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

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.

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], as desired.

• Clarification: when I posted the answer the question was to prove that the inequality was strict.
– Deif
Commented Jul 25, 2023 at 20:02
• Sorry, yes, I made a typo. Commented Jul 25, 2023 at 20:03
• @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 $H^g=H$.
– Deif
Commented Jul 25, 2023 at 21:07