Mixing
Conditions for cyclic quotient group
Clash Royale CLAN TAG#URR8PPP
up vote
8
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Let $G$ be an arbitrary finite group and $H$ a normal subgroup.
What are some good conditions on $H$ that make the quotient $G/H$ cyclic?
I want to avoid any further restriction on $G$.
group-theory finite-groups cyclic-groups quotient-group
asked Aug 3 at 18:10
Samuel Plath
673213
add a comment |Â
up vote
8
down vote
favorite
Let $G$ be an arbitrary finite group and $H$ a normal subgroup.
What are some good conditions on $H$ that make the quotient $G/H$ cyclic?
I want to avoid any further restriction on $G$.
group-theory finite-groups cyclic-groups quotient-group
asked Aug 3 at 18:10
Samuel Plath
673213
3
$G$ is abelian and $H$ is the center of $G$ :p
– leibnewtz
Aug 3 at 18:11
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Let $G$ be an arbitrary finite group and $H$ a normal subgroup.
What are some good conditions on $H$ that make the quotient $G/H$ cyclic?
I want to avoid any further restriction on $G$.
group-theory finite-groups cyclic-groups quotient-group
asked Aug 3 at 18:10
Samuel Plath
673213
Let $G$ be an arbitrary finite group and $H$ a normal subgroup.
What are some good conditions on $H$ that make the quotient $G/H$ cyclic?
I want to avoid any further restriction on $G$.
group-theory finite-groups cyclic-groups quotient-group
asked Aug 3 at 18:10
Samuel Plath
673213
edited Aug 3 at 18:20
asked Aug 3 at 18:10
Samuel Plath
673213
asked Aug 3 at 18:10
Samuel Plath
673213
asked Aug 3 at 18:10
Samuel Plath
673213
673213
3
$G$ is abelian and $H$ is the center of $G$ :p
– leibnewtz
Aug 3 at 18:11
add a comment |Â
3
$G$ is abelian and $H$ is the center of $G$ :p
– leibnewtz
Aug 3 at 18:11
3
3
$G$ is abelian and $H$ is the center of $G$ :p
– leibnewtz
Aug 3 at 18:11
$G$ is abelian and $H$ is the center of $G$ :p
– leibnewtz
Aug 3 at 18:11
add a comment |Â
7 Answers
7
active
oldest
votes
up vote
5
down vote
This is largely the wrong question. Let $H,K$ be groups and let $G = H times K$ so that $H trianglelefteq G$. (This is not the only way to get to the upcoming punchline.)
What sort of conditions can you apply to $H$ to force $K cong G/H$ cyclic? (Notice that $K$ was chosen entirely independently of $H$.)
There are no such conditions generally. You say you don't want to constrain $G$, but $G$ is the only thing that controls the cosets of $H$. $H$ is stuck in one of its cosets and has no control over the rest of $G$.
Notice that no one is giving you constraints on $H$. They constrain the number of cosets of $H$ (via $|G|/|H|$) or they require $G$ special (see especially the finite version of special). This is because constraining a tiny chunk of $G$ is too weak to control the cosets of that chunk.
answered Aug 3 at 23:18
Eric Towers
30.4k22264
add a comment |Â
up vote
3
down vote
Proposition Let $G$ be a finite group, $H$ a subgroup with $|G:H|=p$, a prime. If gcd$(|H|,p-1)=1$, or $p$ is the smallest prime dividing $|G|$, then $H$ is normal and hence $G/H cong C_p$.
answered Aug 3 at 18:39
Nicky Hekster
26.7k52952
add a comment |Â
up vote
3
down vote
Whenever $fracH$ has order $p$ with $p$ prime, or has order $pq$ with primes $p<q$ and $pnmid q-1$, then $G/H$ is cyclic.
If $H$ is special, then $G/H$ is cyclic, e.g., if $H=G$.
answered Aug 3 at 20:11
Dietrich Burde
74.5k64084
What is a special subgroup?
– lhf
Aug 3 at 21:18
I wanted to say, if we chose a special subgroup, e.g., some specific example like $H=G$, as a "good condition on $H$".
– Dietrich Burde
2 days ago
add a comment |Â
up vote
3
down vote
Another sufficient condition that captures some of those already mentioned is that $gcd([G:H], phi([G:H]))= 1$, where $phi$ is Euler's function. This is a condition that guarantees that every group of order $[G:H]$ is cyclic, and includes numbers, such as $255$, that are products of more than two primes.
answered Aug 3 at 22:56
James
6,68721426
add a comment |Â
up vote
1
down vote
A sufficient condition (although not a necessary one) is that $fracH$ is a prime number.
answered Aug 3 at 18:15
José Carlos Santos
112k1696172
add a comment |Â
up vote
1
down vote
Let $G$ be a group and $H$ be a normal subgroup, the easiest condition that I can think of is that $fracH$ is a prime number.
If $G$ is not a solvable group and $H$ is solvable, then $G/H$ must not be cyclic.
Another equivalent condition that you may find useful is that, $G/H$ is cyclic if and only if for any $gin Gbackslash H$, the minimal number $n$ such that $g^nin H$ must satisfy $n=fracH$.
answered Aug 3 at 18:18
ElfHog
50214
add a comment |Â
up vote
1
down vote
It is of course a necessary condition that $H$ contains the derived subgruop $[G,G]$ of $G$, as $G/H$ is commutative. Suppose that you have a set $X$ of generators for $G$. You could then check that, for each prime divisor $p$ of the index $[G:H]$, the subgroup $langle x^pmid xin Xrangle H$ has index $p$ in $G$. (This just amounts to saying that the $p$-th powers of a generating set for the quotient group $G/H$ generate an index $p$ subgroup.)
answered Aug 4 at 0:19
James
6,68721426
add a comment |Â
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
This is largely the wrong question. Let $H,K$ be groups and let $G = H times K$ so that $H trianglelefteq G$. (This is not the only way to get to the upcoming punchline.)
What sort of conditions can you apply to $H$ to force $K cong G/H$ cyclic? (Notice that $K$ was chosen entirely independently of $H$.)
There are no such conditions generally. You say you don't want to constrain $G$, but $G$ is the only thing that controls the cosets of $H$. $H$ is stuck in one of its cosets and has no control over the rest of $G$.
Notice that no one is giving you constraints on $H$. They constrain the number of cosets of $H$ (via $|G|/|H|$) or they require $G$ special (see especially the finite version of special). This is because constraining a tiny chunk of $G$ is too weak to control the cosets of that chunk.
answered Aug 3 at 23:18
Eric Towers
30.4k22264
add a comment |Â
up vote
5
down vote
This is largely the wrong question. Let $H,K$ be groups and let $G = H times K$ so that $H trianglelefteq G$. (This is not the only way to get to the upcoming punchline.)
What sort of conditions can you apply to $H$ to force $K cong G/H$ cyclic? (Notice that $K$ was chosen entirely independently of $H$.)
There are no such conditions generally. You say you don't want to constrain $G$, but $G$ is the only thing that controls the cosets of $H$. $H$ is stuck in one of its cosets and has no control over the rest of $G$.
Notice that no one is giving you constraints on $H$. They constrain the number of cosets of $H$ (via $|G|/|H|$) or they require $G$ special (see especially the finite version of special). This is because constraining a tiny chunk of $G$ is too weak to control the cosets of that chunk.
answered Aug 3 at 23:18
Eric Towers
30.4k22264
add a comment |Â
up vote
5
down vote
up vote
5
down vote
This is largely the wrong question. Let $H,K$ be groups and let $G = H times K$ so that $H trianglelefteq G$. (This is not the only way to get to the upcoming punchline.)
What sort of conditions can you apply to $H$ to force $K cong G/H$ cyclic? (Notice that $K$ was chosen entirely independently of $H$.)
There are no such conditions generally. You say you don't want to constrain $G$, but $G$ is the only thing that controls the cosets of $H$. $H$ is stuck in one of its cosets and has no control over the rest of $G$.
Notice that no one is giving you constraints on $H$. They constrain the number of cosets of $H$ (via $|G|/|H|$) or they require $G$ special (see especially the finite version of special). This is because constraining a tiny chunk of $G$ is too weak to control the cosets of that chunk.
answered Aug 3 at 23:18
Eric Towers
30.4k22264
This is largely the wrong question. Let $H,K$ be groups and let $G = H times K$ so that $H trianglelefteq G$. (This is not the only way to get to the upcoming punchline.)
What sort of conditions can you apply to $H$ to force $K cong G/H$ cyclic? (Notice that $K$ was chosen entirely independently of $H$.)
There are no such conditions generally. You say you don't want to constrain $G$, but $G$ is the only thing that controls the cosets of $H$. $H$ is stuck in one of its cosets and has no control over the rest of $G$.
Notice that no one is giving you constraints on $H$. They constrain the number of cosets of $H$ (via $|G|/|H|$) or they require $G$ special (see especially the finite version of special). This is because constraining a tiny chunk of $G$ is too weak to control the cosets of that chunk.
answered Aug 3 at 23:18
Eric Towers
30.4k22264
answered Aug 3 at 23:18
Eric Towers
30.4k22264
answered Aug 3 at 23:18
Eric Towers
30.4k22264
answered Aug 3 at 23:18
Eric Towers
30.4k22264
30.4k22264
add a comment |Â
add a comment |Â
up vote
3
down vote
Proposition Let $G$ be a finite group, $H$ a subgroup with $|G:H|=p$, a prime. If gcd$(|H|,p-1)=1$, or $p$ is the smallest prime dividing $|G|$, then $H$ is normal and hence $G/H cong C_p$.
answered Aug 3 at 18:39
Nicky Hekster
26.7k52952
add a comment |Â
up vote
3
down vote
Proposition Let $G$ be a finite group, $H$ a subgroup with $|G:H|=p$, a prime. If gcd$(|H|,p-1)=1$, or $p$ is the smallest prime dividing $|G|$, then $H$ is normal and hence $G/H cong C_p$.
answered Aug 3 at 18:39
Nicky Hekster
26.7k52952
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Proposition Let $G$ be a finite group, $H$ a subgroup with $|G:H|=p$, a prime. If gcd$(|H|,p-1)=1$, or $p$ is the smallest prime dividing $|G|$, then $H$ is normal and hence $G/H cong C_p$.
answered Aug 3 at 18:39
Nicky Hekster
26.7k52952
Proposition Let $G$ be a finite group, $H$ a subgroup with $|G:H|=p$, a prime. If gcd$(|H|,p-1)=1$, or $p$ is the smallest prime dividing $|G|$, then $H$ is normal and hence $G/H cong C_p$.
answered Aug 3 at 18:39
Nicky Hekster
26.7k52952
answered Aug 3 at 18:39
Nicky Hekster
26.7k52952
answered Aug 3 at 18:39
Nicky Hekster
26.7k52952
answered Aug 3 at 18:39
Nicky Hekster
26.7k52952
26.7k52952
add a comment |Â
add a comment |Â
up vote
3
down vote
Whenever $fracH$ has order $p$ with $p$ prime, or has order $pq$ with primes $p<q$ and $pnmid q-1$, then $G/H$ is cyclic.
If $H$ is special, then $G/H$ is cyclic, e.g., if $H=G$.
answered Aug 3 at 20:11
Dietrich Burde
74.5k64084
What is a special subgroup?
– lhf
Aug 3 at 21:18
I wanted to say, if we chose a special subgroup, e.g., some specific example like $H=G$, as a "good condition on $H$".
– Dietrich Burde
2 days ago
add a comment |Â
up vote
3
down vote
Whenever $fracH$ has order $p$ with $p$ prime, or has order $pq$ with primes $p<q$ and $pnmid q-1$, then $G/H$ is cyclic.
If $H$ is special, then $G/H$ is cyclic, e.g., if $H=G$.
answered Aug 3 at 20:11
Dietrich Burde
74.5k64084
What is a special subgroup?
– lhf
Aug 3 at 21:18
I wanted to say, if we chose a special subgroup, e.g., some specific example like $H=G$, as a "good condition on $H$".
– Dietrich Burde
2 days ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Whenever $fracH$ has order $p$ with $p$ prime, or has order $pq$ with primes $p<q$ and $pnmid q-1$, then $G/H$ is cyclic.
If $H$ is special, then $G/H$ is cyclic, e.g., if $H=G$.
answered Aug 3 at 20:11
Dietrich Burde
74.5k64084
Whenever $fracH$ has order $p$ with $p$ prime, or has order $pq$ with primes $p<q$ and $pnmid q-1$, then $G/H$ is cyclic.
If $H$ is special, then $G/H$ is cyclic, e.g., if $H=G$.
answered Aug 3 at 20:11
Dietrich Burde
74.5k64084
answered Aug 3 at 20:11
Dietrich Burde
74.5k64084
answered Aug 3 at 20:11
Dietrich Burde
74.5k64084
answered Aug 3 at 20:11
Dietrich Burde
74.5k64084
74.5k64084
What is a special subgroup?
– lhf
Aug 3 at 21:18
I wanted to say, if we chose a special subgroup, e.g., some specific example like $H=G$, as a "good condition on $H$".
– Dietrich Burde
2 days ago
add a comment |Â
What is a special subgroup?
– lhf
Aug 3 at 21:18
I wanted to say, if we chose a special subgroup, e.g., some specific example like $H=G$, as a "good condition on $H$".
– Dietrich Burde
2 days ago
What is a special subgroup?
– lhf
Aug 3 at 21:18
What is a special subgroup?
– lhf
Aug 3 at 21:18
I wanted to say, if we chose a special subgroup, e.g., some specific example like $H=G$, as a "good condition on $H$".
– Dietrich Burde
2 days ago
I wanted to say, if we chose a special subgroup, e.g., some specific example like $H=G$, as a "good condition on $H$".
– Dietrich Burde
2 days ago
add a comment |Â
up vote
3
down vote
Another sufficient condition that captures some of those already mentioned is that $gcd([G:H], phi([G:H]))= 1$, where $phi$ is Euler's function. This is a condition that guarantees that every group of order $[G:H]$ is cyclic, and includes numbers, such as $255$, that are products of more than two primes.
answered Aug 3 at 22:56
James
6,68721426
add a comment |Â
up vote
3
down vote
Another sufficient condition that captures some of those already mentioned is that $gcd([G:H], phi([G:H]))= 1$, where $phi$ is Euler's function. This is a condition that guarantees that every group of order $[G:H]$ is cyclic, and includes numbers, such as $255$, that are products of more than two primes.
answered Aug 3 at 22:56
James
6,68721426
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Another sufficient condition that captures some of those already mentioned is that $gcd([G:H], phi([G:H]))= 1$, where $phi$ is Euler's function. This is a condition that guarantees that every group of order $[G:H]$ is cyclic, and includes numbers, such as $255$, that are products of more than two primes.
answered Aug 3 at 22:56
James
6,68721426
Another sufficient condition that captures some of those already mentioned is that $gcd([G:H], phi([G:H]))= 1$, where $phi$ is Euler's function. This is a condition that guarantees that every group of order $[G:H]$ is cyclic, and includes numbers, such as $255$, that are products of more than two primes.
answered Aug 3 at 22:56
James
6,68721426
answered Aug 3 at 22:56
James
6,68721426
answered Aug 3 at 22:56
James
6,68721426
answered Aug 3 at 22:56
James
6,68721426
6,68721426
add a comment |Â
add a comment |Â
up vote
1
down vote
A sufficient condition (although not a necessary one) is that $fracH$ is a prime number.
answered Aug 3 at 18:15
José Carlos Santos
112k1696172
add a comment |Â
up vote
1
down vote
A sufficient condition (although not a necessary one) is that $fracH$ is a prime number.
answered Aug 3 at 18:15
José Carlos Santos
112k1696172
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A sufficient condition (although not a necessary one) is that $fracH$ is a prime number.
answered Aug 3 at 18:15
José Carlos Santos
112k1696172
A sufficient condition (although not a necessary one) is that $fracH$ is a prime number.
answered Aug 3 at 18:15
José Carlos Santos
112k1696172
answered Aug 3 at 18:15
José Carlos Santos
112k1696172
answered Aug 3 at 18:15
José Carlos Santos
112k1696172
answered Aug 3 at 18:15
José Carlos Santos
112k1696172
112k1696172
add a comment |Â
add a comment |Â
up vote
1
down vote
Let $G$ be a group and $H$ be a normal subgroup, the easiest condition that I can think of is that $fracH$ is a prime number.
If $G$ is not a solvable group and $H$ is solvable, then $G/H$ must not be cyclic.
Another equivalent condition that you may find useful is that, $G/H$ is cyclic if and only if for any $gin Gbackslash H$, the minimal number $n$ such that $g^nin H$ must satisfy $n=fracH$.
answered Aug 3 at 18:18
ElfHog
50214
add a comment |Â
up vote
1
down vote
Let $G$ be a group and $H$ be a normal subgroup, the easiest condition that I can think of is that $fracH$ is a prime number.
If $G$ is not a solvable group and $H$ is solvable, then $G/H$ must not be cyclic.
Another equivalent condition that you may find useful is that, $G/H$ is cyclic if and only if for any $gin Gbackslash H$, the minimal number $n$ such that $g^nin H$ must satisfy $n=fracH$.
answered Aug 3 at 18:18
ElfHog
50214
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $G$ be a group and $H$ be a normal subgroup, the easiest condition that I can think of is that $fracH$ is a prime number.
If $G$ is not a solvable group and $H$ is solvable, then $G/H$ must not be cyclic.
Another equivalent condition that you may find useful is that, $G/H$ is cyclic if and only if for any $gin Gbackslash H$, the minimal number $n$ such that $g^nin H$ must satisfy $n=fracH$.
answered Aug 3 at 18:18
ElfHog
50214
Let $G$ be a group and $H$ be a normal subgroup, the easiest condition that I can think of is that $fracH$ is a prime number.
If $G$ is not a solvable group and $H$ is solvable, then $G/H$ must not be cyclic.
Another equivalent condition that you may find useful is that, $G/H$ is cyclic if and only if for any $gin Gbackslash H$, the minimal number $n$ such that $g^nin H$ must satisfy $n=fracH$.
answered Aug 3 at 18:18
ElfHog
50214
answered Aug 3 at 18:18
ElfHog
50214
answered Aug 3 at 18:18
ElfHog
50214
answered Aug 3 at 18:18
ElfHog
50214
50214
add a comment |Â
add a comment |Â
up vote
1
down vote
It is of course a necessary condition that $H$ contains the derived subgruop $[G,G]$ of $G$, as $G/H$ is commutative. Suppose that you have a set $X$ of generators for $G$. You could then check that, for each prime divisor $p$ of the index $[G:H]$, the subgroup $langle x^pmid xin Xrangle H$ has index $p$ in $G$. (This just amounts to saying that the $p$-th powers of a generating set for the quotient group $G/H$ generate an index $p$ subgroup.)
answered Aug 4 at 0:19
James
6,68721426
add a comment |Â
up vote
1
down vote
It is of course a necessary condition that $H$ contains the derived subgruop $[G,G]$ of $G$, as $G/H$ is commutative. Suppose that you have a set $X$ of generators for $G$. You could then check that, for each prime divisor $p$ of the index $[G:H]$, the subgroup $langle x^pmid xin Xrangle H$ has index $p$ in $G$. (This just amounts to saying that the $p$-th powers of a generating set for the quotient group $G/H$ generate an index $p$ subgroup.)
answered Aug 4 at 0:19
James
6,68721426
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is of course a necessary condition that $H$ contains the derived subgruop $[G,G]$ of $G$, as $G/H$ is commutative. Suppose that you have a set $X$ of generators for $G$. You could then check that, for each prime divisor $p$ of the index $[G:H]$, the subgroup $langle x^pmid xin Xrangle H$ has index $p$ in $G$. (This just amounts to saying that the $p$-th powers of a generating set for the quotient group $G/H$ generate an index $p$ subgroup.)
answered Aug 4 at 0:19
James
6,68721426
It is of course a necessary condition that $H$ contains the derived subgruop $[G,G]$ of $G$, as $G/H$ is commutative. Suppose that you have a set $X$ of generators for $G$. You could then check that, for each prime divisor $p$ of the index $[G:H]$, the subgroup $langle x^pmid xin Xrangle H$ has index $p$ in $G$. (This just amounts to saying that the $p$-th powers of a generating set for the quotient group $G/H$ generate an index $p$ subgroup.)
answered Aug 4 at 0:19
James
6,68721426
answered Aug 4 at 0:19
James
6,68721426
answered Aug 4 at 0:19
James
6,68721426
answered Aug 4 at 0:19
James
6,68721426
6,68721426
add a comment |Â
add a comment |Â
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3
$G$ is abelian and $H$ is the center of $G$ :p
– leibnewtz
Aug 3 at 18:11