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How to show that $H$ is normal in $G$?

how to show that $P$ is normal complement in $H$.Prove H is a normal subgroup of G.Please help me with proving a subgroup is normalShow that $D$ is a normal subgroup.Showing that every subgroup of an abelian group is normalFor all $gin G$, is it true that $gHg^-1 subseteq H $ if H is a normal subgroup of G.An exercise about normal subgroupsOrder of normal subgroupHow to show that a subgroup is normal.$PQ$ is normal implies $P$ and $Q$ are also normal

2

$begingroup$

Let $G$ be a group of order $pq$, where $p$ and $q$ are primes and $p>q$. Ler $ain G$ be of order $p$ and $H=big<a: a^p=1big>.$ Then $H$ is normal in $G$.
I know that $H$ will be normal subgroup of G if $gH=Hg$ or $H=gHg^-1$.
I tried as:

Let $gin G$ and $hin H$ then $h$ can be written in some power of $a$ but I don't, how to write $g$ and proceed to show that $H=ghg^-1$.

group-theory normal-subgroups

share|cite|improve this question

edited Mar 22 at 9:03

Shaun

10.4k113686

asked Mar 22 at 8:51

Noor AslamNoor Aslam

16012

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Are you familiar with Sylow's theorem?
  $endgroup$
  – Thomas Shelby
  Mar 22 at 8:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It suffices to show $g^-1Hgsubseteq H$ for all $gin G$.
  $endgroup$
  – Shaun
  Mar 22 at 9:00
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $G$ be a group of order $pq$, where $p$ and $q$ are primes and $p>q$. Ler $ain G$ be of order $p$ and $H=big<a: a^p=1big>.$ Then $H$ is normal in $G$.
I know that $H$ will be normal subgroup of G if $gH=Hg$ or $H=gHg^-1$.
I tried as:

Let $gin G$ and $hin H$ then $h$ can be written in some power of $a$ but I don't, how to write $g$ and proceed to show that $H=ghg^-1$.

group-theory normal-subgroups

share|cite|improve this question

edited Mar 22 at 9:03

Shaun

10.4k113686

asked Mar 22 at 8:51

Noor AslamNoor Aslam

16012

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Are you familiar with Sylow's theorem?
  $endgroup$
  – Thomas Shelby
  Mar 22 at 8:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It suffices to show $g^-1Hgsubseteq H$ for all $gin G$.
  $endgroup$
  – Shaun
  Mar 22 at 9:00
  

  
  
  
  

add a comment | 

$begingroup$

Let $G$ be a group of order $pq$, where $p$ and $q$ are primes and $p>q$. Ler $ain G$ be of order $p$ and $H=big<a: a^p=1big>.$ Then $H$ is normal in $G$.
I know that $H$ will be normal subgroup of G if $gH=Hg$ or $H=gHg^-1$.
I tried as:

Let $gin G$ and $hin H$ then $h$ can be written in some power of $a$ but I don't, how to write $g$ and proceed to show that $H=ghg^-1$.

group-theory normal-subgroups

share|cite|improve this question

edited Mar 22 at 9:03

Shaun

10.4k113686

asked Mar 22 at 8:51

Noor AslamNoor Aslam

16012

$endgroup$

share|cite|improve this question

edited Mar 22 at 9:03

Shaun

10.4k113686

asked Mar 22 at 8:51

Noor AslamNoor Aslam

16012

share|cite|improve this question

edited Mar 22 at 9:03

Shaun

10.4k113686

asked Mar 22 at 8:51

Noor AslamNoor Aslam

16012

edited Mar 22 at 9:03

Shaun

10.4k113686

edited Mar 22 at 9:03

Shaun

10.4k113686

asked Mar 22 at 8:51

Noor AslamNoor Aslam

16012

asked Mar 22 at 8:51

Noor AslamNoor Aslam

16012

2 Answers
2

active

oldest

votes

0

$begingroup$

The proof above very nearly works, so let's finish it up.

Lemma: If $x^p=1, text then forall g in G (g^-1xg)^p=1,$

Proof: Since consecutive pairs $g^-1g=1$ cancel each other, we have $(g^-1xg)^p=g^-1x^pg=g^-1g=1.$

Now assume $a in H$. Then $exists x_1, x_2, ldots x_n in H text such that x_k^p=1 text and a=x_1x_2cdots x_n$. Then $$g^-1ag= prod_k=1^n g^-1x_kg.$$

By the Lemma, each $g^-1x_kg$ is a generator of $H$ because it has order $p$, so $g^-1ag in H$ and since $a in H, g in G$ were arbitrary, that shows $forall g in G~g^-1Hg subseteq H$ so $H triangleleft G$ and we are done.

share|cite|improve this answer

answered Mar 22 at 9:41

Robert ShoreRobert Shore

3,611324

$endgroup$

add a comment | 

2

$begingroup$

Let's take any $a$ from $H$. Let's take some other element $g$ from $G$ and examine what is $x = gag^-1$. But

$$x^p = g a g^-1 g a g^-1...g a g^-1 = g a^p g^-1 = g e g^-1 = e$$

So, $x$ also belongs to $H$.

Conjugating by any element of the group $G$ leaves any element of $H$ inside $H$. That's actually a definition of normal subgroup. So, $H$ is normal in $G$.

UPDATE

As Tobias Kildetoft pointed out, this is not a proof yet. It is only a proof that conjugating leaves the set of all elements $S = a: a^p=1$ in place. The subgroup generated by these elements is a different thing!

Let's take some element $h$ of $H$. It is generated by elements of $S$:
$$ h = s_1 s_2... s_n $$

It's conjugated element:
$$ ghg^-1 = gs_1 s_2... s_ng^-1 = gs_1g^-1 gs_2g^-1... gs_ng^-1 $$

Is generated by $gs_ig^-1$. So, it is generated by by the elements of the same set $S$. So, it also belongs to $H$, which means $H$ is normal in $G$.

share|cite|improve this answer

edited Mar 22 at 9:54

answered Mar 22 at 9:01

lesniklesnik

1,495712

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The idea is correct, but there is a small detail missing: $H$ does not necessarily just consist of elements of order dividing $p$, but is generated by those.
  $endgroup$
  – Tobias Kildetoft
  Mar 22 at 9:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasShelby No, it is the subgroup generated by all elements of order $p$.
  $endgroup$
  – Tobias Kildetoft
  Mar 22 at 9:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasShelby I guess clarification is needed from the author. $H=big<a: a^p=1big>.$ - doesn't it read as "all $a$ such that ..."?
  $endgroup$
  – lesnik
  Mar 22 at 9:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasShelby Until you have shown that the subgroup in question is in fact of order $p$, you can't use Sylow. And the conclusion that it is of order $p$ is the same as there being just one Sylow $p$-subgroup (the subgroup in question is always normal, regardless of the group, it just happens to be of a nice form in this case).
  $endgroup$
  – Tobias Kildetoft
  Mar 22 at 9:19
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasShelby Sure, but that is not the subgroup we care about here (except that it turns out to be the same due to the order of the group).
  $endgroup$
  – Tobias Kildetoft
  Mar 22 at 9:30
  

  
  
  
  

 | 
show 3 more comments

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0

$begingroup$

The proof above very nearly works, so let's finish it up.

Lemma: If $x^p=1, text then forall g in G (g^-1xg)^p=1,$

Proof: Since consecutive pairs $g^-1g=1$ cancel each other, we have $(g^-1xg)^p=g^-1x^pg=g^-1g=1.$

Now assume $a in H$. Then $exists x_1, x_2, ldots x_n in H text such that x_k^p=1 text and a=x_1x_2cdots x_n$. Then $$g^-1ag= prod_k=1^n g^-1x_kg.$$

By the Lemma, each $g^-1x_kg$ is a generator of $H$ because it has order $p$, so $g^-1ag in H$ and since $a in H, g in G$ were arbitrary, that shows $forall g in G~g^-1Hg subseteq H$ so $H triangleleft G$ and we are done.

share|cite|improve this answer

answered Mar 22 at 9:41

Robert ShoreRobert Shore

3,611324

$endgroup$

add a comment | 

0

$begingroup$

The proof above very nearly works, so let's finish it up.

Lemma: If $x^p=1, text then forall g in G (g^-1xg)^p=1,$

Proof: Since consecutive pairs $g^-1g=1$ cancel each other, we have $(g^-1xg)^p=g^-1x^pg=g^-1g=1.$

Now assume $a in H$. Then $exists x_1, x_2, ldots x_n in H text such that x_k^p=1 text and a=x_1x_2cdots x_n$. Then $$g^-1ag= prod_k=1^n g^-1x_kg.$$

By the Lemma, each $g^-1x_kg$ is a generator of $H$ because it has order $p$, so $g^-1ag in H$ and since $a in H, g in G$ were arbitrary, that shows $forall g in G~g^-1Hg subseteq H$ so $H triangleleft G$ and we are done.

share|cite|improve this answer

answered Mar 22 at 9:41

Robert ShoreRobert Shore

3,611324

$endgroup$

add a comment | 

$begingroup$

The proof above very nearly works, so let's finish it up.

Lemma: If $x^p=1, text then forall g in G (g^-1xg)^p=1,$

Proof: Since consecutive pairs $g^-1g=1$ cancel each other, we have $(g^-1xg)^p=g^-1x^pg=g^-1g=1.$

Now assume $a in H$. Then $exists x_1, x_2, ldots x_n in H text such that x_k^p=1 text and a=x_1x_2cdots x_n$. Then $$g^-1ag= prod_k=1^n g^-1x_kg.$$

By the Lemma, each $g^-1x_kg$ is a generator of $H$ because it has order $p$, so $g^-1ag in H$ and since $a in H, g in G$ were arbitrary, that shows $forall g in G~g^-1Hg subseteq H$ so $H triangleleft G$ and we are done.

share|cite|improve this answer

answered Mar 22 at 9:41

Robert ShoreRobert Shore

3,611324

$endgroup$

share|cite|improve this answer

answered Mar 22 at 9:41

Robert ShoreRobert Shore

3,611324

answered Mar 22 at 9:41

Robert ShoreRobert Shore

3,611324

answered Mar 22 at 9:41

Robert ShoreRobert Shore

3,611324

2

$begingroup$

Let's take any $a$ from $H$. Let's take some other element $g$ from $G$ and examine what is $x = gag^-1$. But

$$x^p = g a g^-1 g a g^-1...g a g^-1 = g a^p g^-1 = g e g^-1 = e$$

So, $x$ also belongs to $H$.

Conjugating by any element of the group $G$ leaves any element of $H$ inside $H$. That's actually a definition of normal subgroup. So, $H$ is normal in $G$.

UPDATE

As Tobias Kildetoft pointed out, this is not a proof yet. It is only a proof that conjugating leaves the set of all elements $S = a: a^p=1$ in place. The subgroup generated by these elements is a different thing!

Let's take some element $h$ of $H$. It is generated by elements of $S$:
$$ h = s_1 s_2... s_n $$

It's conjugated element:
$$ ghg^-1 = gs_1 s_2... s_ng^-1 = gs_1g^-1 gs_2g^-1... gs_ng^-1 $$

Is generated by $gs_ig^-1$. So, it is generated by by the elements of the same set $S$. So, it also belongs to $H$, which means $H$ is normal in $G$.

share|cite|improve this answer

edited Mar 22 at 9:54

answered Mar 22 at 9:01

lesniklesnik

1,495712

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The idea is correct, but there is a small detail missing: $H$ does not necessarily just consist of elements of order dividing $p$, but is generated by those.
  $endgroup$
  – Tobias Kildetoft
  Mar 22 at 9:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasShelby No, it is the subgroup generated by all elements of order $p$.
  $endgroup$
  – Tobias Kildetoft
  Mar 22 at 9:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasShelby I guess clarification is needed from the author. $H=big<a: a^p=1big>.$ - doesn't it read as "all $a$ such that ..."?
  $endgroup$
  – lesnik
  Mar 22 at 9:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasShelby Until you have shown that the subgroup in question is in fact of order $p$, you can't use Sylow. And the conclusion that it is of order $p$ is the same as there being just one Sylow $p$-subgroup (the subgroup in question is always normal, regardless of the group, it just happens to be of a nice form in this case).
  $endgroup$
  – Tobias Kildetoft
  Mar 22 at 9:19
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasShelby Sure, but that is not the subgroup we care about here (except that it turns out to be the same due to the order of the group).
  $endgroup$
  – Tobias Kildetoft
  Mar 22 at 9:30
  

  
  
  
  

 | 
show 3 more comments

2

$begingroup$

Let's take any $a$ from $H$. Let's take some other element $g$ from $G$ and examine what is $x = gag^-1$. But

$$x^p = g a g^-1 g a g^-1...g a g^-1 = g a^p g^-1 = g e g^-1 = e$$

So, $x$ also belongs to $H$.

Conjugating by any element of the group $G$ leaves any element of $H$ inside $H$. That's actually a definition of normal subgroup. So, $H$ is normal in $G$.

UPDATE

As Tobias Kildetoft pointed out, this is not a proof yet. It is only a proof that conjugating leaves the set of all elements $S = a: a^p=1$ in place. The subgroup generated by these elements is a different thing!

Let's take some element $h$ of $H$. It is generated by elements of $S$:
$$ h = s_1 s_2... s_n $$

It's conjugated element:
$$ ghg^-1 = gs_1 s_2... s_ng^-1 = gs_1g^-1 gs_2g^-1... gs_ng^-1 $$

Is generated by $gs_ig^-1$. So, it is generated by by the elements of the same set $S$. So, it also belongs to $H$, which means $H$ is normal in $G$.

share|cite|improve this answer

edited Mar 22 at 9:54

answered Mar 22 at 9:01

lesniklesnik

1,495712

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The idea is correct, but there is a small detail missing: $H$ does not necessarily just consist of elements of order dividing $p$, but is generated by those.
  $endgroup$
  – Tobias Kildetoft
  Mar 22 at 9:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasShelby No, it is the subgroup generated by all elements of order $p$.
  $endgroup$
  – Tobias Kildetoft
  Mar 22 at 9:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasShelby I guess clarification is needed from the author. $H=big<a: a^p=1big>.$ - doesn't it read as "all $a$ such that ..."?
  $endgroup$
  – lesnik
  Mar 22 at 9:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasShelby Until you have shown that the subgroup in question is in fact of order $p$, you can't use Sylow. And the conclusion that it is of order $p$ is the same as there being just one Sylow $p$-subgroup (the subgroup in question is always normal, regardless of the group, it just happens to be of a nice form in this case).
  $endgroup$
  – Tobias Kildetoft
  Mar 22 at 9:19
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasShelby Sure, but that is not the subgroup we care about here (except that it turns out to be the same due to the order of the group).
  $endgroup$
  – Tobias Kildetoft
  Mar 22 at 9:30
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

Let's take any $a$ from $H$. Let's take some other element $g$ from $G$ and examine what is $x = gag^-1$. But

$$x^p = g a g^-1 g a g^-1...g a g^-1 = g a^p g^-1 = g e g^-1 = e$$

So, $x$ also belongs to $H$.

Conjugating by any element of the group $G$ leaves any element of $H$ inside $H$. That's actually a definition of normal subgroup. So, $H$ is normal in $G$.

UPDATE

As Tobias Kildetoft pointed out, this is not a proof yet. It is only a proof that conjugating leaves the set of all elements $S = a: a^p=1$ in place. The subgroup generated by these elements is a different thing!

Let's take some element $h$ of $H$. It is generated by elements of $S$:
$$ h = s_1 s_2... s_n $$

It's conjugated element:
$$ ghg^-1 = gs_1 s_2... s_ng^-1 = gs_1g^-1 gs_2g^-1... gs_ng^-1 $$

Is generated by $gs_ig^-1$. So, it is generated by by the elements of the same set $S$. So, it also belongs to $H$, which means $H$ is normal in $G$.

share|cite|improve this answer

edited Mar 22 at 9:54

answered Mar 22 at 9:01

lesniklesnik

1,495712

$endgroup$

share|cite|improve this answer

edited Mar 22 at 9:54

answered Mar 22 at 9:01

lesniklesnik

1,495712

answered Mar 22 at 9:01

lesniklesnik

1,495712

answered Mar 22 at 9:01

lesniklesnik

1,495712