Basic idea of generators of a group











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Started studying group theory and I couldn't understand the following:
Given group $G = langle a,b rangle$ with binary operation marked $*$ where $a,b$ are the generators of $G$ I understand that $a,a*a,a*b,b*ain G$
but does it mean that also $a*b*ain G$?
I think that it's true.










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  • Yes, it is the minimal group which contains both $a$ and $b$. If $G$ is a group and $a,bin G$ then $abin G$ and $abain G$.
    – Yanko
    Nov 18 at 17:33















up vote
0
down vote

favorite












Started studying group theory and I couldn't understand the following:
Given group $G = langle a,b rangle$ with binary operation marked $*$ where $a,b$ are the generators of $G$ I understand that $a,a*a,a*b,b*ain G$
but does it mean that also $a*b*ain G$?
I think that it's true.










share|cite|improve this question
























  • Yes, it is the minimal group which contains both $a$ and $b$. If $G$ is a group and $a,bin G$ then $abin G$ and $abain G$.
    – Yanko
    Nov 18 at 17:33













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Started studying group theory and I couldn't understand the following:
Given group $G = langle a,b rangle$ with binary operation marked $*$ where $a,b$ are the generators of $G$ I understand that $a,a*a,a*b,b*ain G$
but does it mean that also $a*b*ain G$?
I think that it's true.










share|cite|improve this question















Started studying group theory and I couldn't understand the following:
Given group $G = langle a,b rangle$ with binary operation marked $*$ where $a,b$ are the generators of $G$ I understand that $a,a*a,a*b,b*ain G$
but does it mean that also $a*b*ain G$?
I think that it's true.







group-theory






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edited Nov 24 at 21:59









the_fox

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asked Nov 18 at 17:29









Yaron Scherf

1187




1187












  • Yes, it is the minimal group which contains both $a$ and $b$. If $G$ is a group and $a,bin G$ then $abin G$ and $abain G$.
    – Yanko
    Nov 18 at 17:33


















  • Yes, it is the minimal group which contains both $a$ and $b$. If $G$ is a group and $a,bin G$ then $abin G$ and $abain G$.
    – Yanko
    Nov 18 at 17:33
















Yes, it is the minimal group which contains both $a$ and $b$. If $G$ is a group and $a,bin G$ then $abin G$ and $abain G$.
– Yanko
Nov 18 at 17:33




Yes, it is the minimal group which contains both $a$ and $b$. If $G$ is a group and $a,bin G$ then $abin G$ and $abain G$.
– Yanko
Nov 18 at 17:33










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Yes, it does; it means all finite products of $a$, $b$, and their inverses, all with respect to $*$, are in $G$ (and, in fact, there are no other elements in $G$).






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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

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    up vote
    4
    down vote



    accepted










    Yes, it does; it means all finite products of $a$, $b$, and their inverses, all with respect to $*$, are in $G$ (and, in fact, there are no other elements in $G$).






    share|cite|improve this answer



























      up vote
      4
      down vote



      accepted










      Yes, it does; it means all finite products of $a$, $b$, and their inverses, all with respect to $*$, are in $G$ (and, in fact, there are no other elements in $G$).






      share|cite|improve this answer

























        up vote
        4
        down vote



        accepted







        up vote
        4
        down vote



        accepted






        Yes, it does; it means all finite products of $a$, $b$, and their inverses, all with respect to $*$, are in $G$ (and, in fact, there are no other elements in $G$).






        share|cite|improve this answer














        Yes, it does; it means all finite products of $a$, $b$, and their inverses, all with respect to $*$, are in $G$ (and, in fact, there are no other elements in $G$).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 24 at 20:59

























        answered Nov 18 at 17:33









        Shaun

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        8,100113577






























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