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Let $H$ be a subgroup of a finite group $G$ and $\rho$ a representation of $G$ such that the restriction of $\rho$ to $H$ is invariant under conjugation in $G$, in the sense that its character is invariant under conjugation in $G$. I have found elsewhere that a representation of $H$ which is invariant under conjugation in $G$ is not necessarily the restriction of a representation of $G$.

I am looking for an example where this happens with $H$ a $p$-Sylow subgroup of $G$, that is, a representation of $H$ which is $G$-invariant but does not extend to $G$.

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  • $\begingroup$ You need $H$ to be a normal subgroup for your statement to be true. $\endgroup$ Commented May 12, 2016 at 0:19
  • $\begingroup$ You are right, I have corrected my statement, thanks. $\endgroup$
    – Goa'uld
    Commented May 12, 2016 at 1:13

1 Answer 1

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Consider the symmetric group $\Sigma_5$ and its $3$-Sylow $ S = \{ 1, (1,2,3), (1,3,2) \}$. Note that $(1,2,3)$ y $(1,3,2)$ are conjugate in $\Sigma_5$.

Let $\omega$ be a primitive third root of unity. The representation of $S$ on $\mathbb{C}^2$ given by $$(1,2,3) \cdot (x,y) = (\omega x, \omega^2 y)$$ is $\Sigma_5$-invariant, since its character has the same value for $(1,2,3)$ and $(1,3,2)$, namely $-1$.

However, this representation does not extend to $\Sigma_5$. The irreducible complex representations of this group along with their characters are listed for example in Section 3.1 of the book Representation theory, a first course by W. Fulton and J. Harris. Let $U$ be the trivial $1$-dimensional complex representation of $\Sigma_5$ and $U'$ the alternating representation, which is also $1$-dimensional. The only $2$-dimensional complex representations of $\Sigma_5$ up to isomorphism are $U \oplus U$, $U \oplus U'$ and $ U' \oplus U'$, and their characters take the value $2$ on $(1,2,3)$.

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