Proof for order of cyclic groups. why does it need to continue?












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In Abstract Algebra by Dummit and Foote, page 57, I have a question regarding the proof of:



Proposition 5 Let $G$ be a group, $x in G$. Let $a in mathbb{Z} - { 0 }$. If $|x| = n < infty$, $|x^{a}| = frac{n}{(n,a)}$.



Proof: Let $y =x^{a}, (n,a) = d : and : write: n = db, a=dc$. Note since $d = (n,a)$ $b$ and $c$ are relatively prime $(b,c) = 1$. To establish our prop, we must show $|y| = b$. Note that $y^{b} = x^{ab} = x^{dcb} = (x^{n})^c = 1^c = 1$



my note: I thought that the proof would end here...since by def of order, $y^{b} =1$ implies $|y| = b$ and we let $y = x^{a}$ so $|x^{a}| = b$ and since we let $n = db$, $b = frac{n}{d}$ so we have $|x^{a}| = frac{n}{d}$, which is what we needed to prove. But the proof in the book goes on.end my note



Proof continuted: By previous prop applied to $<y>$, we have that $|y|$ divides $b$. Let $k = |y|$. Then $x^{ak} = y^{k} = 1$. So by the same previous prop applied to $<x>$, $n | ak$, or using our substitutions, $db | dck$. Thus $b | ck$. Since $(b,c) = 1$. we must have that $b |k$. Since $b, k$ are positive ints that divide each other, $b=k$.



My question: Why do we need to continue after my previous note?






abstract-algebra group-theory







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  • 1




    $begingroup$
    No, $y^b =1$ only implies $|y|$ divides $b$. For example, you know that $(-1)^4 =1$, but also note that $(-1)^2 =1$, and the order of $-1$ is $2$ not $4$ [the group could be ${1, -1}$].
    $endgroup$
    – xbh
    Nov 8 '18 at 16:17


















0












$begingroup$


In Abstract Algebra by Dummit and Foote, page 57, I have a question regarding the proof of:



Proposition 5 Let $G$ be a group, $x in G$. Let $a in mathbb{Z} - { 0 }$. If $|x| = n < infty$, $|x^{a}| = frac{n}{(n,a)}$.



Proof: Let $y =x^{a}, (n,a) = d : and : write: n = db, a=dc$. Note since $d = (n,a)$ $b$ and $c$ are relatively prime $(b,c) = 1$. To establish our prop, we must show $|y| = b$. Note that $y^{b} = x^{ab} = x^{dcb} = (x^{n})^c = 1^c = 1$



my note: I thought that the proof would end here...since by def of order, $y^{b} =1$ implies $|y| = b$ and we let $y = x^{a}$ so $|x^{a}| = b$ and since we let $n = db$, $b = frac{n}{d}$ so we have $|x^{a}| = frac{n}{d}$, which is what we needed to prove. But the proof in the book goes on.end my note



Proof continuted: By previous prop applied to $<y>$, we have that $|y|$ divides $b$. Let $k = |y|$. Then $x^{ak} = y^{k} = 1$. So by the same previous prop applied to $<x>$, $n | ak$, or using our substitutions, $db | dck$. Thus $b | ck$. Since $(b,c) = 1$. we must have that $b |k$. Since $b, k$ are positive ints that divide each other, $b=k$.



My question: Why do we need to continue after my previous note?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    No, $y^b =1$ only implies $|y|$ divides $b$. For example, you know that $(-1)^4 =1$, but also note that $(-1)^2 =1$, and the order of $-1$ is $2$ not $4$ [the group could be ${1, -1}$].
    $endgroup$
    – xbh
    Nov 8 '18 at 16:17
















0












0








0





$begingroup$


In Abstract Algebra by Dummit and Foote, page 57, I have a question regarding the proof of:



Proposition 5 Let $G$ be a group, $x in G$. Let $a in mathbb{Z} - { 0 }$. If $|x| = n < infty$, $|x^{a}| = frac{n}{(n,a)}$.



Proof: Let $y =x^{a}, (n,a) = d : and : write: n = db, a=dc$. Note since $d = (n,a)$ $b$ and $c$ are relatively prime $(b,c) = 1$. To establish our prop, we must show $|y| = b$. Note that $y^{b} = x^{ab} = x^{dcb} = (x^{n})^c = 1^c = 1$



my note: I thought that the proof would end here...since by def of order, $y^{b} =1$ implies $|y| = b$ and we let $y = x^{a}$ so $|x^{a}| = b$ and since we let $n = db$, $b = frac{n}{d}$ so we have $|x^{a}| = frac{n}{d}$, which is what we needed to prove. But the proof in the book goes on.end my note



Proof continuted: By previous prop applied to $<y>$, we have that $|y|$ divides $b$. Let $k = |y|$. Then $x^{ak} = y^{k} = 1$. So by the same previous prop applied to $<x>$, $n | ak$, or using our substitutions, $db | dck$. Thus $b | ck$. Since $(b,c) = 1$. we must have that $b |k$. Since $b, k$ are positive ints that divide each other, $b=k$.



My question: Why do we need to continue after my previous note?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$




In Abstract Algebra by Dummit and Foote, page 57, I have a question regarding the proof of:



Proposition 5 Let $G$ be a group, $x in G$. Let $a in mathbb{Z} - { 0 }$. If $|x| = n < infty$, $|x^{a}| = frac{n}{(n,a)}$.



Proof: Let $y =x^{a}, (n,a) = d : and : write: n = db, a=dc$. Note since $d = (n,a)$ $b$ and $c$ are relatively prime $(b,c) = 1$. To establish our prop, we must show $|y| = b$. Note that $y^{b} = x^{ab} = x^{dcb} = (x^{n})^c = 1^c = 1$



my note: I thought that the proof would end here...since by def of order, $y^{b} =1$ implies $|y| = b$ and we let $y = x^{a}$ so $|x^{a}| = b$ and since we let $n = db$, $b = frac{n}{d}$ so we have $|x^{a}| = frac{n}{d}$, which is what we needed to prove. But the proof in the book goes on.end my note



Proof continuted: By previous prop applied to $<y>$, we have that $|y|$ divides $b$. Let $k = |y|$. Then $x^{ak} = y^{k} = 1$. So by the same previous prop applied to $<x>$, $n | ak$, or using our substitutions, $db | dck$. Thus $b | ck$. Since $(b,c) = 1$. we must have that $b |k$. Since $b, k$ are positive ints that divide each other, $b=k$.



My question: Why do we need to continue after my previous note?






abstract-algebra group-theory




abstract-algebra group-theory






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edited Dec 1 '18 at 7:46









José Carlos Santos

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asked Nov 8 '18 at 16:12









JasuusJasuus

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  • 1




    $begingroup$
    No, $y^b =1$ only implies $|y|$ divides $b$. For example, you know that $(-1)^4 =1$, but also note that $(-1)^2 =1$, and the order of $-1$ is $2$ not $4$ [the group could be ${1, -1}$].
    $endgroup$
    – xbh
    Nov 8 '18 at 16:17
















  • 1




    $begingroup$
    No, $y^b =1$ only implies $|y|$ divides $b$. For example, you know that $(-1)^4 =1$, but also note that $(-1)^2 =1$, and the order of $-1$ is $2$ not $4$ [the group could be ${1, -1}$].
    $endgroup$
    – xbh
    Nov 8 '18 at 16:17










1




1




$begingroup$
No, $y^b =1$ only implies $|y|$ divides $b$. For example, you know that $(-1)^4 =1$, but also note that $(-1)^2 =1$, and the order of $-1$ is $2$ not $4$ [the group could be ${1, -1}$].
$endgroup$
– xbh
Nov 8 '18 at 16:17






$begingroup$
No, $y^b =1$ only implies $|y|$ divides $b$. For example, you know that $(-1)^4 =1$, but also note that $(-1)^2 =1$, and the order of $-1$ is $2$ not $4$ [the group could be ${1, -1}$].
$endgroup$
– xbh
Nov 8 '18 at 16:17












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By definition, the order of $y$ is the smallest $k$ such that $y^k=e$. So, from the fact that $y^b=e$, all you can deduce is that $operatorname{ord}yleqslant b$, not that $operatorname{ord}y=b$.






share|cite|improve this answer









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    0












    $begingroup$

    You need to be careful with minimality! The order of $x$ is the minimal $n$ s.t. $x^n=1$, However, what you just found is one $n$ s.t. $x^n=1$. Hence ones till needs to work through the minimality. as an example: take $1 in mathbb{Z}/2mathbb{Z}$ (where we work additively, i.e. $x^n=n*x$)then you have $2k*1=0$, but the order is $2$ since this is the MINIMAL number such that $2*1=0$. in your proof above in this case, you might have found the "order" $16$, but you then need to reduce it to $2$






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      2 Answers
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      $begingroup$

      By definition, the order of $y$ is the smallest $k$ such that $y^k=e$. So, from the fact that $y^b=e$, all you can deduce is that $operatorname{ord}yleqslant b$, not that $operatorname{ord}y=b$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        By definition, the order of $y$ is the smallest $k$ such that $y^k=e$. So, from the fact that $y^b=e$, all you can deduce is that $operatorname{ord}yleqslant b$, not that $operatorname{ord}y=b$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          By definition, the order of $y$ is the smallest $k$ such that $y^k=e$. So, from the fact that $y^b=e$, all you can deduce is that $operatorname{ord}yleqslant b$, not that $operatorname{ord}y=b$.






          share|cite|improve this answer









          $endgroup$



          By definition, the order of $y$ is the smallest $k$ such that $y^k=e$. So, from the fact that $y^b=e$, all you can deduce is that $operatorname{ord}yleqslant b$, not that $operatorname{ord}y=b$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 8 '18 at 16:17









          José Carlos SantosJosé Carlos Santos

          153k22123226




          153k22123226























              0












              $begingroup$

              You need to be careful with minimality! The order of $x$ is the minimal $n$ s.t. $x^n=1$, However, what you just found is one $n$ s.t. $x^n=1$. Hence ones till needs to work through the minimality. as an example: take $1 in mathbb{Z}/2mathbb{Z}$ (where we work additively, i.e. $x^n=n*x$)then you have $2k*1=0$, but the order is $2$ since this is the MINIMAL number such that $2*1=0$. in your proof above in this case, you might have found the "order" $16$, but you then need to reduce it to $2$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You need to be careful with minimality! The order of $x$ is the minimal $n$ s.t. $x^n=1$, However, what you just found is one $n$ s.t. $x^n=1$. Hence ones till needs to work through the minimality. as an example: take $1 in mathbb{Z}/2mathbb{Z}$ (where we work additively, i.e. $x^n=n*x$)then you have $2k*1=0$, but the order is $2$ since this is the MINIMAL number such that $2*1=0$. in your proof above in this case, you might have found the "order" $16$, but you then need to reduce it to $2$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You need to be careful with minimality! The order of $x$ is the minimal $n$ s.t. $x^n=1$, However, what you just found is one $n$ s.t. $x^n=1$. Hence ones till needs to work through the minimality. as an example: take $1 in mathbb{Z}/2mathbb{Z}$ (where we work additively, i.e. $x^n=n*x$)then you have $2k*1=0$, but the order is $2$ since this is the MINIMAL number such that $2*1=0$. in your proof above in this case, you might have found the "order" $16$, but you then need to reduce it to $2$






                  share|cite|improve this answer









                  $endgroup$



                  You need to be careful with minimality! The order of $x$ is the minimal $n$ s.t. $x^n=1$, However, what you just found is one $n$ s.t. $x^n=1$. Hence ones till needs to work through the minimality. as an example: take $1 in mathbb{Z}/2mathbb{Z}$ (where we work additively, i.e. $x^n=n*x$)then you have $2k*1=0$, but the order is $2$ since this is the MINIMAL number such that $2*1=0$. in your proof above in this case, you might have found the "order" $16$, but you then need to reduce it to $2$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 8 '18 at 16:18









                  EnkiduEnkidu

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