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Reduced word is uniquely written

0

$begingroup$

Given a family of groups $G_i:iin I$ we may assume that $G_i$ are pairwise disjoint. Let $X=bigcup^_iin IG_i$ and let $1$ be one element disjoint from $X.$ A word on on $X$ is any sequence $(a_1, a_2, ...)$ such that $a_i ∈ X cup 1$ and for some $n ∈ N, a_i = 1$ for all $i ≥ n.$
A word $(a_1, a_2,...)$ is reduced if:

$(a)$ no $a_i ∈ X$ is the identity element in its group $G_i;$

$(b)$ for all $i ≥ 1, a_i$ and $a_i+1$ are not in the same group $G_i;$

$(c)$ $a_k = 1$ implies $a_i = 1$ for all $i ≥ k.$

In particular $1=(1,1,...,1)$ is reduced. Show that every reduced word can be written uniquely as $a_1cdot a_2 cdot ... cdot a_n=(a_1,a_2,...,a_n,1,1,...),$where $a_i in X.$

I dont even know if I understand definition, but here is my attempt:

Assume some word can't be written uniquely, so:

$a_1cdot a_2 cdot ... cdot a_n$=$b_1cdot b_2 cdot ... cdot b_n$

$b_1^-1cdot a_1cdot a_2 cdot ... cdot a_n = e_1 cdot b_2 cdot ... cdot b_n$.

$e_1$ is indentity element in $G_1$. Contradiction with $(a)$ point.

abstract-algebra group-theory formal-languages combinatorial-group-theory

share|cite|improve this question

edited Apr 2 at 11:18

user1729

17.7k64294

asked Mar 31 at 14:59

ToidiToidi

226

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  May you can tell us where you found this problem/definition?
  $endgroup$
  – Dominik
  Mar 31 at 15:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dominik This is problem given by my professor for homework. Why you asking?
  $endgroup$
  – Toidi
  Mar 31 at 15:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Firstly, your attempt does not work, because "multiplication" has not been defined on this set. This problem looks to be setting up a "normal form" for free products of groups. It also seems to be extremely simple, and not really worthy of a proof, unless I am missing something. This is because there is no concept of "equality" in the sequences $a_1cdot a_2cdotsldots cdot a_n$, apart from the one inherited via their definition. That is, by their definition, two of these sequences are equal if and only if their corresponding reduced words are equal.
  $endgroup$
  – user1729
  Apr 2 at 11:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  (In summary, the question is either missing a definition of "equality" for the sequences $a_1cdot a_2cdotsldots cdot a_n$, or is elementary.)
  $endgroup$
  – user1729
  Apr 2 at 11:14
  

  
  
  
  

add a comment | 

0

$begingroup$

Given a family of groups $G_i:iin I$ we may assume that $G_i$ are pairwise disjoint. Let $X=bigcup^_iin IG_i$ and let $1$ be one element disjoint from $X.$ A word on on $X$ is any sequence $(a_1, a_2, ...)$ such that $a_i ∈ X cup 1$ and for some $n ∈ N, a_i = 1$ for all $i ≥ n.$
A word $(a_1, a_2,...)$ is reduced if:

$(a)$ no $a_i ∈ X$ is the identity element in its group $G_i;$

$(b)$ for all $i ≥ 1, a_i$ and $a_i+1$ are not in the same group $G_i;$

$(c)$ $a_k = 1$ implies $a_i = 1$ for all $i ≥ k.$

In particular $1=(1,1,...,1)$ is reduced. Show that every reduced word can be written uniquely as $a_1cdot a_2 cdot ... cdot a_n=(a_1,a_2,...,a_n,1,1,...),$where $a_i in X.$

I dont even know if I understand definition, but here is my attempt:

Assume some word can't be written uniquely, so:

$a_1cdot a_2 cdot ... cdot a_n$=$b_1cdot b_2 cdot ... cdot b_n$

$b_1^-1cdot a_1cdot a_2 cdot ... cdot a_n = e_1 cdot b_2 cdot ... cdot b_n$.

$e_1$ is indentity element in $G_1$. Contradiction with $(a)$ point.

abstract-algebra group-theory formal-languages combinatorial-group-theory

share|cite|improve this question

edited Apr 2 at 11:18

user1729

17.7k64294

asked Mar 31 at 14:59

ToidiToidi

226

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  May you can tell us where you found this problem/definition?
  $endgroup$
  – Dominik
  Mar 31 at 15:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dominik This is problem given by my professor for homework. Why you asking?
  $endgroup$
  – Toidi
  Mar 31 at 15:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Firstly, your attempt does not work, because "multiplication" has not been defined on this set. This problem looks to be setting up a "normal form" for free products of groups. It also seems to be extremely simple, and not really worthy of a proof, unless I am missing something. This is because there is no concept of "equality" in the sequences $a_1cdot a_2cdotsldots cdot a_n$, apart from the one inherited via their definition. That is, by their definition, two of these sequences are equal if and only if their corresponding reduced words are equal.
  $endgroup$
  – user1729
  Apr 2 at 11:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  (In summary, the question is either missing a definition of "equality" for the sequences $a_1cdot a_2cdotsldots cdot a_n$, or is elementary.)
  $endgroup$
  – user1729
  Apr 2 at 11:14
  

  
  
  
  

add a comment | 

$begingroup$

Given a family of groups $G_i:iin I$ we may assume that $G_i$ are pairwise disjoint. Let $X=bigcup^_iin IG_i$ and let $1$ be one element disjoint from $X.$ A word on on $X$ is any sequence $(a_1, a_2, ...)$ such that $a_i ∈ X cup 1$ and for some $n ∈ N, a_i = 1$ for all $i ≥ n.$
A word $(a_1, a_2,...)$ is reduced if:

$(a)$ no $a_i ∈ X$ is the identity element in its group $G_i;$

$(b)$ for all $i ≥ 1, a_i$ and $a_i+1$ are not in the same group $G_i;$

$(c)$ $a_k = 1$ implies $a_i = 1$ for all $i ≥ k.$

In particular $1=(1,1,...,1)$ is reduced. Show that every reduced word can be written uniquely as $a_1cdot a_2 cdot ... cdot a_n=(a_1,a_2,...,a_n,1,1,...),$where $a_i in X.$

I dont even know if I understand definition, but here is my attempt:

Assume some word can't be written uniquely, so:

$a_1cdot a_2 cdot ... cdot a_n$=$b_1cdot b_2 cdot ... cdot b_n$

$b_1^-1cdot a_1cdot a_2 cdot ... cdot a_n = e_1 cdot b_2 cdot ... cdot b_n$.

$e_1$ is indentity element in $G_1$. Contradiction with $(a)$ point.

abstract-algebra group-theory formal-languages combinatorial-group-theory

share|cite|improve this question

edited Apr 2 at 11:18

user1729

17.7k64294

asked Mar 31 at 14:59

ToidiToidi

226

$endgroup$

share|cite|improve this question

edited Apr 2 at 11:18

user1729

17.7k64294

asked Mar 31 at 14:59

ToidiToidi

226

share|cite|improve this question

edited Apr 2 at 11:18

user1729

17.7k64294

asked Mar 31 at 14:59

ToidiToidi

226

edited Apr 2 at 11:18

user1729

17.7k64294

edited Apr 2 at 11:18

user1729

17.7k64294

asked Mar 31 at 14:59

ToidiToidi

226

asked Mar 31 at 14:59

ToidiToidi

226