Peter

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visits member for 2 years, 3 months
seen Jun 2 '12 at 18:59

Jul
2
awarded  Curious
Jul
8
awarded  Teacher
May
29
comment Why do meta-abelian groups contain no free subgroup of rank two?
Nice. Thank you.
May
29
answered Nontrivial Splitting of a subgroup $H$ of a free product $G=A*B$
May
29
revised Nontrivial Splitting of a subgroup $H$ of a free product $G=A*B$
edited title
May
29
asked Nontrivial Splitting of a subgroup $H$ of a free product $G=A*B$
May
28
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.
May
28
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.
May
28
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.
May
28
asked Baumslag-Solitar Group $G=\langle a,t \mid tat^{-1}=a^k\rangle\cong\mathbb{Z}[1/k]\rtimes\langle t\rangle$?
May
28
awarded  Commentator
May
28
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.
May
28
accepted $U\neq H< G$ infinite groups, $p$ prime, $|G:U|=p=|G:H|$; is $|G:H\cap U|= p^2$?
May
28
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?
May
28
asked $U\neq H< G$ infinite groups, $p$ prime, $|G:U|=p=|G:H|$; is $|G:H\cap U|= p^2$?
May
25
revised When does the Commensurator of a subgroup of a group $G$ not equal $G$?
added 16 characters in body
May
25
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 $1<\text{comm}_G(H)$<G$.
May
25
comment When does the Commensurator of a subgroup of a group $G$ not equal $G$?
Nice. Thank you.
May
25
revised When does the Commensurator of a subgroup of a group $G$ not equal $G$?
deleted 10 characters in body
May
25
asked When does the Commensurator of a subgroup of a group $G$ not equal $G$?