Showing that $o(phi(g))|o(g)$ if $phi:G_1mapsto G_2$ is a homomorphism and $gin G_1$












2












$begingroup$


$o(g)$ denotes the order of $g$.



This is how I think I proved it:



Let $m:=o(g)$



Let $d=o(phi(g))$. $~dle m$ because $langlephi (g)rangle={phi(g),phi(g)^2=phi(g^2),...,phi(g^m)=e_{G_2}}$



If $o(phi(g))=d$ by euclidean division there exist $qinBbb N$ and $0le r<d$ such that $$m=dq+r~text{ and }~phi(g)^d=e=phi(g^d)\ e=g^m=g^{dq+r}=g^{dq}g^r implies g^r=g^{-dq}\ phi(g)^r=phi(g^r)=phi(g^{-dq})=(phi(g)^d)^{-q}=e$$
but $d>r$ is the order of $phi(g)$ so $r$ must be zero.



I'm not sure if this is all correct






group-theory proof-verification group-homomorphism







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It looks fine to me. Well done! :)
    $endgroup$
    – Shaun
    Jan 3 at 19:37










  • $begingroup$
    It's considered bad form to start a sentence with the symbols one uses in the mathematics. That's all I can say at a glance.
    $endgroup$
    – Shaun
    Jan 3 at 19:38








  • 1




    $begingroup$
    The use of $$begin{align} x &=y\ &=zend{align}$$ for $$begin{align} x &=y\ &=zend{align}$$ wouldn't go amiss.
    $endgroup$
    – Shaun
    Jan 3 at 19:41


















2












$begingroup$


$o(g)$ denotes the order of $g$.



This is how I think I proved it:



Let $m:=o(g)$



Let $d=o(phi(g))$. $~dle m$ because $langlephi (g)rangle={phi(g),phi(g)^2=phi(g^2),...,phi(g^m)=e_{G_2}}$



If $o(phi(g))=d$ by euclidean division there exist $qinBbb N$ and $0le r<d$ such that $$m=dq+r~text{ and }~phi(g)^d=e=phi(g^d)\ e=g^m=g^{dq+r}=g^{dq}g^r implies g^r=g^{-dq}\ phi(g)^r=phi(g^r)=phi(g^{-dq})=(phi(g)^d)^{-q}=e$$
but $d>r$ is the order of $phi(g)$ so $r$ must be zero.



I'm not sure if this is all correct






group-theory proof-verification group-homomorphism







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It looks fine to me. Well done! :)
    $endgroup$
    – Shaun
    Jan 3 at 19:37










  • $begingroup$
    It's considered bad form to start a sentence with the symbols one uses in the mathematics. That's all I can say at a glance.
    $endgroup$
    – Shaun
    Jan 3 at 19:38








  • 1




    $begingroup$
    The use of $$begin{align} x &=y\ &=zend{align}$$ for $$begin{align} x &=y\ &=zend{align}$$ wouldn't go amiss.
    $endgroup$
    – Shaun
    Jan 3 at 19:41
















2












2








2





$begingroup$


$o(g)$ denotes the order of $g$.



This is how I think I proved it:



Let $m:=o(g)$



Let $d=o(phi(g))$. $~dle m$ because $langlephi (g)rangle={phi(g),phi(g)^2=phi(g^2),...,phi(g^m)=e_{G_2}}$



If $o(phi(g))=d$ by euclidean division there exist $qinBbb N$ and $0le r<d$ such that $$m=dq+r~text{ and }~phi(g)^d=e=phi(g^d)\ e=g^m=g^{dq+r}=g^{dq}g^r implies g^r=g^{-dq}\ phi(g)^r=phi(g^r)=phi(g^{-dq})=(phi(g)^d)^{-q}=e$$
but $d>r$ is the order of $phi(g)$ so $r$ must be zero.



I'm not sure if this is all correct






group-theory proof-verification group-homomorphism







share|cite|improve this question











$endgroup$




$o(g)$ denotes the order of $g$.



This is how I think I proved it:



Let $m:=o(g)$



Let $d=o(phi(g))$. $~dle m$ because $langlephi (g)rangle={phi(g),phi(g)^2=phi(g^2),...,phi(g^m)=e_{G_2}}$



If $o(phi(g))=d$ by euclidean division there exist $qinBbb N$ and $0le r<d$ such that $$m=dq+r~text{ and }~phi(g)^d=e=phi(g^d)\ e=g^m=g^{dq+r}=g^{dq}g^r implies g^r=g^{-dq}\ phi(g)^r=phi(g^r)=phi(g^{-dq})=(phi(g)^d)^{-q}=e$$
but $d>r$ is the order of $phi(g)$ so $r$ must be zero.



I'm not sure if this is all correct






group-theory proof-verification group-homomorphism




group-theory proof-verification group-homomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 20:47









Shaun

10.1k113685




10.1k113685










asked Jan 3 at 19:18









John CataldoJohn Cataldo

1,1831316




1,1831316








  • 1




    $begingroup$
    It looks fine to me. Well done! :)
    $endgroup$
    – Shaun
    Jan 3 at 19:37










  • $begingroup$
    It's considered bad form to start a sentence with the symbols one uses in the mathematics. That's all I can say at a glance.
    $endgroup$
    – Shaun
    Jan 3 at 19:38








  • 1




    $begingroup$
    The use of $$begin{align} x &=y\ &=zend{align}$$ for $$begin{align} x &=y\ &=zend{align}$$ wouldn't go amiss.
    $endgroup$
    – Shaun
    Jan 3 at 19:41
















  • 1




    $begingroup$
    It looks fine to me. Well done! :)
    $endgroup$
    – Shaun
    Jan 3 at 19:37










  • $begingroup$
    It's considered bad form to start a sentence with the symbols one uses in the mathematics. That's all I can say at a glance.
    $endgroup$
    – Shaun
    Jan 3 at 19:38








  • 1




    $begingroup$
    The use of $$begin{align} x &=y\ &=zend{align}$$ for $$begin{align} x &=y\ &=zend{align}$$ wouldn't go amiss.
    $endgroup$
    – Shaun
    Jan 3 at 19:41










1




1




$begingroup$
It looks fine to me. Well done! :)
$endgroup$
– Shaun
Jan 3 at 19:37




$begingroup$
It looks fine to me. Well done! :)
$endgroup$
– Shaun
Jan 3 at 19:37












$begingroup$
It's considered bad form to start a sentence with the symbols one uses in the mathematics. That's all I can say at a glance.
$endgroup$
– Shaun
Jan 3 at 19:38






$begingroup$
It's considered bad form to start a sentence with the symbols one uses in the mathematics. That's all I can say at a glance.
$endgroup$
– Shaun
Jan 3 at 19:38






1




1




$begingroup$
The use of $$begin{align} x &=y\ &=zend{align}$$ for $$begin{align} x &=y\ &=zend{align}$$ wouldn't go amiss.
$endgroup$
– Shaun
Jan 3 at 19:41






$begingroup$
The use of $$begin{align} x &=y\ &=zend{align}$$ for $$begin{align} x &=y\ &=zend{align}$$ wouldn't go amiss.
$endgroup$
– Shaun
Jan 3 at 19:41












2 Answers
2






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oldest

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$begingroup$

You can get this more quickly using the First Isomorphism Theorem. Recall that $o(g) = |langle g rangle|$. Now $o(phi(g)) = |phi( langle g rangle)|$. Taking $G = langle g rangle$, the First Isomorphism theorem tells us that $|G| = |Im(phi)| cdot |ker(phi)|$. Note that $|Im(phi)| = o(phi(g))$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    I would just write $phi (g)^r=e$, with $rlt d$. So $r=0$. (It's pretty clear, since $e=phi(g)^m=phi(g)^d$).



    But it seems to be correct.



    Actually, once $phi (g)^m=e$ you're done (by the little argument you essentially gave). That is, $g^a=eimplies o(g)mid a$, generally.






    share|cite|improve this answer











    $endgroup$














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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      2












      $begingroup$

      You can get this more quickly using the First Isomorphism Theorem. Recall that $o(g) = |langle g rangle|$. Now $o(phi(g)) = |phi( langle g rangle)|$. Taking $G = langle g rangle$, the First Isomorphism theorem tells us that $|G| = |Im(phi)| cdot |ker(phi)|$. Note that $|Im(phi)| = o(phi(g))$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        You can get this more quickly using the First Isomorphism Theorem. Recall that $o(g) = |langle g rangle|$. Now $o(phi(g)) = |phi( langle g rangle)|$. Taking $G = langle g rangle$, the First Isomorphism theorem tells us that $|G| = |Im(phi)| cdot |ker(phi)|$. Note that $|Im(phi)| = o(phi(g))$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          You can get this more quickly using the First Isomorphism Theorem. Recall that $o(g) = |langle g rangle|$. Now $o(phi(g)) = |phi( langle g rangle)|$. Taking $G = langle g rangle$, the First Isomorphism theorem tells us that $|G| = |Im(phi)| cdot |ker(phi)|$. Note that $|Im(phi)| = o(phi(g))$.






          share|cite|improve this answer









          $endgroup$



          You can get this more quickly using the First Isomorphism Theorem. Recall that $o(g) = |langle g rangle|$. Now $o(phi(g)) = |phi( langle g rangle)|$. Taking $G = langle g rangle$, the First Isomorphism theorem tells us that $|G| = |Im(phi)| cdot |ker(phi)|$. Note that $|Im(phi)| = o(phi(g))$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 3 at 19:48









          ml0105ml0105

          11.5k21638




          11.5k21638























              0












              $begingroup$

              I would just write $phi (g)^r=e$, with $rlt d$. So $r=0$. (It's pretty clear, since $e=phi(g)^m=phi(g)^d$).



              But it seems to be correct.



              Actually, once $phi (g)^m=e$ you're done (by the little argument you essentially gave). That is, $g^a=eimplies o(g)mid a$, generally.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                I would just write $phi (g)^r=e$, with $rlt d$. So $r=0$. (It's pretty clear, since $e=phi(g)^m=phi(g)^d$).



                But it seems to be correct.



                Actually, once $phi (g)^m=e$ you're done (by the little argument you essentially gave). That is, $g^a=eimplies o(g)mid a$, generally.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I would just write $phi (g)^r=e$, with $rlt d$. So $r=0$. (It's pretty clear, since $e=phi(g)^m=phi(g)^d$).



                  But it seems to be correct.



                  Actually, once $phi (g)^m=e$ you're done (by the little argument you essentially gave). That is, $g^a=eimplies o(g)mid a$, generally.






                  share|cite|improve this answer











                  $endgroup$



                  I would just write $phi (g)^r=e$, with $rlt d$. So $r=0$. (It's pretty clear, since $e=phi(g)^m=phi(g)^d$).



                  But it seems to be correct.



                  Actually, once $phi (g)^m=e$ you're done (by the little argument you essentially gave). That is, $g^a=eimplies o(g)mid a$, generally.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 4 at 18:20

























                  answered Jan 3 at 21:00









                  Chris CusterChris Custer

                  14.3k3827




                  14.3k3827






























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