to clarify. – Tobias Kildetoft Feb 1 '13 at 16:52
In particular, isn't it true that $\operatorname{Ker}\beta\leq \operatorname{Ker}(\alpha)^1\leq H$? – user60572 Feb 1 '13 at 18:08
Sorry, I somehow reversed the inclusion in my head. Yes, the kernel of the induced representation is indeed contained in the kernel of the representation of the smaller group. – Tobias Kildetoft Feb 1 '13 at 18:09

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