Mixing
Show that every nonzero element of $(mathbb{Z}/pmathbb{Z})$ has order $p$
$begingroup$
Consider $G = (mathbb{Z}/pmathbb{Z})$, where $p$ is prime. Show that every non-zero element has order $p$.
I have seen the proof for this in Dummit and Foote using multiplicative notation, but I'm having trouble solving this using additive notation. What I have so far:
Let $(mathbb{Z}/pmathbb{Z}) = {bar{0}, bar{1}, bar{2}, ..., overline{p-1}}$, and let $bar{a} in { bar{1}, bar{2}, ..., overline{p-1}}$. Suppose ord$(bar{a}) = n$, that is, $na equiv 0$ (mod $p$). We want to show $n=p$.
Since $na equiv 0$ (mod $p$), then $p mid na$. Given that $p$ is prime and $a < p$, we have gcd$(a,p) = 1$. These facts together imply that $p mid n$.
I still need to show that $p=n$. I know by the definition of order that $n$ must be the least positive integer such that $nbar{a} = bar{0}$, i.e. $na = p$. Is this enough to say also that $n mid p$, and thus $n=p$?
I think I am very close to solving this but I can't quite put the pieces together. Any help would be appreciated.
group-theory elementary-number-theory
asked Feb 2 at 16:35
PawneePawnee
365
$endgroup$
add a comment |
$begingroup$
Consider $G = (mathbb{Z}/pmathbb{Z})$, where $p$ is prime. Show that every non-zero element has order $p$.
I have seen the proof for this in Dummit and Foote using multiplicative notation, but I'm having trouble solving this using additive notation. What I have so far:
Let $(mathbb{Z}/pmathbb{Z}) = {bar{0}, bar{1}, bar{2}, ..., overline{p-1}}$, and let $bar{a} in { bar{1}, bar{2}, ..., overline{p-1}}$. Suppose ord$(bar{a}) = n$, that is, $na equiv 0$ (mod $p$). We want to show $n=p$.
Since $na equiv 0$ (mod $p$), then $p mid na$. Given that $p$ is prime and $a < p$, we have gcd$(a,p) = 1$. These facts together imply that $p mid n$.
I still need to show that $p=n$. I know by the definition of order that $n$ must be the least positive integer such that $nbar{a} = bar{0}$, i.e. $na = p$. Is this enough to say also that $n mid p$, and thus $n=p$?
I think I am very close to solving this but I can't quite put the pieces together. Any help would be appreciated.
group-theory elementary-number-theory
asked Feb 2 at 16:35
PawneePawnee
365
$endgroup$
$begingroup$
Yes, since $pbar a=bar 0$, so either $p$ is the order or by definition of order of an element, $p$ is divisible by the order, so indeed $nmid p$. Using the previous inference that $pmid n$, your proof is complete.
$endgroup$
– learner
Feb 2 at 16:56
$begingroup$
I am confused about this comment. You assume $pbar{a} = bar{0}$, but this is what we're trying to prove. Is this a typo or am I misunderstanding something?
$endgroup$
– Pawnee
Feb 2 at 19:45
$begingroup$
I'm not assuming it. Note that $pbar a={pxmid xinbar a}$ and since $pmid px$ for all integers $x$, so $pmid y$ for all $yin pbar a$, thus implying that $pbar a=bar 0$ by definition of $bar 0$ in $Bbb Z/pBbb Z$. Also, note that we're not trying to prove that $pbar a=bar 0$, we're trying to prove that the smallest positive integer $n$ such that $nbar a=bar 0$ is $p$
$endgroup$
– learner
Feb 3 at 14:14
add a comment |
$begingroup$
Consider $G = (mathbb{Z}/pmathbb{Z})$, where $p$ is prime. Show that every non-zero element has order $p$.
I have seen the proof for this in Dummit and Foote using multiplicative notation, but I'm having trouble solving this using additive notation. What I have so far:
Let $(mathbb{Z}/pmathbb{Z}) = {bar{0}, bar{1}, bar{2}, ..., overline{p-1}}$, and let $bar{a} in { bar{1}, bar{2}, ..., overline{p-1}}$. Suppose ord$(bar{a}) = n$, that is, $na equiv 0$ (mod $p$). We want to show $n=p$.
Since $na equiv 0$ (mod $p$), then $p mid na$. Given that $p$ is prime and $a < p$, we have gcd$(a,p) = 1$. These facts together imply that $p mid n$.
I still need to show that $p=n$. I know by the definition of order that $n$ must be the least positive integer such that $nbar{a} = bar{0}$, i.e. $na = p$. Is this enough to say also that $n mid p$, and thus $n=p$?
I think I am very close to solving this but I can't quite put the pieces together. Any help would be appreciated.
group-theory elementary-number-theory
asked Feb 2 at 16:35
PawneePawnee
365
$endgroup$
Consider $G = (mathbb{Z}/pmathbb{Z})$, where $p$ is prime. Show that every non-zero element has order $p$.
I have seen the proof for this in Dummit and Foote using multiplicative notation, but I'm having trouble solving this using additive notation. What I have so far:
Let $(mathbb{Z}/pmathbb{Z}) = {bar{0}, bar{1}, bar{2}, ..., overline{p-1}}$, and let $bar{a} in { bar{1}, bar{2}, ..., overline{p-1}}$. Suppose ord$(bar{a}) = n$, that is, $na equiv 0$ (mod $p$). We want to show $n=p$.
Since $na equiv 0$ (mod $p$), then $p mid na$. Given that $p$ is prime and $a < p$, we have gcd$(a,p) = 1$. These facts together imply that $p mid n$.
I still need to show that $p=n$. I know by the definition of order that $n$ must be the least positive integer such that $nbar{a} = bar{0}$, i.e. $na = p$. Is this enough to say also that $n mid p$, and thus $n=p$?
I think I am very close to solving this but I can't quite put the pieces together. Any help would be appreciated.
group-theory elementary-number-theory
group-theory elementary-number-theory
asked Feb 2 at 16:35
PawneePawnee
365
asked Feb 2 at 16:35
PawneePawnee
365
asked Feb 2 at 16:35
PawneePawnee
365
asked Feb 2 at 16:35
PawneePawnee
365
asked Feb 2 at 16:35
PawneePawnee
365
365
$begingroup$
Yes, since $pbar a=bar 0$, so either $p$ is the order or by definition of order of an element, $p$ is divisible by the order, so indeed $nmid p$. Using the previous inference that $pmid n$, your proof is complete.
$endgroup$
– learner
Feb 2 at 16:56
$begingroup$
I am confused about this comment. You assume $pbar{a} = bar{0}$, but this is what we're trying to prove. Is this a typo or am I misunderstanding something?
$endgroup$
– Pawnee
Feb 2 at 19:45
$begingroup$
I'm not assuming it. Note that $pbar a={pxmid xinbar a}$ and since $pmid px$ for all integers $x$, so $pmid y$ for all $yin pbar a$, thus implying that $pbar a=bar 0$ by definition of $bar 0$ in $Bbb Z/pBbb Z$. Also, note that we're not trying to prove that $pbar a=bar 0$, we're trying to prove that the smallest positive integer $n$ such that $nbar a=bar 0$ is $p$
$endgroup$
– learner
Feb 3 at 14:14
add a comment |
$begingroup$
Yes, since $pbar a=bar 0$, so either $p$ is the order or by definition of order of an element, $p$ is divisible by the order, so indeed $nmid p$. Using the previous inference that $pmid n$, your proof is complete.
$endgroup$
– learner
Feb 2 at 16:56
$begingroup$
I am confused about this comment. You assume $pbar{a} = bar{0}$, but this is what we're trying to prove. Is this a typo or am I misunderstanding something?
$endgroup$
– Pawnee
Feb 2 at 19:45
$begingroup$
I'm not assuming it. Note that $pbar a={pxmid xinbar a}$ and since $pmid px$ for all integers $x$, so $pmid y$ for all $yin pbar a$, thus implying that $pbar a=bar 0$ by definition of $bar 0$ in $Bbb Z/pBbb Z$. Also, note that we're not trying to prove that $pbar a=bar 0$, we're trying to prove that the smallest positive integer $n$ such that $nbar a=bar 0$ is $p$
$endgroup$
– learner
Feb 3 at 14:14
$begingroup$
Yes, since $pbar a=bar 0$, so either $p$ is the order or by definition of order of an element, $p$ is divisible by the order, so indeed $nmid p$. Using the previous inference that $pmid n$, your proof is complete.
$endgroup$
– learner
Feb 2 at 16:56
$begingroup$
Yes, since $pbar a=bar 0$, so either $p$ is the order or by definition of order of an element, $p$ is divisible by the order, so indeed $nmid p$. Using the previous inference that $pmid n$, your proof is complete.
$endgroup$
– learner
Feb 2 at 16:56
$begingroup$
I am confused about this comment. You assume $pbar{a} = bar{0}$, but this is what we're trying to prove. Is this a typo or am I misunderstanding something?
$endgroup$
– Pawnee
Feb 2 at 19:45
$begingroup$
I am confused about this comment. You assume $pbar{a} = bar{0}$, but this is what we're trying to prove. Is this a typo or am I misunderstanding something?
$endgroup$
– Pawnee
Feb 2 at 19:45
$begingroup$
I'm not assuming it. Note that $pbar a={pxmid xinbar a}$ and since $pmid px$ for all integers $x$, so $pmid y$ for all $yin pbar a$, thus implying that $pbar a=bar 0$ by definition of $bar 0$ in $Bbb Z/pBbb Z$. Also, note that we're not trying to prove that $pbar a=bar 0$, we're trying to prove that the smallest positive integer $n$ such that $nbar a=bar 0$ is $p$
$endgroup$
– learner
Feb 3 at 14:14
$begingroup$
I'm not assuming it. Note that $pbar a={pxmid xinbar a}$ and since $pmid px$ for all integers $x$, so $pmid y$ for all $yin pbar a$, thus implying that $pbar a=bar 0$ by definition of $bar 0$ in $Bbb Z/pBbb Z$. Also, note that we're not trying to prove that $pbar a=bar 0$, we're trying to prove that the smallest positive integer $n$ such that $nbar a=bar 0$ is $p$
$endgroup$
– learner
Feb 3 at 14:14
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Observe that order of $g in G$ is same as the cyclic group generated by $g$. By Lagrange's theorem, the order of the subgroup divides the order of the group.
Since the order of $G$ is prime, the order of subgroup has to be $1$ or $p$. So every non-trivial element must have order $p$.
answered Feb 2 at 16:40
Exp ikxExp ikx
4389
$endgroup$
$begingroup$
This problem is quite easy to solve with Lagrange's theorem, but we haven't covered it yet. All we have covered are basic notions of group axioms, subgroups, cyclic groups, and orders.
$endgroup$
– Pawnee
Feb 2 at 16:41
$begingroup$
How is it possible that $n mid p$, because p is prime? I think my proof is still wrong.
$endgroup$
– Pawnee
Feb 2 at 19:53
add a comment |
$begingroup$
In $(mathbb{Z}/pmathbb{Z})$, for each element $overline a$ you have$$overbrace{overline a+overline a+cdots+overline a}^{ptext{ times}}=overline{pa}=0.$$So, no element has order greater than $p$.
answered Feb 2 at 16:38
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Observe that order of $g in G$ is same as the cyclic group generated by $g$. By Lagrange's theorem, the order of the subgroup divides the order of the group.
Since the order of $G$ is prime, the order of subgroup has to be $1$ or $p$. So every non-trivial element must have order $p$.
answered Feb 2 at 16:40
Exp ikxExp ikx
4389
$endgroup$
$begingroup$
This problem is quite easy to solve with Lagrange's theorem, but we haven't covered it yet. All we have covered are basic notions of group axioms, subgroups, cyclic groups, and orders.
$endgroup$
– Pawnee
Feb 2 at 16:41
$begingroup$
How is it possible that $n mid p$, because p is prime? I think my proof is still wrong.
$endgroup$
– Pawnee
Feb 2 at 19:53
add a comment |
$begingroup$
Observe that order of $g in G$ is same as the cyclic group generated by $g$. By Lagrange's theorem, the order of the subgroup divides the order of the group.
Since the order of $G$ is prime, the order of subgroup has to be $1$ or $p$. So every non-trivial element must have order $p$.
answered Feb 2 at 16:40
Exp ikxExp ikx
4389
$endgroup$
$begingroup$
This problem is quite easy to solve with Lagrange's theorem, but we haven't covered it yet. All we have covered are basic notions of group axioms, subgroups, cyclic groups, and orders.
$endgroup$
– Pawnee
Feb 2 at 16:41
$begingroup$
How is it possible that $n mid p$, because p is prime? I think my proof is still wrong.
$endgroup$
– Pawnee
Feb 2 at 19:53
add a comment |
$begingroup$
Observe that order of $g in G$ is same as the cyclic group generated by $g$. By Lagrange's theorem, the order of the subgroup divides the order of the group.
Since the order of $G$ is prime, the order of subgroup has to be $1$ or $p$. So every non-trivial element must have order $p$.
answered Feb 2 at 16:40
Exp ikxExp ikx
4389
$endgroup$
Observe that order of $g in G$ is same as the cyclic group generated by $g$. By Lagrange's theorem, the order of the subgroup divides the order of the group.
Since the order of $G$ is prime, the order of subgroup has to be $1$ or $p$. So every non-trivial element must have order $p$.
answered Feb 2 at 16:40
Exp ikxExp ikx
4389
answered Feb 2 at 16:40
Exp ikxExp ikx
4389
answered Feb 2 at 16:40
Exp ikxExp ikx
4389
answered Feb 2 at 16:40
Exp ikxExp ikx
4389
4389
$begingroup$
This problem is quite easy to solve with Lagrange's theorem, but we haven't covered it yet. All we have covered are basic notions of group axioms, subgroups, cyclic groups, and orders.
$endgroup$
– Pawnee
Feb 2 at 16:41
$begingroup$
How is it possible that $n mid p$, because p is prime? I think my proof is still wrong.
$endgroup$
– Pawnee
Feb 2 at 19:53
add a comment |
$begingroup$
This problem is quite easy to solve with Lagrange's theorem, but we haven't covered it yet. All we have covered are basic notions of group axioms, subgroups, cyclic groups, and orders.
$endgroup$
– Pawnee
Feb 2 at 16:41
$begingroup$
How is it possible that $n mid p$, because p is prime? I think my proof is still wrong.
$endgroup$
– Pawnee
Feb 2 at 19:53
$begingroup$
This problem is quite easy to solve with Lagrange's theorem, but we haven't covered it yet. All we have covered are basic notions of group axioms, subgroups, cyclic groups, and orders.
$endgroup$
– Pawnee
Feb 2 at 16:41
$begingroup$
This problem is quite easy to solve with Lagrange's theorem, but we haven't covered it yet. All we have covered are basic notions of group axioms, subgroups, cyclic groups, and orders.
$endgroup$
– Pawnee
Feb 2 at 16:41
$begingroup$
How is it possible that $n mid p$, because p is prime? I think my proof is still wrong.
$endgroup$
– Pawnee
Feb 2 at 19:53
$begingroup$
How is it possible that $n mid p$, because p is prime? I think my proof is still wrong.
$endgroup$
– Pawnee
Feb 2 at 19:53
add a comment |
$begingroup$
In $(mathbb{Z}/pmathbb{Z})$, for each element $overline a$ you have$$overbrace{overline a+overline a+cdots+overline a}^{ptext{ times}}=overline{pa}=0.$$So, no element has order greater than $p$.
answered Feb 2 at 16:38
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
add a comment |
$begingroup$
In $(mathbb{Z}/pmathbb{Z})$, for each element $overline a$ you have$$overbrace{overline a+overline a+cdots+overline a}^{ptext{ times}}=overline{pa}=0.$$So, no element has order greater than $p$.
answered Feb 2 at 16:38
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
add a comment |
$begingroup$
In $(mathbb{Z}/pmathbb{Z})$, for each element $overline a$ you have$$overbrace{overline a+overline a+cdots+overline a}^{ptext{ times}}=overline{pa}=0.$$So, no element has order greater than $p$.
answered Feb 2 at 16:38
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
In $(mathbb{Z}/pmathbb{Z})$, for each element $overline a$ you have$$overbrace{overline a+overline a+cdots+overline a}^{ptext{ times}}=overline{pa}=0.$$So, no element has order greater than $p$.
answered Feb 2 at 16:38
José Carlos SantosJosé Carlos Santos
174k23134243
answered Feb 2 at 16:38
José Carlos SantosJosé Carlos Santos
174k23134243
answered Feb 2 at 16:38
José Carlos SantosJosé Carlos Santos
174k23134243
answered Feb 2 at 16:38
José Carlos SantosJosé Carlos Santos
174k23134243
174k23134243
add a comment |
add a comment |
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$begingroup$
Yes, since $pbar a=bar 0$, so either $p$ is the order or by definition of order of an element, $p$ is divisible by the order, so indeed $nmid p$. Using the previous inference that $pmid n$, your proof is complete.
$endgroup$
– learner
Feb 2 at 16:56
$begingroup$
I am confused about this comment. You assume $pbar{a} = bar{0}$, but this is what we're trying to prove. Is this a typo or am I misunderstanding something?
$endgroup$
– Pawnee
Feb 2 at 19:45
$begingroup$
I'm not assuming it. Note that $pbar a={pxmid xinbar a}$ and since $pmid px$ for all integers $x$, so $pmid y$ for all $yin pbar a$, thus implying that $pbar a=bar 0$ by definition of $bar 0$ in $Bbb Z/pBbb Z$. Also, note that we're not trying to prove that $pbar a=bar 0$, we're trying to prove that the smallest positive integer $n$ such that $nbar a=bar 0$ is $p$
$endgroup$
– learner
Feb 3 at 14:14