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I think this question has been asked in some form on MO before, though I can't remember where or when. Anyway an extreme example for sufficiently large primes $p$ is a Tarski Monster $G$, by which I mean a simple infinite group in which every proper non-identity subgroup has order $p$. Then when $H$ is a proper subgroup of $G$, we certainly have $H \cap g^{-1}Hg = 1$ for every $g \in G \backslash H$ whenever $H$ is a proper non-identity subgroup of $G$.

There is a dichotomy however : if $p$ is sufficiently large, there are Tarski monsters in which all proper non-identity subgroups are conjugate. In such a Tarski monster $G$, we have $\{1_{G}\} = \{1_{G}\} \cup \left( G\backslash \bigcup_{g \in G} g^{-1}Hg \right)$, so the analogue of the Frobenius kernel is the identity subgroup.

On the other hand, if we have a (simple) Tarski monster $G$ in which not all non-identity proper subgroups of $G$ are conjugate ( I believe there are such groups, but I haven't double checked), then for every proper non-identity subgroup $H$ of $G$, see that the non-identity normal subset $\{1_{G} \} \cup \left( G \backslash \bigcup_{g \in G} g^{-1}Hg \right)$ is neither $G$ nor $\{1_{G} \}$, so can't be a subgroup ( if it were a subgroup it would be normal).

I think this question has been asked in some form on MO before, though I can't remember where or when. Anyway an extreme example for sufficiently large primes $p$ is a Tarski Monster $G$, by which I mean a simple infinite group in which every proper non-identity subgroup has order $p$. Then when $H$ is a proper subgroup of $G$, we certainly have $H \cap g^{-1}Hg = 1$ for every $g \in G \backslash H$ whenever $H$ is a proper non-identity subgroup of $G$.

There is a dichotomy however : if $p$ is sufficiently large, there are Tarski monsters in which all proper non-identity subgroups are conjugate. In such a Tarski monster $G$, we have $\{1_{G}\} = \{1_{G}\} \cup \left( G\backslash \bigcup_{g \in G} g^{-1}Hg \right)$, so the analogue of the Frobenius kernel is the identity subgroup.

On the other hand, if we have a (simple) Tarski monster $G$ in which not all non-identity proper subgroups of $G$ are conjugate ( I believe there are such groups, but I haven't double checked), then for every proper non-identity subgroup $H$ of $G$, we see that the non-identity normal subset $\{1_{G} \} \cup \left( G \backslash \bigcup_{g \in G} g^{-1}Hg \right)$ is neither $G$ nor $\{1_{G} \}$, so can't be a subgroup ( if it were a subgroup it would be normal).

I think this question has been asked in some form on MO before, though I can't remember where or when. Anyway an extreme example for sufficiently large primes $p$ is a Tarski Monster $G$, by which I mean a simple infinite group in which every proper non-identity subgroup has order $p$. Then when $H$ is a proper subgroup of $G$, we certainly have $H \cap g^{-1}Hg = 1$ for every $g \in G \backslash H$ whenever $H$ is a proper non-identity subgroup of $G$.

There is a dichotomy however : if $p$ is sufficiently large, there are Tarski monsters in which all proper non-identity subgroups are conjugate. In such a Tarski monster $G$, we have $\{1_{G}\} = \{1_{G}\} \cup \backslash \left( \bigcup_{g \in G} g^{-1}Hg \right)$, so the analogue of the Frobenius kernel is the identity subgroup.

On the other hand, if we have a (simple) Tarski monster $G$ in which not all non-identity proper subgroups of $G$ are conjugate ( I believe there are such groups, but I haven't double checked), then for every proper non-identity subgroup $H$ of $G$, see that the non-identity normal subset $\{1_{G} \} \cup \left( G \backslash \left( \bigcup_{g \in G} g^{-1}Hg \right) \right)$ is neither $G$ nor $\{1_{G} \}$, so can't be a subgroup ( if it were a subgroup it would be normal).

I think this question has been asked in some form on MO before, though I can't remember where or when. Anyway an extreme example for sufficiently large primes $p$ is a Tarski Monster $G$, by which I mean a simple infinite group in which every proper non-identity subgroup has order $p$. Then when $H$ is a proper subgroup of $G$, we certainly have $H \cap g^{-1}Hg = 1$ for every $g \in G \backslash H$ whenever $H$ is a proper non-identity subgroup of $G$.

There is a dichotomy however : if $p$ is sufficiently large, there are Tarski monsters in which all proper non-identity subgroups are conjugate. In such a Tarski monster $G$, we have $\{1_{G}\} = \{1_{G}\} \cup \left( G\backslash \bigcup_{g \in G} g^{-1}Hg \right)$, so the analogue of the Frobenius kernel is the identity subgroup.

On the other hand, if we have a (simple) Tarski monster $G$ in which not all non-identity proper subgroups of $G$ are conjugate ( I believe there are such groups, but I haven't double checked), then for every proper non-identity subgroup $H$ of $G$, see that the non-identity normal subset $\{1_{G} \} \cup \left( G \backslash \bigcup_{g \in G} g^{-1}Hg \right)$ is neither $G$ nor $\{1_{G} \}$, so can't be a subgroup ( if it were a subgroup it would be normal).

I think this question has been asked in some form on MO before, though I can't remember where or when. Anyway an extreme example for sufficiently large primes $p$ is a Tarski Monster $G$, by which I mean a simple infinite group in which every proper non-identity subgroup has order $p$. Then when $H$ is a proper subgroup of $G$, we certainly have $H \cap g^{-1}Hg = 1$ for every $g \in G \backslash H$ whenever $H$ is a proper non-identity subgroup of $G$.

There is a dichotomy however : if $p$ is sufficiently large, there are Tarski monsters in which all proper non-identity subgroups are conjugate. In such a Tarski monster $G$, we have $\{1\} = G \backslash \left( \bigcup_{g \in G} g^{-1}Hg \right)$, so the analogue of the Frobenius kernel is the identity subgroup.

On the other hand, if we have a (simple) Tarski monster $G$ in which not all non-identity proper subgroups of $G$ are conjugate ( I believe there are such groups, but I haven't double checked), then for every proper non-identity subgroup $H$ of $G$, see that the non-identity normal subset $G \backslash \left( \bigcup_{g \in G} g^{-1}Hg \right)$ is neither $G$ nor $\{1_{G} \}$, so can't be a subgroup ( if it were a subgroup it would be normal).

I think this question has been asked in some form on MO before, though I can't remember where or when. Anyway an extreme example for sufficiently large primes $p$ is a Tarski Monster $G$, by which I mean a simple infinite group in which every proper non-identity subgroup has order $p$. Then when $H$ is a proper subgroup of $G$, we certainly have $H \cap g^{-1}Hg = 1$ for every $g \in G \backslash H$ whenever $H$ is a proper non-identity subgroup of $G$.

There is a dichotomy however : if $p$ is sufficiently large, there are Tarski monsters in which all proper non-identity subgroups are conjugate. In such a Tarski monster $G$, we have $\{1_{G}\} = \{1_{G}\} \cup \backslash \left( \bigcup_{g \in G} g^{-1}Hg \right)$, so the analogue of the Frobenius kernel is the identity subgroup.

On the other hand, if we have a (simple) Tarski monster $G$ in which not all non-identity proper subgroups of $G$ are conjugate ( I believe there are such groups, but I haven't double checked), then for every proper non-identity subgroup $H$ of $G$, see that the non-identity normal subset $\{1_{G} \} \cup \left( G \backslash \left( \bigcup_{g \in G} g^{-1}Hg \right) \right)$ is neither $G$ nor $\{1_{G} \}$, so can't be a subgroup ( if it were a subgroup it would be normal).