I have the following definitions which I want to compare:
Definition 1: Let $G$ be a group and $H$ be a subgroup. $H$ is called normalish if for any finite sequence $g_1,\dots,g_n\in G$ the intersection $\bigcap_n g_nHg_n^{-1}$ is infinite.
Definition 2: Let $G$ be a group and $H$ be a subgroup. $H$ is called recurrent if there exists a finite subset $F\subset G\setminus\{1\}$ such that $F\cap gHg^{-1}\neq\emptyset$ for all $g\in G$.
Or equivalently for Definition 2, if $G$ is discrete, for any sequence $(g_n)_n$ in $G$ there is a subsequence $(g_{n_k})_k$ such that $\bigcap_k g_{n_k}Hg_{n_k}^{-1}\neq\{1\}$.
Definition 3: Let $G$ be a group and $H$ be a subgroup. H has the $\star$-property if $\bigcap_{t\in F} tHt^{-1}\neq \{1\}$ for every finite subset $F\subset G$. (I don't know if there is a name for such subgroups)
I want to know if these definitions are related, i.e. that from one definitions follow the other. It is clear to me that Def. 2 doesn't imply Def. 1. Furthermore, it is clear that Def. 3 doesn't imply Def. 1. Could it be that Def. 1 (sometimes) implies Def. 2 and Def. 1 (sometimes) implies Def.3? (How) are definition 2 and definition 3 related?
Thank you.
To expand on YCor's comment: Of the six possible implications between these properties, only the implication 1$\implies$3 holds in general.
This can be easily seen by translating every definition into plain english:
Def. 1="Every finite intersection of conjugates of $H$ is infinite"
Def. 2="Every conjugate of $H$ contains a non-trivial (cyclic) subgroup from a given, finite list".
Def. 3="Every finite intersection of conjugates of $H$ is non-trivial".
Counterexamples:
1.$\not\Rightarrow$2. and 3.$\not\Rightarrow$2.: Consider $G=Sym_{fin}(\mathbb{Z})=\{\pi\in Sym(\mathbb{Z}) \mid \pi(k)=k \text{ for all but finitely many } k\in\mathbb{Z}\}$ and $H:=\{\pi\in G \mid \pi_{|\mathbb{N}}=id\}$ the subgroup of elements which only move negative integers. Then ${^{g_1} H}\cap\ldots\cap{^{g_n} H}=\{\pi\in G \mid \pi_{|g_1\mathbb{N} \cup \ldots \cup g_n \mathbb{N}} = id\}$ contains the subgroup which moves only integers $<M$ for some constant $M$ and is therefore infinite. Similarly if $F\subseteq G\setminus\{1\}$ were such that $\forall g\in G: F\cap{^g H}\neq\emptyset$, then for all $g\in G$ there would exist a $f_g\in F$ and some $h\in H$ with $Fix(f_g)=Fix(ghg^{-1})=g Fix(h) \supseteq g\mathbb{N}$. Consider the following sequence $(g_n)$ of permutations $$g_n(x) := \begin{cases} n-x & -n\leq x\leq 2n \\ x &\text{otherwise}\end{cases}$$ Because $F$ is finite, there exists a subsequence $(g_{n_k})$ such that $f:=f_{g_{n_k}}$ is constant. For this particular $f$ we get $\forall k: Fix(f) \supseteq g_{n_k}\mathbb{N}\supseteq g_{n_k}[0,2n_k] = [-n_k,+n_k]$ and therefore $Fix(f)\supseteq \mathbb{Z}$ so that $f=1$ contrary to the assumption $F\subseteq G\setminus\{1\}$.
2.$\not\Rightarrow$1. and 3.$\not\Rightarrow$1.: For the converse consider $G_0=Aff(\mathbb{R})=\{x\mapsto ax+b \mid a\in\mathbb{R}^\times, b\in\mathbb{R}\}$. This is a Frobenius group with the translation group $\{x\mapsto x+b\}$ as kernel and $H_0:=\{x\mapsto ax\}$ as complement, i.e. $H_0\cap{^g H_0} = \{1\}$ for all $g\in G_0\setminus H_0$. In other words two conjugates are either equal (iff the conjugating elements lie in the same $H$-coset) or intersect trivially. Now let $G:=G_0\times X$ and $H:=H_0\times X$. Therefore $${^{g_1} H}\cap\ldots\cap{^{g_n} H} = \begin{cases} H & g_1H=\ldots=g_nH \\ \{1\}\times X &\text{otherwise}\end{cases}$$ so that both 2. and 3. hold in this case. If we choose $X$ to be non-trivial but finite, then $H$ is not normalish in $G$.
2.$\not\Rightarrow$3.: Finite groups yield many counterexamples. If $G$ is finite, then every $H\neq 1$ satisfies condition 2 because we can choose $F=G\setminus\{1\}$ itself. And $F=G$ is also finite so that 3 claims $core_G(H)=\bigcap_{g\in G} gHg^{-1} \neq 1$ which of course need not be true. Choose for example $G=S_3, H=\langle(12)\rangle$.