The "guess" is wrong; here is a counterexample. Take $G$ to be dihedral of order 16. Let $H$ be one of the two copies of the dihedral group of order 8 in $G$, and let $N$ be one of the two copies of the Klein fours group in $H$. Let $\pi$ be the unique irreducible character of degree $2$ of $H$, and let $\rho$ be one of the two linear constituents of the restriction $\pi_N$.

Now $\rho^H = \pi$ is irreducible, and since $\pi$ is unique, it is invariant in $G$. We argue, however, that $G \ne HT$, where $T = I_G(\rho)$. By definition, $T$ is contained in the normalizer in $G$ of $N$, and of course, $N$ is also normal in $H$. Thus if $G = HT$, it would follow that $N$ is normal in $G$. This is not the case, however. One way to see that
$N \not\kern -2pt\triangleleft\ G$ is to observe that otherwise $G/C$ would be embedded in
${\rm Aut}(G)$, where $C$ is the centralizer of $N$. This would force $|C| \ge 8$, and that would imply that $N$ is contained in an abelian subgroup of $G$ of order $8$. The only abelian subgroup of oirder $8$ in $G$, however, is cyclic, so does not contain $N$.

`\lbrace`

and`\rbrace`

instead of`\{`

and`\}`

, respectively. – José Figueroa-O'Farrill Nov 10 '11 at 1:36