Mixing
Power of a generator generates the group iff this power is coprime to the group order.
$begingroup$
Let $langle grangle$ be a cyclic group of order $n$. Suppose $1leq q leq n-1$, I want to show that $g^q$ generates $langle grangle$ if and only if $gcd(n,q)=1$.
Suppose $g^q$ generates $langle grangle$, then
$$
1,g^q,g^{2q},...,g^{(n-1)q}
$$
are all distinct and $g^{nq}=1$, since $n$ is the order of the group. Note that this is the lowest multiple of $q$ for which this is the case. Therefore $operatorname{lcm}(q,n)=qn$ and it follows that $q$ and $n$ are coprime.
On the other hand, if $gcd(q,n)=1$, then $operatorname{lcm}(q,n)=qn$, we have $g^{nq}=1$ and hence all of
$$
1,g^q,g^{2q},...,g^{(n-1)q}
$$
are distinct again and $g^q$ is a generator.
Is my proof correct at all and is there maybe a more elegant argument?
group-theory
asked Oct 27 '13 at 14:34
Jimmy RJimmy R
1,200619
$endgroup$
add a comment |
$begingroup$
Let $langle grangle$ be a cyclic group of order $n$. Suppose $1leq q leq n-1$, I want to show that $g^q$ generates $langle grangle$ if and only if $gcd(n,q)=1$.
Suppose $g^q$ generates $langle grangle$, then
$$
1,g^q,g^{2q},...,g^{(n-1)q}
$$
are all distinct and $g^{nq}=1$, since $n$ is the order of the group. Note that this is the lowest multiple of $q$ for which this is the case. Therefore $operatorname{lcm}(q,n)=qn$ and it follows that $q$ and $n$ are coprime.
On the other hand, if $gcd(q,n)=1$, then $operatorname{lcm}(q,n)=qn$, we have $g^{nq}=1$ and hence all of
$$
1,g^q,g^{2q},...,g^{(n-1)q}
$$
are distinct again and $g^q$ is a generator.
Is my proof correct at all and is there maybe a more elegant argument?
group-theory
asked Oct 27 '13 at 14:34
Jimmy RJimmy R
1,200619
$endgroup$
add a comment |
$begingroup$
Let $langle grangle$ be a cyclic group of order $n$. Suppose $1leq q leq n-1$, I want to show that $g^q$ generates $langle grangle$ if and only if $gcd(n,q)=1$.
Suppose $g^q$ generates $langle grangle$, then
$$
1,g^q,g^{2q},...,g^{(n-1)q}
$$
are all distinct and $g^{nq}=1$, since $n$ is the order of the group. Note that this is the lowest multiple of $q$ for which this is the case. Therefore $operatorname{lcm}(q,n)=qn$ and it follows that $q$ and $n$ are coprime.
On the other hand, if $gcd(q,n)=1$, then $operatorname{lcm}(q,n)=qn$, we have $g^{nq}=1$ and hence all of
$$
1,g^q,g^{2q},...,g^{(n-1)q}
$$
are distinct again and $g^q$ is a generator.
Is my proof correct at all and is there maybe a more elegant argument?
group-theory
asked Oct 27 '13 at 14:34
Jimmy RJimmy R
1,200619
$endgroup$
Let $langle grangle$ be a cyclic group of order $n$. Suppose $1leq q leq n-1$, I want to show that $g^q$ generates $langle grangle$ if and only if $gcd(n,q)=1$.
Suppose $g^q$ generates $langle grangle$, then
$$
1,g^q,g^{2q},...,g^{(n-1)q}
$$
are all distinct and $g^{nq}=1$, since $n$ is the order of the group. Note that this is the lowest multiple of $q$ for which this is the case. Therefore $operatorname{lcm}(q,n)=qn$ and it follows that $q$ and $n$ are coprime.
On the other hand, if $gcd(q,n)=1$, then $operatorname{lcm}(q,n)=qn$, we have $g^{nq}=1$ and hence all of
$$
1,g^q,g^{2q},...,g^{(n-1)q}
$$
are distinct again and $g^q$ is a generator.
Is my proof correct at all and is there maybe a more elegant argument?
group-theory
group-theory
asked Oct 27 '13 at 14:34
Jimmy RJimmy R
1,200619
asked Oct 27 '13 at 14:34
Jimmy RJimmy R
1,200619
asked Oct 27 '13 at 14:34
Jimmy RJimmy R
1,200619
asked Oct 27 '13 at 14:34
Jimmy RJimmy R
1,200619
asked Oct 27 '13 at 14:34
Jimmy RJimmy R
1,200619
1,200619
add a comment |
add a comment |
1 Answer
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$begingroup$
Your proof seems correct to me.
answered Oct 27 '13 at 14:39
BIS HDBIS HD
1,9371127
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your proof seems correct to me.
answered Oct 27 '13 at 14:39
BIS HDBIS HD
1,9371127
$endgroup$
add a comment |
$begingroup$
Your proof seems correct to me.
answered Oct 27 '13 at 14:39
BIS HDBIS HD
1,9371127
$endgroup$
add a comment |
$begingroup$
Your proof seems correct to me.
answered Oct 27 '13 at 14:39
BIS HDBIS HD
1,9371127
$endgroup$
Your proof seems correct to me.
answered Oct 27 '13 at 14:39
BIS HDBIS HD
1,9371127
answered Oct 27 '13 at 14:39
BIS HDBIS HD
1,9371127
answered Oct 27 '13 at 14:39
BIS HDBIS HD
1,9371127
answered Oct 27 '13 at 14:39
BIS HDBIS HD
1,9371127
1,9371127
add a comment |
add a comment |
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