Prove that a cyclic group with only one generator can have at most 2 elements
$begingroup$
Prove that a cyclic group that has only one generator has at most $2$ elements.
I want to know if my proof would be valid:
Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.
I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.
abstract-algebra greatest-common-divisor cyclic-groups
New contributor
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$begingroup$
Prove that a cyclic group that has only one generator has at most $2$ elements.
I want to know if my proof would be valid:
Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.
I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.
abstract-algebra greatest-common-divisor cyclic-groups
New contributor
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add a comment |
$begingroup$
Prove that a cyclic group that has only one generator has at most $2$ elements.
I want to know if my proof would be valid:
Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.
I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.
abstract-algebra greatest-common-divisor cyclic-groups
New contributor
$endgroup$
Prove that a cyclic group that has only one generator has at most $2$ elements.
I want to know if my proof would be valid:
Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.
I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.
abstract-algebra greatest-common-divisor cyclic-groups
abstract-algebra greatest-common-divisor cyclic-groups
New contributor
New contributor
edited 3 hours ago
darij grinberg
11.3k33164
11.3k33164
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asked 4 hours ago
PabloPablo
361
361
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3 Answers
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$begingroup$
Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).
$endgroup$
add a comment |
$begingroup$
Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.
$endgroup$
add a comment |
$begingroup$
Here is another take.
The number of generators is $phi(n)$, where $phi$ is Euler's function.
Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.
In the first case, we have $phi(n) ge phi(p)=p-1ge2$.
In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.
Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.
(The key fact is this: if $d$ divides $n$, then $phi(n) ge phi(d)$, because if $x$ is coprime with $n$, then $x$ is coprime with $d$.)
$endgroup$
$begingroup$
The argument in the very last sentence is wrong (though the claim is true).
$endgroup$
– darij grinberg
1 hour ago
add a comment |
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3 Answers
3
active
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3 Answers
3
active
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$begingroup$
Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).
$endgroup$
add a comment |
$begingroup$
Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).
$endgroup$
add a comment |
$begingroup$
Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).
$endgroup$
Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).
answered 3 hours ago
MPWMPW
30.1k12057
30.1k12057
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$begingroup$
Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.
$endgroup$
add a comment |
$begingroup$
Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.
$endgroup$
add a comment |
$begingroup$
Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.
$endgroup$
Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.
answered 3 hours ago
LBJFSLBJFS
1257
1257
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$begingroup$
Here is another take.
The number of generators is $phi(n)$, where $phi$ is Euler's function.
Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.
In the first case, we have $phi(n) ge phi(p)=p-1ge2$.
In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.
Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.
(The key fact is this: if $d$ divides $n$, then $phi(n) ge phi(d)$, because if $x$ is coprime with $n$, then $x$ is coprime with $d$.)
$endgroup$
$begingroup$
The argument in the very last sentence is wrong (though the claim is true).
$endgroup$
– darij grinberg
1 hour ago
add a comment |
$begingroup$
Here is another take.
The number of generators is $phi(n)$, where $phi$ is Euler's function.
Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.
In the first case, we have $phi(n) ge phi(p)=p-1ge2$.
In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.
Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.
(The key fact is this: if $d$ divides $n$, then $phi(n) ge phi(d)$, because if $x$ is coprime with $n$, then $x$ is coprime with $d$.)
$endgroup$
$begingroup$
The argument in the very last sentence is wrong (though the claim is true).
$endgroup$
– darij grinberg
1 hour ago
add a comment |
$begingroup$
Here is another take.
The number of generators is $phi(n)$, where $phi$ is Euler's function.
Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.
In the first case, we have $phi(n) ge phi(p)=p-1ge2$.
In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.
Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.
(The key fact is this: if $d$ divides $n$, then $phi(n) ge phi(d)$, because if $x$ is coprime with $n$, then $x$ is coprime with $d$.)
$endgroup$
Here is another take.
The number of generators is $phi(n)$, where $phi$ is Euler's function.
Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.
In the first case, we have $phi(n) ge phi(p)=p-1ge2$.
In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.
Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.
(The key fact is this: if $d$ divides $n$, then $phi(n) ge phi(d)$, because if $x$ is coprime with $n$, then $x$ is coprime with $d$.)
edited 2 hours ago
answered 3 hours ago
lhflhf
165k10171396
165k10171396
$begingroup$
The argument in the very last sentence is wrong (though the claim is true).
$endgroup$
– darij grinberg
1 hour ago
add a comment |
$begingroup$
The argument in the very last sentence is wrong (though the claim is true).
$endgroup$
– darij grinberg
1 hour ago
$begingroup$
The argument in the very last sentence is wrong (though the claim is true).
$endgroup$
– darij grinberg
1 hour ago
$begingroup$
The argument in the very last sentence is wrong (though the claim is true).
$endgroup$
– darij grinberg
1 hour ago
add a comment |
Pablo is a new contributor. Be nice, and check out our Code of Conduct.
Pablo is a new contributor. Be nice, and check out our Code of Conduct.
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