Showing that $o(phi(g))|o(g)$ if $phi:G_1mapsto G_2$ is a homomorphism and $gin G_1$
$o(g)$ 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
add a comment |
$o(g)$ 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
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
add a comment |
$o(g)$ 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
$o(g)$ 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
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
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 $|Im(phi)| = o(phi(g))$.
add a comment |
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.
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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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 $|Im(phi)| = o(phi(g))$.
add a comment |
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 $|Im(phi)| = o(phi(g))$.
add a comment |
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 $|Im(phi)| = o(phi(g))$.
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 $|Im(phi)| = o(phi(g))$.
answered 2 days ago
ml0105
11.4k21538
11.4k21538
add a comment |
add a comment |
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.
add a comment |
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.
add a comment |
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.
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.
edited yesterday
answered 2 days ago
Chris Custer
11k3824
11k3824
add a comment |
add a comment |
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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