Upper central series, semidirect products and a lifting property (argument verification)

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Let $G$ be a profinite group, and define the lifting property: $(*_p)$ For every extension $1 to P to E to W to 1$ where $E$ is finite and $P$ is a $p$-group and for every surjective morphism $fcolon G to W$ there is a surjective extension $f'colon G to E$.

Suppose $G$ satisfies $(*_p)$ in the following two cases:

• (i) Every subgroup of $E$ which projects onto $W$ is equal to $E$.


• (ii) $P$ is abelian and killed by $p$, and $E$ is isomorphic to the semidirect product $P rtimes W$

Then, I want to prove that $G$ satisfies the general case. My progress so far was taking the long upper central series $P_i$ for $P$ (with $P_0 = 1$ and $P_n = P$ for some $n in mathbbN$), and noticing that we have surjections $$E = E/P_0 to E/P_1 to cdots to E/P_n-1 to E/P_n = E/P = W to 1$$
with kernels $P_i/P_i-1$ that are abelian groups killed by $p$. If we have the lifting property for the extension $$1 to P_n/P_n-1 to E/P_n-1 to W to 1$$ then we have it for the general case. I'm not sure how can I use (i) and (ii) now.. If $p notmid |W|$ than (ii) easily applies, but that need not be the case.

For context, this is part of exercise 1 in section 3.4 of chapter 1 in Serre's Galois Cohomology.

---

PROGRESS I know how to reduce the problem a little using (i). Using the reductions as before, we can suppose that $P$ is abelian and killed by $p$. It is a known fact that $P$ has a minimal subgroup complement $K$ in $E$: $PK = E$ and for any $L leq K$ s.t. $PL = E$ we have $L = K$. But this means that the extension $$1 to P cap K to K to W to 1$$ satisfies (i), and thus we have a surjective extension of $fcolon G to W$ to $tildefcolon G to K$. How can we use (ii) now to lift it to $E$?

FINAL STRETCH We can suppose (WLOG by the remarks above, I guess) that $P = mathbbZ/pmathbbZ$. We have the minimal complement $K$ and a lifting $tildefcolon G to K$ of $fcolon G to W$. If $K neq E$ then the intersection $K cap P$ cannot equal $P$ and we must therefore have $K cap P = 1$, proving that there is an isomorphism $E = P rtimes K$. Then, we apply (ii) to get a lifting $f'colon G to E$. I am just unsure about reducing to $mathbbZ/pmathbbZ$...

Actually, maybe the reduction isn't even necessary... if $K neq E$ then $K cap P$ must be a proper subgroup of $P$. But it is a normal subgroup of $E$, and by the minimality of $P$ it must be trivial!

proof-verification p-groups semidirect-product profinite-groups group-extensions

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edited Jul 31 at 16:30

asked Jul 31 at 13:13

Henrique Augusto Souza

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  The saga continues on: math.stackexchange.com/questions/2868366/…
  – Henrique Augusto Souza
  Jul 31 at 18:57
  

  
  
  
  

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Let $G$ be a profinite group, and define the lifting property: $(*_p)$ For every extension $1 to P to E to W to 1$ where $E$ is finite and $P$ is a $p$-group and for every surjective morphism $fcolon G to W$ there is a surjective extension $f'colon G to E$.

Suppose $G$ satisfies $(*_p)$ in the following two cases:

• (i) Every subgroup of $E$ which projects onto $W$ is equal to $E$.


• (ii) $P$ is abelian and killed by $p$, and $E$ is isomorphic to the semidirect product $P rtimes W$

Then, I want to prove that $G$ satisfies the general case. My progress so far was taking the long upper central series $P_i$ for $P$ (with $P_0 = 1$ and $P_n = P$ for some $n in mathbbN$), and noticing that we have surjections $$E = E/P_0 to E/P_1 to cdots to E/P_n-1 to E/P_n = E/P = W to 1$$
with kernels $P_i/P_i-1$ that are abelian groups killed by $p$. If we have the lifting property for the extension $$1 to P_n/P_n-1 to E/P_n-1 to W to 1$$ then we have it for the general case. I'm not sure how can I use (i) and (ii) now.. If $p notmid |W|$ than (ii) easily applies, but that need not be the case.

For context, this is part of exercise 1 in section 3.4 of chapter 1 in Serre's Galois Cohomology.

---

PROGRESS I know how to reduce the problem a little using (i). Using the reductions as before, we can suppose that $P$ is abelian and killed by $p$. It is a known fact that $P$ has a minimal subgroup complement $K$ in $E$: $PK = E$ and for any $L leq K$ s.t. $PL = E$ we have $L = K$. But this means that the extension $$1 to P cap K to K to W to 1$$ satisfies (i), and thus we have a surjective extension of $fcolon G to W$ to $tildefcolon G to K$. How can we use (ii) now to lift it to $E$?

FINAL STRETCH We can suppose (WLOG by the remarks above, I guess) that $P = mathbbZ/pmathbbZ$. We have the minimal complement $K$ and a lifting $tildefcolon G to K$ of $fcolon G to W$. If $K neq E$ then the intersection $K cap P$ cannot equal $P$ and we must therefore have $K cap P = 1$, proving that there is an isomorphism $E = P rtimes K$. Then, we apply (ii) to get a lifting $f'colon G to E$. I am just unsure about reducing to $mathbbZ/pmathbbZ$...

Actually, maybe the reduction isn't even necessary... if $K neq E$ then $K cap P$ must be a proper subgroup of $P$. But it is a normal subgroup of $E$, and by the minimality of $P$ it must be trivial!

proof-verification p-groups semidirect-product profinite-groups group-extensions

share|cite|improve this question

edited Jul 31 at 16:30

asked Jul 31 at 13:13

Henrique Augusto Souza

1,226314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The saga continues on: math.stackexchange.com/questions/2868366/…
  – Henrique Augusto Souza
  Jul 31 at 18:57
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

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up vote
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down vote

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1

Let $G$ be a profinite group, and define the lifting property: $(*_p)$ For every extension $1 to P to E to W to 1$ where $E$ is finite and $P$ is a $p$-group and for every surjective morphism $fcolon G to W$ there is a surjective extension $f'colon G to E$.

Suppose $G$ satisfies $(*_p)$ in the following two cases:

• (i) Every subgroup of $E$ which projects onto $W$ is equal to $E$.


• (ii) $P$ is abelian and killed by $p$, and $E$ is isomorphic to the semidirect product $P rtimes W$

Then, I want to prove that $G$ satisfies the general case. My progress so far was taking the long upper central series $P_i$ for $P$ (with $P_0 = 1$ and $P_n = P$ for some $n in mathbbN$), and noticing that we have surjections $$E = E/P_0 to E/P_1 to cdots to E/P_n-1 to E/P_n = E/P = W to 1$$
with kernels $P_i/P_i-1$ that are abelian groups killed by $p$. If we have the lifting property for the extension $$1 to P_n/P_n-1 to E/P_n-1 to W to 1$$ then we have it for the general case. I'm not sure how can I use (i) and (ii) now.. If $p notmid |W|$ than (ii) easily applies, but that need not be the case.

For context, this is part of exercise 1 in section 3.4 of chapter 1 in Serre's Galois Cohomology.

---

PROGRESS I know how to reduce the problem a little using (i). Using the reductions as before, we can suppose that $P$ is abelian and killed by $p$. It is a known fact that $P$ has a minimal subgroup complement $K$ in $E$: $PK = E$ and for any $L leq K$ s.t. $PL = E$ we have $L = K$. But this means that the extension $$1 to P cap K to K to W to 1$$ satisfies (i), and thus we have a surjective extension of $fcolon G to W$ to $tildefcolon G to K$. How can we use (ii) now to lift it to $E$?

FINAL STRETCH We can suppose (WLOG by the remarks above, I guess) that $P = mathbbZ/pmathbbZ$. We have the minimal complement $K$ and a lifting $tildefcolon G to K$ of $fcolon G to W$. If $K neq E$ then the intersection $K cap P$ cannot equal $P$ and we must therefore have $K cap P = 1$, proving that there is an isomorphism $E = P rtimes K$. Then, we apply (ii) to get a lifting $f'colon G to E$. I am just unsure about reducing to $mathbbZ/pmathbbZ$...

Actually, maybe the reduction isn't even necessary... if $K neq E$ then $K cap P$ must be a proper subgroup of $P$. But it is a normal subgroup of $E$, and by the minimality of $P$ it must be trivial!

proof-verification p-groups semidirect-product profinite-groups group-extensions

share|cite|improve this question

edited Jul 31 at 16:30

asked Jul 31 at 13:13

Henrique Augusto Souza

1,226314

Let $G$ be a profinite group, and define the lifting property: $(*_p)$ For every extension $1 to P to E to W to 1$ where $E$ is finite and $P$ is a $p$-group and for every surjective morphism $fcolon G to W$ there is a surjective extension $f'colon G to E$.

Suppose $G$ satisfies $(*_p)$ in the following two cases:

• (i) Every subgroup of $E$ which projects onto $W$ is equal to $E$.


• (ii) $P$ is abelian and killed by $p$, and $E$ is isomorphic to the semidirect product $P rtimes W$

Then, I want to prove that $G$ satisfies the general case. My progress so far was taking the long upper central series $P_i$ for $P$ (with $P_0 = 1$ and $P_n = P$ for some $n in mathbbN$), and noticing that we have surjections $$E = E/P_0 to E/P_1 to cdots to E/P_n-1 to E/P_n = E/P = W to 1$$
with kernels $P_i/P_i-1$ that are abelian groups killed by $p$. If we have the lifting property for the extension $$1 to P_n/P_n-1 to E/P_n-1 to W to 1$$ then we have it for the general case. I'm not sure how can I use (i) and (ii) now.. If $p notmid |W|$ than (ii) easily applies, but that need not be the case.

For context, this is part of exercise 1 in section 3.4 of chapter 1 in Serre's Galois Cohomology.

---

PROGRESS I know how to reduce the problem a little using (i). Using the reductions as before, we can suppose that $P$ is abelian and killed by $p$. It is a known fact that $P$ has a minimal subgroup complement $K$ in $E$: $PK = E$ and for any $L leq K$ s.t. $PL = E$ we have $L = K$. But this means that the extension $$1 to P cap K to K to W to 1$$ satisfies (i), and thus we have a surjective extension of $fcolon G to W$ to $tildefcolon G to K$. How can we use (ii) now to lift it to $E$?

FINAL STRETCH We can suppose (WLOG by the remarks above, I guess) that $P = mathbbZ/pmathbbZ$. We have the minimal complement $K$ and a lifting $tildefcolon G to K$ of $fcolon G to W$. If $K neq E$ then the intersection $K cap P$ cannot equal $P$ and we must therefore have $K cap P = 1$, proving that there is an isomorphism $E = P rtimes K$. Then, we apply (ii) to get a lifting $f'colon G to E$. I am just unsure about reducing to $mathbbZ/pmathbbZ$...

Actually, maybe the reduction isn't even necessary... if $K neq E$ then $K cap P$ must be a proper subgroup of $P$. But it is a normal subgroup of $E$, and by the minimality of $P$ it must be trivial!

proof-verification p-groups semidirect-product profinite-groups group-extensions

share|cite|improve this question

edited Jul 31 at 16:30

asked Jul 31 at 13:13

Henrique Augusto Souza

1,226314

share|cite|improve this question

share|cite|improve this question

edited Jul 31 at 16:30

edited Jul 31 at 16:30

edited Jul 31 at 16:30

asked Jul 31 at 13:13

Henrique Augusto Souza

1,226314

asked Jul 31 at 13:13

Henrique Augusto Souza

1,226314

asked Jul 31 at 13:13

Henrique Augusto Souza

1,226314

1,226314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The saga continues on: math.stackexchange.com/questions/2868366/…
  – Henrique Augusto Souza
  Jul 31 at 18:57
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The saga continues on: math.stackexchange.com/questions/2868366/…
  – Henrique Augusto Souza
  Jul 31 at 18:57
  

  
  
  
  

The saga continues on: math.stackexchange.com/questions/2868366/…
– Henrique Augusto Souza
Jul 31 at 18:57

The saga continues on: math.stackexchange.com/questions/2868366/…
– Henrique Augusto Souza
Jul 31 at 18:57

add a comment | 

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