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Let $G$ be a finite group, $H$ a subgroup of $G$ with index $p$, where $p$ is a prime number. If $K \leq G$, then either $K \leq H$ or $[K: K \cap H]=p$.

I don't get it to show that always one of the statements is true. I think it's a short proof or is it more complicate?

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    It's not true in general. Let $K$ and $H$ be two subgroups of order $2$ in $G=S_3$. – Derek Holt Mar 03 '18 at 20:49
  • Then please have a look on this answer, there the author uses this fact. Is the answer there wrong? https://math.stackexchange.com/a/229187/529121 –  Mar 03 '18 at 20:54
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    It's true if $H$ is a normal subgroup of $G$, which it is in the application in that answer. – Derek Holt Mar 03 '18 at 20:57
  • Does $K \le G$ mean normal subgroup, or just subgroup? – Robert Lewis Mar 03 '18 at 20:57
  • Ok, in the application $A_G$ is normal. I think it is sufficient that $K$ is a subgroup. But how to go on with the proof? –  Mar 03 '18 at 21:03

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I have found this exercise in the textbook I use:

3. Prove that if $H$ is a normal subgroup of $G$ of prime index $p$ then for all $K\leq G$ either (i) $K\leq H$ or (ii) $G=HK$ and $|K:K\cap H|=p$.

Proof. Assume $K\not\leq H$. Since $H\unlhd G$, we have $HK/H\cong K/H\cap K$. Also, we have $H\leq HK\leq G$. Since $|G:H|=p$, we have $HK=H$ or $HK=G$. Suppose $HK=H$. Then $K\leq HK=H$, contradiction. So $HK=G$. Hence $G/H\cong K/H\cap K$. So $|K:H\cap K|=p$.

The exercise is from Dummit and Foote page 101.

In the link, $H=A_{G}$ is a normal subgroup of $S_{G}$, so we can still use the fact.

Delong
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