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A $p$-group of order $p^n$ has a normal subgroup of order $p^k$ for each $0le k le n$

$p$-groups have normal subgroups of each ordernormal subgroup of a group $G$Let $G$ be a group of order $p^n$ where $p$ is a prime and $n in mathbb{N}$. Prove that exist normal subgroupsLet be $G$ a group if order $p^n$ where p is a prime and n a natural number. Prove that there exist normal subgroups…Groups with “few” subgroupsProve that : A group $G$ of order $p^n$ has normal subgroup of order $p^k$ , for all $0≤k≤n$Let $G$ be a group of order $p^4$ for $p$ prime. Prove that $G'$ is abelianEvery group of order $567$ has normal subgroup of order $27$.Prove that $G$ is finite and $|G|$ is prime.Prove that there exists subgroup of any order of any power of $p$ in a $p$-groupFor any finite group, there is a homomorphism whose image is simpleProve that a group of order 30 has at least three different normal subgroupsAny group of order $12$ must contain a normal Sylow subgroupShow that if $G$ is a group of order $168$ that has a normal subgroup of order $4$ , then $G$ has a normal subgroup of order $28$Group of order $|G|=pqr$, $p,q,r$ primes has a normal subgroup of orderA group of order $pqr$ (primes $p > q > r$) has a subgroup of order $qr$Every group of order $567$ has normal subgroup of order $27$.A group of order 2010 has a normal subgroup of order 5Prove that a group of order $24$ has a normal subgroup of order $4$ or $8$Order of an arbitrary (finite) product of subgroups $|N_1cdots N_k|$

28

14

$begingroup$

This is problem 3 from Hungerford's section about the Sylow theorems.

I have already read hints saying to use induction and that $p$-groups always have non-trivial centres, but I'm still confused. This is what I have so far:

Suppose $|G| = p^n$. For $k = 0$, ${e}$ is a normal subgroup of order $p^0$. Suppose $N_1, ..., N_k$ are normal subgroups of $G$ with orders as described (ie. $|N_i| = p^i$) and $k < n$.

$G/N_k$ is a $p$-group, so it has a non-trivial centre $C(G/N_k)$. $pi^{-1}(C(G/N_k))$ is a subgroup of $G$, where $pi : G to G/N_k$ is the quotient map. It is normal in $G$ because $pi$ is a homomorphism and $C(G/N_k)$ is a normal subgroup of $G/N_k$.

I know it contains $N_k$ and some stuff not in $N_k$, so it has to have order at least $p^{k+1}$, but I do not know how to argue it must be equal, or indeed if it even is equal.

abstract-algebra group-theory finite-groups p-groups

share|cite|improve this question

edited Nov 3 '13 at 17:38

user104986

asked Nov 3 '13 at 0:57

user104986user104986

20028

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Certainly you mean $k leq log_p(|G|)$.
  $endgroup$
  – Yassine Guerboussa
  Nov 3 '13 at 8:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are correct, I fixed it.
  $endgroup$
  – user104986
  Nov 3 '13 at 17:38
  

  
  
  
  

add a comment | 

28

14

$begingroup$

This is problem 3 from Hungerford's section about the Sylow theorems.

I have already read hints saying to use induction and that $p$-groups always have non-trivial centres, but I'm still confused. This is what I have so far:

Suppose $|G| = p^n$. For $k = 0$, ${e}$ is a normal subgroup of order $p^0$. Suppose $N_1, ..., N_k$ are normal subgroups of $G$ with orders as described (ie. $|N_i| = p^i$) and $k < n$.

$G/N_k$ is a $p$-group, so it has a non-trivial centre $C(G/N_k)$. $pi^{-1}(C(G/N_k))$ is a subgroup of $G$, where $pi : G to G/N_k$ is the quotient map. It is normal in $G$ because $pi$ is a homomorphism and $C(G/N_k)$ is a normal subgroup of $G/N_k$.

I know it contains $N_k$ and some stuff not in $N_k$, so it has to have order at least $p^{k+1}$, but I do not know how to argue it must be equal, or indeed if it even is equal.

abstract-algebra group-theory finite-groups p-groups

share|cite|improve this question

edited Nov 3 '13 at 17:38

user104986

asked Nov 3 '13 at 0:57

user104986user104986

20028

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Certainly you mean $k leq log_p(|G|)$.
  $endgroup$
  – Yassine Guerboussa
  Nov 3 '13 at 8:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are correct, I fixed it.
  $endgroup$
  – user104986
  Nov 3 '13 at 17:38
  

  
  
  
  

add a comment | 

$begingroup$

This is problem 3 from Hungerford's section about the Sylow theorems.

I have already read hints saying to use induction and that $p$-groups always have non-trivial centres, but I'm still confused. This is what I have so far:

Suppose $|G| = p^n$. For $k = 0$, ${e}$ is a normal subgroup of order $p^0$. Suppose $N_1, ..., N_k$ are normal subgroups of $G$ with orders as described (ie. $|N_i| = p^i$) and $k < n$.

$G/N_k$ is a $p$-group, so it has a non-trivial centre $C(G/N_k)$. $pi^{-1}(C(G/N_k))$ is a subgroup of $G$, where $pi : G to G/N_k$ is the quotient map. It is normal in $G$ because $pi$ is a homomorphism and $C(G/N_k)$ is a normal subgroup of $G/N_k$.

I know it contains $N_k$ and some stuff not in $N_k$, so it has to have order at least $p^{k+1}$, but I do not know how to argue it must be equal, or indeed if it even is equal.

abstract-algebra group-theory finite-groups p-groups

share|cite|improve this question

edited Nov 3 '13 at 17:38

user104986

asked Nov 3 '13 at 0:57

user104986user104986

20028

$endgroup$

share|cite|improve this question

edited Nov 3 '13 at 17:38

user104986

asked Nov 3 '13 at 0:57

user104986user104986

20028

share|cite|improve this question

edited Nov 3 '13 at 17:38

user104986

asked Nov 3 '13 at 0:57

user104986user104986

20028

asked Nov 3 '13 at 0:57

user104986user104986

20028

asked Nov 3 '13 at 0:57

user104986user104986

20028

1 Answer
1

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12

$begingroup$

Hint: Apply Cauchy's theorem to $C(G/N_k)$. This will give you a normal subgroup of $G$ of order $p^{k+1}$.

share|cite|improve this answer

answered Nov 3 '13 at 1:20

Ayman HouriehAyman Hourieh

31.2k466119

$endgroup$

add a comment | 

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12

$begingroup$

Hint: Apply Cauchy's theorem to $C(G/N_k)$. This will give you a normal subgroup of $G$ of order $p^{k+1}$.

share|cite|improve this answer

answered Nov 3 '13 at 1:20

Ayman HouriehAyman Hourieh

31.2k466119

$endgroup$

add a comment | 

12

$begingroup$

Hint: Apply Cauchy's theorem to $C(G/N_k)$. This will give you a normal subgroup of $G$ of order $p^{k+1}$.

share|cite|improve this answer

answered Nov 3 '13 at 1:20

Ayman HouriehAyman Hourieh

31.2k466119

$endgroup$

add a comment | 

$begingroup$

Hint: Apply Cauchy's theorem to $C(G/N_k)$. This will give you a normal subgroup of $G$ of order $p^{k+1}$.

share|cite|improve this answer

answered Nov 3 '13 at 1:20

Ayman HouriehAyman Hourieh

31.2k466119

$endgroup$

share|cite|improve this answer

answered Nov 3 '13 at 1:20

Ayman HouriehAyman Hourieh

31.2k466119

answered Nov 3 '13 at 1:20

Ayman HouriehAyman Hourieh

31.2k466119

answered Nov 3 '13 at 1:20

Ayman HouriehAyman Hourieh

31.2k466119