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I would like to know whether this is known or not. We start with some setting: let $G$ be a non-abelian finite simple group of order $N$. Let us denote

$$G=\{1=g_1,g_2,\ldots,g_N\}.$$

For any non-trivial proper subgroup $H$ of $G$, we define

$$H_i:=g_iHg_i^{-1}$$

(obviously $H_1=H$). With this setting, here are the questions.

Question 1: Is there a non-trivial proper subgroup $H$ such that $H\cap H_i\gneq 1$ for all $i\geq2$?

Question 2: Can we expect more? - For any non-trivial proper subgroup $H$ of $G$, $H\cap H_i\gneq 1$ for all $i\geq2$.

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  • $\begingroup$ What does $H \cap H_i \ge 1$ mean? The intersection contains the identity element so it has size at least $1$. $\endgroup$ – Derek Holt Oct 19 '16 at 9:56
  • $\begingroup$ @DerekHolt I think the meaning is obvious... It means that the intersection is non-trivial... $\endgroup$ – User0829 Oct 19 '16 at 9:57
  • $\begingroup$ @DerekHolt Also, I think that the notation $N\lneq K$ is well-known in group theory - which means $N$ is a proper subgroup of $K$. $\endgroup$ – User0829 Oct 19 '16 at 9:58
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    $\begingroup$ Sorry my eyesight is not good enough! I misread the symbol as $\ge$. I would just write $N < K$ for proper containment. $\endgroup$ – Derek Holt Oct 19 '16 at 10:36
  • $\begingroup$ I think the answer to Question 1 is yes. Any subgroup $H$ with $|H|^2 >|G|$ would work, and I think there always exiists such a subgroup. This might be known already, but it would be possible to check it from known results and data. $\endgroup$ – Derek Holt Oct 19 '16 at 10:48
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Question 1: Here is an example that such H can exist (my guess would be that there always is one): Let $G = A_5$ and $H \cong A_4$. Since all its conjugates are isomorphic, $H$ and $H_i$ each are the even permutations on 4 of the 5 elements that $G$ permutes. Then for any $i$, there are $a,b,c,d,e$ such that $H = S_{\lbrace a,b,c,d \rbrace}$ and $H_i = S_{\lbrace a,b,c,e \rbrace}$ and then $(abc) \in H\cap H_i$.

Question 2: Not true for any G. Every finite simple group has a subgroup of prime order (Cauchy's Theorem) whose distinct conjugates all have pairwise trivial intersection.

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As I said in my comment, to show that the answer to Question 1 is yes, it is sufficeint to show that all finite simpleg roups $G$ have a subgroup $H$ with $|H|^2 > |G|$.

There is a complete list of the indexes of the maximal subgroups of the simple groups of Lie type on page 16 of this paper, and you can use that to verify the property. For $G=A_n$ you can take $H=A_{n-1}$, and for $G$ sporadic use theinformation you need is in the ATLAS of Finite Groups.

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