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Let $H\leq N\leq G$ be groups with finite indexes in each other, and with $N$ normal in $G$. An exercise often given to students is to show that ${\rm core}_G(H)$ (the largest normal subgroup of $G$ contained in $H$), has index in $G$ dividing $[G:H]!$. However, this doesn't take into account the information coming from the existence of $N$. Can the number $[G:N]$ be used to improve that upper bound?

I'm particularly interested in the case when $[G:N]=3$ and $[N:H]=2$ (so that $H$ is in fact normal in $N$, but perhaps not in $G$). In that case, I believe ${\rm core}_G(H)$ is forced to be much, much smaller than $6!$.

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We can modify the standard argument. Observe that $\mathrm{core}_G(H)$ is also the kernel of the action of $N$ on the coset space $G/H$ by left translation. We also know that the action of $N$ is trivial on the coset space $G/N$ since $N$ is a normal subgroup and the orbits of the $N$-action on $G/H$ are the fibers of the map $G/H \to G/N$.

From here, we see that $N/\mathrm{core}_G(H)$ injects into the product of the symmetric groups on each fiber of the map $G/H \to G/N$. Thus, we have the inequality: $$[G:\mathrm{core}_G(H)] = [G: N] \cdot [N : \mathrm{core}_G(H)] \leqslant [G: N] \cdot \left([N: H]!\right)^{[G: N]}$$

Note that this inequality is also sharp. Indeed, we can achieve equality above by letting $G$ be the permutation group on $[G: H]$ objects, consisting of $[G: N]$ blocks of $[N: H]$ objects, where the permutations are required to permute the blocks in a cyclic fashion. We can then let $N$ be the stabilizer of a block and $H$ be the stabilizer of an object.

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  • $\begingroup$ This was a very helpful and well-written answer. Thank you! $\endgroup$ Commented 7 hours ago

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