Timeline for Is the free product $\mathbb{Z}*\mathbb{Z}/n\mathbb{Z}$ linear over $\mathbb{Z}$?
Current License: CC BY-SA 3.0
16 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
|
|
Sep 14, 2015 at 16:39 | comment | added | Edgar Ndie | If I consider a word $p_0^{k_0}\cdots p_{n-1}^{k_{n-1}}$ of $N,$ with $k_0,\dots,k_{n-1}\in\mathbb{Z}$ I could rewrite it $p_0^{l_0}\cdots p_{n-1}^{l_{n-1}}$ with $l_0,\dots,l_{n-1}\in\{0,\dots,n-1\}.$ I can simplify it as $q^jp^{l_j}qp^{l_{j+1}}q\cdots p^{l_{n-1}}q.$ where $j$ is the smallest non negative integer such that $p_j=q^jp^{l_j}q^{-j}\neq0.$ I don't see how it is possible to write this word in terms of $p,p^q,p^{q^2},\ldots,p^{q^{n-1}}.$ | |
Sep 14, 2015 at 6:19 | comment | added | Edgar Ndie | I tried to prove this statement but I am stucked. I set for $k\in\{0,\dots,n-1\},p_k=q^kpq^{-k}=q^kpq^{n-k}.$ To prove that $N$ is free on $p,p^q,p^{q^2},\ldots,p^{q^{n-1}},$ is it sufficient to prove that any word $p_0^{k_0}\cdots p_{n-1}^{k_{n-1}}$ where $k_0,\dots,k_{n-1}\in\mathbb{Z}$ can be written in words in $p,p^q,p^{q^2},\ldots,p^{q^{n-1}}?$ | |
Sep 13, 2015 at 7:27 | comment | added | Derek Holt | Also, it's not very hard to show directly (without using the Kurosh theorem) that $N$ is free on $p,p^q,p^{q^2},\ldots,p^{q^{n-1}}$. | |
Sep 13, 2015 at 0:39 | comment | added | Benjamin Steinberg | In a free product any element of finite order is conjugate to an element of a free factor. So the kernel of the projection to Z/N has no element of finite order so is free. | |
Sep 13, 2015 at 0:36 | comment | added | Benjamin Steinberg | Induction of a faithful representation is always faithful. | |
Sep 12, 2015 at 22:34 | comment | added | Edgar Ndie | It's ok for the index of $N.$ My bad | |
Sep 12, 2015 at 22:28 | comment | added | Edgar Ndie | Is the induction of a faithful representation over $\mathbb{Z}$ always faithful? | |
Sep 12, 2015 at 22:24 | comment | added | Edgar Ndie | I have never heard of Kurosh theorem before. I read an article on Wikipedia and according to it, we have that normal closure of p say $N:=\langle q^kpq^{−k}|k∈{0,…,n−1}\rangle$ is the a product $F(X)*(*_{i\in I}A_i)*(*_{j\in J}B_j)$ where $X$ is a subset of $H$, $(A_i)_{i\in I}$ and $(B_j)_{j\in I}$ are respectively families of subgroups of $\mathbb{Z}$ and $\mathbb{Z}/n\mathbb{Z}$ But I don't see why we can get rid of the last two components to say $N$ is free and I don't see why $N$ has index n | |
Sep 12, 2015 at 22:03 | comment | added | Edgar Ndie | where can I find a detailed proof of the fact that a free group has a faithful representation over $\mathbb{Z}$ of degree $2?$ | |
Sep 12, 2015 at 12:38 | vote | accept | Edgar Ndie | ||
Sep 12, 2015 at 11:51 | comment | added | Benjamin Steinberg | I might as well make this an answer. | |
Sep 12, 2015 at 11:50 | answer | added | Benjamin Steinberg | timeline score: 14 | |
Sep 12, 2015 at 10:55 | comment | added | Derek Holt | @BenjaminSteinberg Yes that sounds good! The normal closure of $\langle p \rangle$ is a free subgroup of rank $n$ and of index $n$ in $H$. Since ${\rm GL}_2({\mathbb Z})$ contains free subgroups, we get a faithful representation of $H$ of degree $2n$. I should have thought of that myself - but I was trying to find a representation of degree $n$. | |
Sep 12, 2015 at 10:39 | comment | added | Benjamin Steinberg | Can't you use that it has a free subgroup of finite index? This subgroup has a faithful Z-linear representation and you can induce it to the whole group. Or am I missing something? | |
Sep 12, 2015 at 10:09 | history | asked | Edgar Ndie | CC BY-SA 3.0 |