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Hi I'm confused about the outline of the proof given here. Show that $G\neq\bigcup_{g\in G}gHg^{-1}$ for any $H\subsetneq G$ I don't know if I should comment there or ask a new question so I guess I ask a new one. Specifically, I understand why $gHg^{-1}=fHf^{-1} \Rightarrow gN(H)=fN(H)$ and that $[G:N(H)]\leq [G:H]$. However, I'm struggling to piece together the exact inequality leading to the desired result and the reasons behind them. I think one of the things I'm confused about is why 1 is in every coset of $N(H)$. But since $N(H)$ is a group, shouldn't they all be disjoint? From what I understand we have that $|\bigcup_\limits{g\in G}gHg^{-1}|= |H|[G:N(H)]\leq |H|[G:H] = |G|$. How exactly does $\leq$ turn into <?

Nicky Hekster
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Goob
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    See here: https://math.stackexchange.com/questions/121526/union-of-the-conjugates-of-a-proper-subgroup – Nicky Hekster Oct 11 '22 at 11:06
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    Cosets form a partition, so of course "$1$ is in every coset of $N(H)$" is a false statement (assuming $N(H)\ne G$, of course). The correct statement is that the identity is in every conjugate $gHg^{-1}$ of $H$, and the conjugates of $H$ are parametrized by cosets of $N(H)$. It is true that $\bigcup_{g\in G}gHg^{-1}$ is a union of $[G:H]$ cosets each of size $H$, but that only implies the union has size $|G|$ if it is a disjoint union, which it isn't (read the comments in the question you linked). – anon Oct 11 '22 at 11:56
  • Each of $gHg^{-1}$ has exactly $|H|$ elements, and they all contain $1$. There are exactly $[G:N_G(H)]$ distinct subgroups of the form $gHg^{-1}$. So the union is a union of exactly $[G:N_G(H)]$ subgroups, each of order exactly $|H|$. So the union has at most (in fact, strictly less than) $|H|[G:N_G(H)]$ elements; in fact, it has fewer (unless $H$ is normal), because the union is not disjoint. – Arturo Magidin Oct 11 '22 at 14:09
  • Could you please explain why $|gHg^{-1} = |H|$. Other than that I think I understand. You use $[G:N(H)]$ to understand how many sets you have in the union, not how to describe those sets. In fact you see that the actual sets are not disjoint so that's were you get the < sign, right? Thanks – Goob Oct 11 '22 at 18:29
  • Conjugation is an automorphism, hence preserves the order. So $|gHg^{-1}|=|H|$. – Dietrich Burde Oct 12 '22 at 13:16

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