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












2














$o(g)$ 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




















  • 1




    It looks fine to me. Well done! :)
    – Shaun
    2 days ago










  • 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.
    – Shaun
    2 days ago








  • 1




    The use of $$begin{align} x &=y\ &=zend{align}$$ for $$begin{align} x &=y\ &=zend{align}$$ wouldn't go amiss.
    – Shaun
    2 days ago


















2














$o(g)$ 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




















  • 1




    It looks fine to me. Well done! :)
    – Shaun
    2 days ago










  • 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.
    – Shaun
    2 days ago








  • 1




    The use of $$begin{align} x &=y\ &=zend{align}$$ for $$begin{align} x &=y\ &=zend{align}$$ wouldn't go amiss.
    – Shaun
    2 days ago
















2












2








2







$o(g)$ 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















$o(g)$ 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 2 days ago









Shaun

8,810113680




8,810113680










asked 2 days ago









John Cataldo

1,0911216




1,0911216








  • 1




    It looks fine to me. Well done! :)
    – Shaun
    2 days ago










  • 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.
    – Shaun
    2 days ago








  • 1




    The use of $$begin{align} x &=y\ &=zend{align}$$ for $$begin{align} x &=y\ &=zend{align}$$ wouldn't go amiss.
    – Shaun
    2 days ago
















  • 1




    It looks fine to me. Well done! :)
    – Shaun
    2 days ago










  • 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.
    – Shaun
    2 days ago








  • 1




    The use of $$begin{align} x &=y\ &=zend{align}$$ for $$begin{align} x &=y\ &=zend{align}$$ wouldn't go amiss.
    – Shaun
    2 days ago










1




1




It looks fine to me. Well done! :)
– Shaun
2 days ago




It looks fine to me. Well done! :)
– Shaun
2 days ago












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.
– Shaun
2 days ago






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.
– Shaun
2 days ago






1




1




The use of $$begin{align} x &=y\ &=zend{align}$$ for $$begin{align} x &=y\ &=zend{align}$$ wouldn't go amiss.
– Shaun
2 days ago






The use of $$begin{align} x &=y\ &=zend{align}$$ for $$begin{align} x &=y\ &=zend{align}$$ wouldn't go amiss.
– Shaun
2 days ago












2 Answers
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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





























    0














    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























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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

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      2














      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


























        2














        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
























          2












          2








          2






          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












          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 2 days ago









          ml0105

          11.4k21538




          11.4k21538























              0














              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




























                0














                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


























                  0












                  0








                  0






                  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














                  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 yesterday

























                  answered 2 days ago









                  Chris Custer

                  11k3824




                  11k3824






























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