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Suppose $G$ is a finite group and $|G:H|=n$ I want to show that $|H:H\cap H^g|\leq n$ for all $g\in G$.

If I let $g\in H$ then we know $H^g \leq N_G(H)$ which then we can apply the first isomorphism theorem and conclude $HH^g \leq G$, $H\trianglelefteq HH^g$ and $H \cap H^g\trianglelefteq H$ and hence $$G/ H \geq HH^g/ H \trianglerighteq H/H\cap H^g$$

But for $g\notin H$ I have no clue. Since $g\notin H$ , I guess I can argue $H^g\leq G$ and hence $H^g\cap H \leq G$ and also $H\cap H^g \leq H$ but then these left cosets arent a group so I can't really say much. Maybe I can argue $H\cap H^g \subset H$ and hence $G/ H\cap H^g \supset H/ H\cap H^g$ and conlude it there?

Any hint would be great appreciated. I want to fill the part knowledge I'm missing to complete this.

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First notice that if $g\in H$ then $H^g=H$ and the result is trivial. Anyway, it can be done without cases.

Hint: It holds that $[H:H\cap H^g]\leq[G:H^g]=n$, with equality if and only if $HH^g=G$ (Hungerford Proposition 4.8).
Using this assert, I deduced the result as it follows.

If $[H:H\cap H^g]=n$ then $HH^g=G$. Now, if $g\in G=HH^g$ then $g=h_1g^{-1}h_2g$ for some $h_1,h_2\in H$. Then $1=h_1g^{-1}h_2$ and $g^{-1}=h_1^{-1}h_2^{-1}$. Since $g=h_2h_1\in H$ we conclude that $G=HH^g=H$, which is a contradiction since $[G:H]=n$. Thus $[H:H\cap H^g]<n$, as desired.

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    $\begingroup$ Clarification: when I posted the answer the question was to prove that the inequality was strict. $\endgroup$
    – Deif
    Commented Jul 25, 2023 at 20:02
  • $\begingroup$ Sorry, yes, I made a typo. $\endgroup$
    – Remu X
    Commented Jul 25, 2023 at 20:03
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    $\begingroup$ @RemuXu No problem, it is always better knowing a stronger result. Note that since the inequality is strict it follows directly the elementary result which states that if $|G:H|=2$ then $H\unlhd G$. Taking $g\in G$ we obtain that $[H:H\cap H^g]=1$ and hence $H^g=H$. $\endgroup$
    – Deif
    Commented Jul 25, 2023 at 21:07

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