Let $(G, cdot)$ be a finite group, with $H triangleleft G$ and $[G:H]$ prime. What are the consequences of...
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I have a problem where I am given these two statements as a fact. However, I am having trouble figuring out what the relevant consequences of these statements would be, since I have only been able to determine one:
$G/H$ is a group, since $H triangleleft G$; furthermore, $left|G/Hright| = [G:H] = p$ where $p$ is prime, so $G/H$ must be cyclic. This means that for all $g in G$:
begin{align*}
&quad; (g cdot H)^{p} = H \
&Leftrightarrow g^{p} cdot H = H
end{align*}
What are some other interesting consequences of these statements? Please do not give me a proof to go along with them, as I would prefer to try and prove them myself.
group-theory finite-groups
asked Dec 1 '18 at 20:03
user89user89
6861647
$endgroup$
closed as unclear what you're asking by Arturo Magidin, Brahadeesh, Rebellos, José Carlos Santos, ancientmathematician Dec 7 '18 at 17:50
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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show 4 more comments
$begingroup$
I have a problem where I am given these two statements as a fact. However, I am having trouble figuring out what the relevant consequences of these statements would be, since I have only been able to determine one:
$G/H$ is a group, since $H triangleleft G$; furthermore, $left|G/Hright| = [G:H] = p$ where $p$ is prime, so $G/H$ must be cyclic. This means that for all $g in G$:
begin{align*}
&quad; (g cdot H)^{p} = H \
&Leftrightarrow g^{p} cdot H = H
end{align*}
What are some other interesting consequences of these statements? Please do not give me a proof to go along with them, as I would prefer to try and prove them myself.
group-theory finite-groups
asked Dec 1 '18 at 20:03
user89user89
6861647
$endgroup$
closed as unclear what you're asking by Arturo Magidin, Brahadeesh, Rebellos, José Carlos Santos, ancientmathematician Dec 7 '18 at 17:50
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Did I distill correctly that you want to prove a) “G/H must be cyclic” and b) the equivalence you mentioned? Or is only the first part interesting to you? They seem rather independent.
$endgroup$
– Luke
Dec 1 '18 at 20:10
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@Luke I want neither. The bullet point I wrote out is a consequence of $H triangleleft G$, and $[G:H]$ being prime. I want to know the statements of some other consequences (but I do not want their proofs!).
$endgroup$
– user89
Dec 1 '18 at 20:23
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Then what are the consequences you want to find? You stated you „have only been able to determine one“. I'm thus not sure what a useful “consequence” constitutes for you.
$endgroup$
– Luke
Dec 1 '18 at 20:26
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@Luke I never asked for "useful"? I am just looking for some simple, interesting consequences, beyond the single one I have thought of. I am hoping that by proving these consequences, I might be able to expand my imagination and be able to figure out the main problem.
$endgroup$
– user89
Dec 1 '18 at 20:27
1
$begingroup$
With $gH=Hg$ a generator of $G/H$ you can look at the properties of the permutation $f : H to H$ such that $gh = f(h) g$. For $p nmid |H|$ it is a semi-direct product.
$endgroup$
– reuns
Dec 1 '18 at 21:22
|
show 4 more comments
$begingroup$
I have a problem where I am given these two statements as a fact. However, I am having trouble figuring out what the relevant consequences of these statements would be, since I have only been able to determine one:
$G/H$ is a group, since $H triangleleft G$; furthermore, $left|G/Hright| = [G:H] = p$ where $p$ is prime, so $G/H$ must be cyclic. This means that for all $g in G$:
begin{align*}
&quad; (g cdot H)^{p} = H \
&Leftrightarrow g^{p} cdot H = H
end{align*}
What are some other interesting consequences of these statements? Please do not give me a proof to go along with them, as I would prefer to try and prove them myself.
group-theory finite-groups
asked Dec 1 '18 at 20:03
user89user89
6861647
$endgroup$
I have a problem where I am given these two statements as a fact. However, I am having trouble figuring out what the relevant consequences of these statements would be, since I have only been able to determine one:
$G/H$ is a group, since $H triangleleft G$; furthermore, $left|G/Hright| = [G:H] = p$ where $p$ is prime, so $G/H$ must be cyclic. This means that for all $g in G$:
begin{align*}
&quad; (g cdot H)^{p} = H \
&Leftrightarrow g^{p} cdot H = H
end{align*}
What are some other interesting consequences of these statements? Please do not give me a proof to go along with them, as I would prefer to try and prove them myself.
group-theory finite-groups
group-theory finite-groups
asked Dec 1 '18 at 20:03
user89user89
6861647
asked Dec 1 '18 at 20:03
user89user89
6861647
edited Dec 1 '18 at 22:56
user89
asked Dec 1 '18 at 20:03
user89user89
6861647
asked Dec 1 '18 at 20:03
user89user89
6861647
asked Dec 1 '18 at 20:03
user89user89
6861647
6861647
closed as unclear what you're asking by Arturo Magidin, Brahadeesh, Rebellos, José Carlos Santos, ancientmathematician Dec 7 '18 at 17:50
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Arturo Magidin, Brahadeesh, Rebellos, José Carlos Santos, ancientmathematician Dec 7 '18 at 17:50
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Did I distill correctly that you want to prove a) “G/H must be cyclic” and b) the equivalence you mentioned? Or is only the first part interesting to you? They seem rather independent.
$endgroup$
– Luke
Dec 1 '18 at 20:10
$begingroup$
@Luke I want neither. The bullet point I wrote out is a consequence of $H triangleleft G$, and $[G:H]$ being prime. I want to know the statements of some other consequences (but I do not want their proofs!).
$endgroup$
– user89
Dec 1 '18 at 20:23
$begingroup$
Then what are the consequences you want to find? You stated you „have only been able to determine one“. I'm thus not sure what a useful “consequence” constitutes for you.
$endgroup$
– Luke
Dec 1 '18 at 20:26
$begingroup$
@Luke I never asked for "useful"? I am just looking for some simple, interesting consequences, beyond the single one I have thought of. I am hoping that by proving these consequences, I might be able to expand my imagination and be able to figure out the main problem.
$endgroup$
– user89
Dec 1 '18 at 20:27
1
$begingroup$
With $gH=Hg$ a generator of $G/H$ you can look at the properties of the permutation $f : H to H$ such that $gh = f(h) g$. For $p nmid |H|$ it is a semi-direct product.
$endgroup$
– reuns
Dec 1 '18 at 21:22
|
show 4 more comments
$begingroup$
Did I distill correctly that you want to prove a) “G/H must be cyclic” and b) the equivalence you mentioned? Or is only the first part interesting to you? They seem rather independent.
$endgroup$
– Luke
Dec 1 '18 at 20:10
$begingroup$
@Luke I want neither. The bullet point I wrote out is a consequence of $H triangleleft G$, and $[G:H]$ being prime. I want to know the statements of some other consequences (but I do not want their proofs!).
$endgroup$
– user89
Dec 1 '18 at 20:23
$begingroup$
Then what are the consequences you want to find? You stated you „have only been able to determine one“. I'm thus not sure what a useful “consequence” constitutes for you.
$endgroup$
– Luke
Dec 1 '18 at 20:26
$begingroup$
@Luke I never asked for "useful"? I am just looking for some simple, interesting consequences, beyond the single one I have thought of. I am hoping that by proving these consequences, I might be able to expand my imagination and be able to figure out the main problem.
$endgroup$
– user89
Dec 1 '18 at 20:27
1
$begingroup$
With $gH=Hg$ a generator of $G/H$ you can look at the properties of the permutation $f : H to H$ such that $gh = f(h) g$. For $p nmid |H|$ it is a semi-direct product.
$endgroup$
– reuns
Dec 1 '18 at 21:22
$begingroup$
Did I distill correctly that you want to prove a) “G/H must be cyclic” and b) the equivalence you mentioned? Or is only the first part interesting to you? They seem rather independent.
$endgroup$
– Luke
Dec 1 '18 at 20:10
$begingroup$
Did I distill correctly that you want to prove a) “G/H must be cyclic” and b) the equivalence you mentioned? Or is only the first part interesting to you? They seem rather independent.
$endgroup$
– Luke
Dec 1 '18 at 20:10
$begingroup$
@Luke I want neither. The bullet point I wrote out is a consequence of $H triangleleft G$, and $[G:H]$ being prime. I want to know the statements of some other consequences (but I do not want their proofs!).
$endgroup$
– user89
Dec 1 '18 at 20:23
$begingroup$
@Luke I want neither. The bullet point I wrote out is a consequence of $H triangleleft G$, and $[G:H]$ being prime. I want to know the statements of some other consequences (but I do not want their proofs!).
$endgroup$
– user89
Dec 1 '18 at 20:23
$begingroup$
Then what are the consequences you want to find? You stated you „have only been able to determine one“. I'm thus not sure what a useful “consequence” constitutes for you.
$endgroup$
– Luke
Dec 1 '18 at 20:26
$begingroup$
Then what are the consequences you want to find? You stated you „have only been able to determine one“. I'm thus not sure what a useful “consequence” constitutes for you.
$endgroup$
– Luke
Dec 1 '18 at 20:26
$begingroup$
@Luke I never asked for "useful"? I am just looking for some simple, interesting consequences, beyond the single one I have thought of. I am hoping that by proving these consequences, I might be able to expand my imagination and be able to figure out the main problem.
$endgroup$
– user89
Dec 1 '18 at 20:27
$begingroup$
@Luke I never asked for "useful"? I am just looking for some simple, interesting consequences, beyond the single one I have thought of. I am hoping that by proving these consequences, I might be able to expand my imagination and be able to figure out the main problem.
$endgroup$
– user89
Dec 1 '18 at 20:27
1
1
$begingroup$
With $gH=Hg$ a generator of $G/H$ you can look at the properties of the permutation $f : H to H$ such that $gh = f(h) g$. For $p nmid |H|$ it is a semi-direct product.
$endgroup$
– reuns
Dec 1 '18 at 21:22
$begingroup$
With $gH=Hg$ a generator of $G/H$ you can look at the properties of the permutation $f : H to H$ such that $gh = f(h) g$. For $p nmid |H|$ it is a semi-direct product.
$endgroup$
– reuns
Dec 1 '18 at 21:22
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
In general, you can't say much.
Ex 1. The group $Q_8$ of quaternions has 3 subgroups of order $4$, that are normal and yet there is no structural implication of their index.
Rem 2. On the other hand, when $|H|$ is coprime with its index you can even prove that $G cong H rtimes mathbb{Z}_p$.
answered Dec 1 '18 at 23:05
Ivan Di LibertiIvan Di Liberti
2,57311122
$endgroup$
2
$begingroup$
Second part is not right. You are only guaranteed a semidirect product.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 23:09
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In general, you can't say much.
Ex 1. The group $Q_8$ of quaternions has 3 subgroups of order $4$, that are normal and yet there is no structural implication of their index.
Rem 2. On the other hand, when $|H|$ is coprime with its index you can even prove that $G cong H rtimes mathbb{Z}_p$.
answered Dec 1 '18 at 23:05
Ivan Di LibertiIvan Di Liberti
2,57311122
$endgroup$
2
$begingroup$
Second part is not right. You are only guaranteed a semidirect product.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 23:09
add a comment |
$begingroup$
In general, you can't say much.
Ex 1. The group $Q_8$ of quaternions has 3 subgroups of order $4$, that are normal and yet there is no structural implication of their index.
Rem 2. On the other hand, when $|H|$ is coprime with its index you can even prove that $G cong H rtimes mathbb{Z}_p$.
answered Dec 1 '18 at 23:05
Ivan Di LibertiIvan Di Liberti
2,57311122
$endgroup$
2
$begingroup$
Second part is not right. You are only guaranteed a semidirect product.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 23:09
add a comment |
$begingroup$
In general, you can't say much.
Ex 1. The group $Q_8$ of quaternions has 3 subgroups of order $4$, that are normal and yet there is no structural implication of their index.
Rem 2. On the other hand, when $|H|$ is coprime with its index you can even prove that $G cong H rtimes mathbb{Z}_p$.
answered Dec 1 '18 at 23:05
Ivan Di LibertiIvan Di Liberti
2,57311122
$endgroup$
In general, you can't say much.
Ex 1. The group $Q_8$ of quaternions has 3 subgroups of order $4$, that are normal and yet there is no structural implication of their index.
Rem 2. On the other hand, when $|H|$ is coprime with its index you can even prove that $G cong H rtimes mathbb{Z}_p$.
answered Dec 1 '18 at 23:05
Ivan Di LibertiIvan Di Liberti
2,57311122
edited Dec 1 '18 at 23:10
answered Dec 1 '18 at 23:05
Ivan Di LibertiIvan Di Liberti
2,57311122
answered Dec 1 '18 at 23:05
Ivan Di LibertiIvan Di Liberti
2,57311122
answered Dec 1 '18 at 23:05
Ivan Di LibertiIvan Di Liberti
2,57311122
2,57311122
2
$begingroup$
Second part is not right. You are only guaranteed a semidirect product.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 23:09
add a comment |
2
$begingroup$
Second part is not right. You are only guaranteed a semidirect product.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 23:09
2
2
$begingroup$
Second part is not right. You are only guaranteed a semidirect product.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 23:09
$begingroup$
Second part is not right. You are only guaranteed a semidirect product.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 23:09
add a comment |
$begingroup$
Did I distill correctly that you want to prove a) “G/H must be cyclic” and b) the equivalence you mentioned? Or is only the first part interesting to you? They seem rather independent.
$endgroup$
– Luke
Dec 1 '18 at 20:10
$begingroup$
@Luke I want neither. The bullet point I wrote out is a consequence of $H triangleleft G$, and $[G:H]$ being prime. I want to know the statements of some other consequences (but I do not want their proofs!).
$endgroup$
– user89
Dec 1 '18 at 20:23
$begingroup$
Then what are the consequences you want to find? You stated you „have only been able to determine one“. I'm thus not sure what a useful “consequence” constitutes for you.
$endgroup$
– Luke
Dec 1 '18 at 20:26
$begingroup$
@Luke I never asked for "useful"? I am just looking for some simple, interesting consequences, beyond the single one I have thought of. I am hoping that by proving these consequences, I might be able to expand my imagination and be able to figure out the main problem.
$endgroup$
– user89
Dec 1 '18 at 20:27
1
$begingroup$
With $gH=Hg$ a generator of $G/H$ you can look at the properties of the permutation $f : H to H$ such that $gh = f(h) g$. For $p nmid |H|$ it is a semi-direct product.
$endgroup$
– reuns
Dec 1 '18 at 21:22