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In "A Book of Abstract Algebra" by Pinter, chapter 16.G is the following exercise:

If H is a subgroup of a group G, let X designate the set of all the left cosets of H in G. For each element $a \in G$, define $ρ_a:X \to X$ as follows: $$ρ_a(xH)=(ax)H$$ 1 Prove that each $ρ_a$ is a permutation of X

2 Prove that $h:G \to S_X$ defined by $h(a)=ρ_a$ is a homomorphism

3 Prove that the set $\{a \in H:xax^{-1} \in H$ for every $x \in G\}$, that is, the set of all elements of H whose conjugates are all in H, is the kernel of h

4 Prove that if H contains no normal subgroup of G except {e} (trivial group), then G is isomorphic to a subgroup of $S_X$

I finished proving these, but I don't understand what the theorem is really saying, and why it's an interesting result and what "sharper" means, since the original Cayley's theorem already said that every group G is isomorphic to a group of permutations, and here I proved that G is isomorphic to "another" group of permutations, but now making extra assumptions.

The wikipedia page on Cayley's theorem has the same result but there instead of making the assuption that H has no normal subgroups except for the trivial group, it just says that the case where H itself is the trivial group produces the original Cayley's theorem, but I still can't "see" it, is the fact that the quotient group of G by the set in point 3 (or the normal core of H in G) is isomorphic to a group of permutations important?.

I understand that Cayley's theorem is one "case" of this new theorem but I'd like to deeply understand what this new one means and how to intuitively think about it.

citadel
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falomir
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    It is sharper because the degree (i.e. cardinality of the set upon the group acts) is smaller if $H\ne 1$. In other words the embedding of $G$ into $S_n$ is tighter. For instance, Caryley theorem says that $S_4$ is a subgroup of $S_{24}$. But taking $H=S_3$, the sharper version states that $S_4$ is a subgroup of $S_4$. – Brauer Suzuki Mar 07 '23 at 05:29
  • I have added some emphasis so that others don't vote to close (as needing focus) by mistake. – Shaun Mar 07 '23 at 08:51
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    @BrauerSuzuki, I think you should add that as an answer. – Mees de Vries Mar 07 '23 at 08:57
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    @Mees I did so. – Brauer Suzuki Mar 07 '23 at 09:06
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    The existence of a "sharper" (than Cayley's) embedding means that the group $G$ acts faithfully on a smaller set than (the set) $G$. Perhaps this has some interesting geometrical interpretation in terms of symmetries? – citadel Mar 07 '23 at 12:22
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    For example, for $D_n$ the dihedral group of order $2n$, we get $D_n\hookrightarrow S_{m_n}$ and $D_n\not\hookrightarrow S_{m_n-1}$, where $m_n$ is the sum of the factors in the prime-power decomposition of $n$. So, e.g., the sharpest embedding of $D_6$ is into $S_{3+2}=S_5$ (which is definitely sharper than Cayley's one into $S_{12}$, but also than the other popular one into $S_{6}$). – citadel Mar 07 '23 at 13:31
  • Thanks for the intuition @Devo! Really interesting, I'll have to read more on it, do you have references for a proof of the fact that the sharpest embedding of $D_n$ is into $S_{m_n}$? – falomir Mar 07 '23 at 16:39
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    For example here, to start with: https://math.stackexchange.com/q/1599089/1092170 Just be careful, therein "$D_{2n}$" is my $D_n$ here. – citadel Mar 07 '23 at 16:45
  • Thanks :-) I hope I'll be able to understand. – falomir Mar 07 '23 at 16:52
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    A classic on the subject is Easdown & Praeger, "On minimal faithful permutation representations of finite groups", 1988. Therein Proposition 2.8 proves that result, and more. – citadel Mar 07 '23 at 16:59

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It is sharper because the degree (i.e. cardinality of the set upon the group acts) is smaller if $H\ne 1$. Cayley's theorem embeds $G$ into $S_n$ where $n=|G|$ while this exercise embeds $G$ into $S_n$ with $n=|G:H|$. For instance, Cayley says that $S_4$ is a subgroup of $S_{24}$. But taking $H=S_3$, the sharper version states that $S_4$ is a subgroup of $S_4$.

Brauer Suzuki
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