Can a degree $p^n$ field extension always be factored in a sequence of prime extensions?












4












$begingroup$


Suppose $L/K$ is a field extension of degree $p^n$ for some prime $p$ (if necessary, assume the characteristic of $K$ is not $p$).



Then, is it always possible to find a sequence of extensions $K = K_0 subset K_1 subset K_2 dots subset K_n = L$ such that $[K_r:K_{r-1}] = p$?



Using Galois theory, this problem translates into the following:



Suppose $G$ is a finite group with a subgroup $H$ such that $[G:H] = p^n$. Is it always possible to find a subgroup $G supset H' supset H$ so that $[H':H] = p$?






group-theory field-theory galois-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, let me correct that, thanks!
    $endgroup$
    – Asvin
    Dec 7 '18 at 23:05






  • 1




    $begingroup$
    This is possible if $L/K$ is Galois. Note that your Galois theory translation only works if $L/K$ is separable.
    $endgroup$
    – Qiaochu Yuan
    Dec 7 '18 at 23:24








  • 1




    $begingroup$
    If the extension is Galois, though, the answer is “yes”, since every group of order $p^n$ has normal subgroups of order $p$
    $endgroup$
    – Arturo Magidin
    Dec 7 '18 at 23:34










  • $begingroup$
    If the extension is purely inseparable the answer is "Yes" since we have for every $xin L$ that $x^{p^j}in K$ for some $j$. Take the $xin Lsetminus K$ and take $k$ such that $x^{p^k}notin K$ but $x^{p^{k+1}}in K$. Then $K(x^{p^k})$ has dimension $p$ over $K$ and now induct.
    $endgroup$
    – C Monsour
    Dec 8 '18 at 3:20


















4












$begingroup$


Suppose $L/K$ is a field extension of degree $p^n$ for some prime $p$ (if necessary, assume the characteristic of $K$ is not $p$).



Then, is it always possible to find a sequence of extensions $K = K_0 subset K_1 subset K_2 dots subset K_n = L$ such that $[K_r:K_{r-1}] = p$?



Using Galois theory, this problem translates into the following:



Suppose $G$ is a finite group with a subgroup $H$ such that $[G:H] = p^n$. Is it always possible to find a subgroup $G supset H' supset H$ so that $[H':H] = p$?






group-theory field-theory galois-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, let me correct that, thanks!
    $endgroup$
    – Asvin
    Dec 7 '18 at 23:05






  • 1




    $begingroup$
    This is possible if $L/K$ is Galois. Note that your Galois theory translation only works if $L/K$ is separable.
    $endgroup$
    – Qiaochu Yuan
    Dec 7 '18 at 23:24








  • 1




    $begingroup$
    If the extension is Galois, though, the answer is “yes”, since every group of order $p^n$ has normal subgroups of order $p$
    $endgroup$
    – Arturo Magidin
    Dec 7 '18 at 23:34










  • $begingroup$
    If the extension is purely inseparable the answer is "Yes" since we have for every $xin L$ that $x^{p^j}in K$ for some $j$. Take the $xin Lsetminus K$ and take $k$ such that $x^{p^k}notin K$ but $x^{p^{k+1}}in K$. Then $K(x^{p^k})$ has dimension $p$ over $K$ and now induct.
    $endgroup$
    – C Monsour
    Dec 8 '18 at 3:20
















4












4








4





$begingroup$


Suppose $L/K$ is a field extension of degree $p^n$ for some prime $p$ (if necessary, assume the characteristic of $K$ is not $p$).



Then, is it always possible to find a sequence of extensions $K = K_0 subset K_1 subset K_2 dots subset K_n = L$ such that $[K_r:K_{r-1}] = p$?



Using Galois theory, this problem translates into the following:



Suppose $G$ is a finite group with a subgroup $H$ such that $[G:H] = p^n$. Is it always possible to find a subgroup $G supset H' supset H$ so that $[H':H] = p$?






group-theory field-theory galois-theory







share|cite|improve this question











$endgroup$




Suppose $L/K$ is a field extension of degree $p^n$ for some prime $p$ (if necessary, assume the characteristic of $K$ is not $p$).



Then, is it always possible to find a sequence of extensions $K = K_0 subset K_1 subset K_2 dots subset K_n = L$ such that $[K_r:K_{r-1}] = p$?



Using Galois theory, this problem translates into the following:



Suppose $G$ is a finite group with a subgroup $H$ such that $[G:H] = p^n$. Is it always possible to find a subgroup $G supset H' supset H$ so that $[H':H] = p$?






group-theory field-theory galois-theory




group-theory field-theory galois-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 23:06







Asvin

















asked Dec 7 '18 at 23:03









AsvinAsvin

3,41321331




3,41321331












  • $begingroup$
    Yes, let me correct that, thanks!
    $endgroup$
    – Asvin
    Dec 7 '18 at 23:05






  • 1




    $begingroup$
    This is possible if $L/K$ is Galois. Note that your Galois theory translation only works if $L/K$ is separable.
    $endgroup$
    – Qiaochu Yuan
    Dec 7 '18 at 23:24








  • 1




    $begingroup$
    If the extension is Galois, though, the answer is “yes”, since every group of order $p^n$ has normal subgroups of order $p$
    $endgroup$
    – Arturo Magidin
    Dec 7 '18 at 23:34










  • $begingroup$
    If the extension is purely inseparable the answer is "Yes" since we have for every $xin L$ that $x^{p^j}in K$ for some $j$. Take the $xin Lsetminus K$ and take $k$ such that $x^{p^k}notin K$ but $x^{p^{k+1}}in K$. Then $K(x^{p^k})$ has dimension $p$ over $K$ and now induct.
    $endgroup$
    – C Monsour
    Dec 8 '18 at 3:20




















  • $begingroup$
    Yes, let me correct that, thanks!
    $endgroup$
    – Asvin
    Dec 7 '18 at 23:05






  • 1




    $begingroup$
    This is possible if $L/K$ is Galois. Note that your Galois theory translation only works if $L/K$ is separable.
    $endgroup$
    – Qiaochu Yuan
    Dec 7 '18 at 23:24








  • 1




    $begingroup$
    If the extension is Galois, though, the answer is “yes”, since every group of order $p^n$ has normal subgroups of order $p$
    $endgroup$
    – Arturo Magidin
    Dec 7 '18 at 23:34










  • $begingroup$
    If the extension is purely inseparable the answer is "Yes" since we have for every $xin L$ that $x^{p^j}in K$ for some $j$. Take the $xin Lsetminus K$ and take $k$ such that $x^{p^k}notin K$ but $x^{p^{k+1}}in K$. Then $K(x^{p^k})$ has dimension $p$ over $K$ and now induct.
    $endgroup$
    – C Monsour
    Dec 8 '18 at 3:20


















$begingroup$
Yes, let me correct that, thanks!
$endgroup$
– Asvin
Dec 7 '18 at 23:05




$begingroup$
Yes, let me correct that, thanks!
$endgroup$
– Asvin
Dec 7 '18 at 23:05




1




1




$begingroup$
This is possible if $L/K$ is Galois. Note that your Galois theory translation only works if $L/K$ is separable.
$endgroup$
– Qiaochu Yuan
Dec 7 '18 at 23:24






$begingroup$
This is possible if $L/K$ is Galois. Note that your Galois theory translation only works if $L/K$ is separable.
$endgroup$
– Qiaochu Yuan
Dec 7 '18 at 23:24






1




1




$begingroup$
If the extension is Galois, though, the answer is “yes”, since every group of order $p^n$ has normal subgroups of order $p$
$endgroup$
– Arturo Magidin
Dec 7 '18 at 23:34




$begingroup$
If the extension is Galois, though, the answer is “yes”, since every group of order $p^n$ has normal subgroups of order $p$
$endgroup$
– Arturo Magidin
Dec 7 '18 at 23:34












$begingroup$
If the extension is purely inseparable the answer is "Yes" since we have for every $xin L$ that $x^{p^j}in K$ for some $j$. Take the $xin Lsetminus K$ and take $k$ such that $x^{p^k}notin K$ but $x^{p^{k+1}}in K$. Then $K(x^{p^k})$ has dimension $p$ over $K$ and now induct.
$endgroup$
– C Monsour
Dec 8 '18 at 3:20






$begingroup$
If the extension is purely inseparable the answer is "Yes" since we have for every $xin L$ that $x^{p^j}in K$ for some $j$. Take the $xin Lsetminus K$ and take $k$ such that $x^{p^k}notin K$ but $x^{p^{k+1}}in K$. Then $K(x^{p^k})$ has dimension $p$ over $K$ and now induct.
$endgroup$
– C Monsour
Dec 8 '18 at 3:20












1 Answer
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5












$begingroup$

No, this is not always possible. For instance, consider $G=A_4$. Then $G$ has a subgroup of index $2^2$ (any subgroup generated by a $3$-cycle) but has no subgroup of index $2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Trying to thwart an eventual follow-up question with the edit :-) The key really is the lack of intermediate groups of prescribed order. For example $G=S_4$ has a subgroup of order $12$, but no intermediate subgroups of order $12$ between $G$ and a point stabilizer $H=S_3$ of index four.
    $endgroup$
    – Jyrki Lahtonen
    Dec 8 '18 at 6:45













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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

No, this is not always possible. For instance, consider $G=A_4$. Then $G$ has a subgroup of index $2^2$ (any subgroup generated by a $3$-cycle) but has no subgroup of index $2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Trying to thwart an eventual follow-up question with the edit :-) The key really is the lack of intermediate groups of prescribed order. For example $G=S_4$ has a subgroup of order $12$, but no intermediate subgroups of order $12$ between $G$ and a point stabilizer $H=S_3$ of index four.
    $endgroup$
    – Jyrki Lahtonen
    Dec 8 '18 at 6:45


















5












$begingroup$

No, this is not always possible. For instance, consider $G=A_4$. Then $G$ has a subgroup of index $2^2$ (any subgroup generated by a $3$-cycle) but has no subgroup of index $2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Trying to thwart an eventual follow-up question with the edit :-) The key really is the lack of intermediate groups of prescribed order. For example $G=S_4$ has a subgroup of order $12$, but no intermediate subgroups of order $12$ between $G$ and a point stabilizer $H=S_3$ of index four.
    $endgroup$
    – Jyrki Lahtonen
    Dec 8 '18 at 6:45
















5












5








5





$begingroup$

No, this is not always possible. For instance, consider $G=A_4$. Then $G$ has a subgroup of index $2^2$ (any subgroup generated by a $3$-cycle) but has no subgroup of index $2$.






share|cite|improve this answer











$endgroup$



No, this is not always possible. For instance, consider $G=A_4$. Then $G$ has a subgroup of index $2^2$ (any subgroup generated by a $3$-cycle) but has no subgroup of index $2$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 6:43









Jyrki Lahtonen

109k13169372




109k13169372










answered Dec 7 '18 at 23:09









Eric WofseyEric Wofsey

185k14214341




185k14214341












  • $begingroup$
    Trying to thwart an eventual follow-up question with the edit :-) The key really is the lack of intermediate groups of prescribed order. For example $G=S_4$ has a subgroup of order $12$, but no intermediate subgroups of order $12$ between $G$ and a point stabilizer $H=S_3$ of index four.
    $endgroup$
    – Jyrki Lahtonen
    Dec 8 '18 at 6:45




















  • $begingroup$
    Trying to thwart an eventual follow-up question with the edit :-) The key really is the lack of intermediate groups of prescribed order. For example $G=S_4$ has a subgroup of order $12$, but no intermediate subgroups of order $12$ between $G$ and a point stabilizer $H=S_3$ of index four.
    $endgroup$
    – Jyrki Lahtonen
    Dec 8 '18 at 6:45


















$begingroup$
Trying to thwart an eventual follow-up question with the edit :-) The key really is the lack of intermediate groups of prescribed order. For example $G=S_4$ has a subgroup of order $12$, but no intermediate subgroups of order $12$ between $G$ and a point stabilizer $H=S_3$ of index four.
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 6:45






$begingroup$
Trying to thwart an eventual follow-up question with the edit :-) The key really is the lack of intermediate groups of prescribed order. For example $G=S_4$ has a subgroup of order $12$, but no intermediate subgroups of order $12$ between $G$ and a point stabilizer $H=S_3$ of index four.
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 6:45




















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