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 Jul2 awarded Curious Jul8 awarded Teacher May29 comment Why do meta-abelian groups contain no free subgroup of rank two? Nice. Thank you. May29 answered Nontrivial Splitting of a subgroup $H$ of a free product $G=A*B$ May29 revised Nontrivial Splitting of a subgroup $H$ of a free product $G=A*B$ edited title May29 asked Nontrivial Splitting of a subgroup $H$ of a free product $G=A*B$ May28 comment Baumslag-Solitar Group $G=\langle a,t \mid tat^{-1}=a^k\rangle\cong\mathbb{Z}[1/k]\rtimes\langle t\rangle$? Thank you very much. May28 comment Baumslag-Solitar Group $G=\langle a,t \mid tat^{-1}=a^k\rangle\cong\mathbb{Z}[1/k]\rtimes\langle t\rangle$? Thank you very much. May28 comment Baumslag-Solitar Group $G=\langle a,t \mid tat^{-1}=a^k\rangle\cong\mathbb{Z}[1/k]\rtimes\langle t\rangle$? Yes. Thank you. May28 asked Baumslag-Solitar Group $G=\langle a,t \mid tat^{-1}=a^k\rangle\cong\mathbb{Z}[1/k]\rtimes\langle t\rangle$? May28 awarded Commentator May28 comment $U\neq H< G$ infinite groups, $p$ prime, $|G:U|=p=|G:H|$; is $|G:H\cap U|= p^2$? Thank you very much. For my field of interest, I wanted to have that the index is a power of $p$ not a multiple of $p$. But I thoughted that it won't work, too. But didn't know how to proof it. May28 accepted $U\neq H< G$ infinite groups, $p$ prime, $|G:U|=p=|G:H|$; is $|G:H\cap U|= p^2$? May28 comment $U\neq H< G$ infinite groups, $p$ prime, $|G:U|=p=|G:H|$; is $|G:H\cap U|= p^2$? Okay. But the index of the interesection divides the product of the indices of the subgroups, so it divides 9. right? May28 asked $U\neq H< G$ infinite groups, $p$ prime, $|G:U|=p=|G:H|$; is $|G:H\cap U|= p^2$? May25 revised When does the Commensurator of a subgroup of a group $G$ not equal $G$? added 16 characters in body May25 comment When does the Commensurator of a subgroup of a group $G$ not equal $G$? But then $\text{comm}_G(H)=1$, right? Since $|G:H|=\infty$, right? So I'm searching for a subgroup where $1<\text{comm}_G(H)$