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Prove that the center of a group is a normal subgroup [closed]

Prove that the centralizer is a normal subgroup.Elements which commute with a given element form a subgroupIs centralizer a normal subgroup?Show that $Z(G) = x in G : gx = xg$ for all $g in G$ is a subgroup of $G$.Finding a normal subgroup 5 (abelian subgroup)Prove that Z(G) which is the center of G is a subgroup of GProve a Group is a SubgroupOn normal subgroup of a finite groupProve that: the center of any group is characteristic subgroup .Normal subgroup in a $p$-groupProve that an automorphism group is a normal subgroupNormal subgroup in center of the groupProve that in a nilpotent group every normal subgroup of prime order is contained in the center.Prove that the center of a group is normal.Let $G$ denote an arbitrary group. Prove: The center of any group $G$ is a normal subgroup of $G$Show that group has a nontrivial normal subgroup.$H$ prime order, normal subgroup of group $G$. Prove $H$ in center $Z(G)$.

14

5

$begingroup$

Let $G$ be a group. We define

$$H=hin Gmid forall gin G: hg=gh,$$

the center of $G$.

Prove that $H$ is a (normal) subgroup of $G$.

abstract-algebra group-theory

share|cite|improve this question

edited Dec 17 '16 at 15:50

user26857

39.5k124283

asked Sep 20 '11 at 12:08

Rok KraljRok Kralj

2251210

$endgroup$

closed as off-topic by user26857, Namaste, астон вілла олоф мэллбэрг, Behrouz Maleki, user223391 Dec 18 '16 at 18:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Namaste, астон вілла олоф мэллбэрг, Behrouz Maleki, Community

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, it was my exam question. Just finnished it. I think I managed it up to part to prove that it is normal. But I couldn't prove that it is a subgroup.
  $endgroup$
  – Rok Kralj
  Sep 20 '11 at 12:15
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  proofwiki.org/wiki/Center_is_a_Normal_Subgroup
  $endgroup$
  – Martin Sleziak
  Sep 20 '11 at 12:16
  

  
  
  
  


• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  Just curious: what does it mean to prove that it is normal without proving that it is a subgroup?
  $endgroup$
  – ShreevatsaR
  Sep 20 '11 at 12:28
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @ShreevatsaR Let $S$ be a subset of a group $G$. We say that $S$ is a normal subset of $G$ if for all $sin S$ and $gin G$, we have $g^-1sgin S$. In particular, a subset $S$ of $G$ is a normal subgroup of $G$ if and only if $S$ is a subgroup of $G$ and $S$ is a normal subset of $G$.
  $endgroup$
  – Amitesh Datta
  Sep 20 '11 at 13:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AmiteshDatta: Thanks.
  $endgroup$
  – ShreevatsaR
  Sep 20 '11 at 13:25
  

  
  
  
  

 | 
show 1 more comment

14

5

$begingroup$

Let $G$ be a group. We define

$$H=hin Gmid forall gin G: hg=gh,$$

the center of $G$.

Prove that $H$ is a (normal) subgroup of $G$.

abstract-algebra group-theory

share|cite|improve this question

edited Dec 17 '16 at 15:50

user26857

39.5k124283

asked Sep 20 '11 at 12:08

Rok KraljRok Kralj

2251210

$endgroup$

closed as off-topic by user26857, Namaste, астон вілла олоф мэллбэрг, Behrouz Maleki, user223391 Dec 18 '16 at 18:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Namaste, астон вілла олоф мэллбэрг, Behrouz Maleki, Community

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, it was my exam question. Just finnished it. I think I managed it up to part to prove that it is normal. But I couldn't prove that it is a subgroup.
  $endgroup$
  – Rok Kralj
  Sep 20 '11 at 12:15
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  proofwiki.org/wiki/Center_is_a_Normal_Subgroup
  $endgroup$
  – Martin Sleziak
  Sep 20 '11 at 12:16
  

  
  
  
  


• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  Just curious: what does it mean to prove that it is normal without proving that it is a subgroup?
  $endgroup$
  – ShreevatsaR
  Sep 20 '11 at 12:28
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @ShreevatsaR Let $S$ be a subset of a group $G$. We say that $S$ is a normal subset of $G$ if for all $sin S$ and $gin G$, we have $g^-1sgin S$. In particular, a subset $S$ of $G$ is a normal subgroup of $G$ if and only if $S$ is a subgroup of $G$ and $S$ is a normal subset of $G$.
  $endgroup$
  – Amitesh Datta
  Sep 20 '11 at 13:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AmiteshDatta: Thanks.
  $endgroup$
  – ShreevatsaR
  Sep 20 '11 at 13:25
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Let $G$ be a group. We define

$$H=hin Gmid forall gin G: hg=gh,$$

the center of $G$.

Prove that $H$ is a (normal) subgroup of $G$.

abstract-algebra group-theory

share|cite|improve this question

edited Dec 17 '16 at 15:50

user26857

39.5k124283

asked Sep 20 '11 at 12:08

Rok KraljRok Kralj

2251210

$endgroup$

share|cite|improve this question

edited Dec 17 '16 at 15:50

user26857

39.5k124283

asked Sep 20 '11 at 12:08

Rok KraljRok Kralj

2251210

share|cite|improve this question

edited Dec 17 '16 at 15:50

user26857

39.5k124283

asked Sep 20 '11 at 12:08

Rok KraljRok Kralj

2251210

edited Dec 17 '16 at 15:50

user26857

39.5k124283

edited Dec 17 '16 at 15:50

user26857

39.5k124283

asked Sep 20 '11 at 12:08

Rok KraljRok Kralj

2251210

asked Sep 20 '11 at 12:08

Rok KraljRok Kralj

2251210

3 Answers
3

active

oldest

votes

17

$begingroup$

As you said in a comment you already showed that it is normal. So I will only show that it is a subgroup.

Clearly it contains $e$, since $eg = ge$.

Now, we will show that it is closed. Let $a,b in H$, we know that $forall g: ag = ga$ and $gb = gb$. Thus, $gab = agb = abg$ and thus $ab in H$.

Now we only have to show that every $h in H$ has an inverse and we are done.
Let $h in H$, we know that $forall g in G: gh = hg$, thus
$$beginalign*h^-1(gh)h^-1 &= h^-1(hg)h^-1\
h^-1g(hh^-1) &= (h^-1h)gh^-1\
h^-1(ge) &= (eg)h^-1\
h^-1g &= gh^-1
endalign*$$

Which implies that $h^-1 in H$.

share|cite|improve this answer

edited Sep 20 '11 at 12:37

answered Sep 20 '11 at 12:30

sxdsxd

2,85831836

$endgroup$

add a comment | 

5

$begingroup$

sxd has shown that it is a subgroup.

That it is normal follows from here:

Let $xin Z(G)$ (center of $G$).

Then for any $gin G$, $gxg^-1=gg^-1x=xin Z(G)$.

This proves $Z(G)$ is a normal subgroup.

share|cite|improve this answer

answered Dec 17 '16 at 15:46

yoyosteinyoyostein

8,155104073

$endgroup$

add a comment | 

1

$begingroup$

i think the kernel argument (Mikko's no 1) is the most elegant approach. but sxd shows with thoroughness that the centre is a group, and it is perhaps worth a passing mention that once we know it is a group its normality is implied by the fact that it consists of complete conjugacy classes, albeit they are all singletons.

share|cite|improve this answer

answered Nov 17 '13 at 2:26

David HoldenDavid Holden

14.9k21225

$endgroup$

add a comment | 

17

$begingroup$

As you said in a comment you already showed that it is normal. So I will only show that it is a subgroup.

Clearly it contains $e$, since $eg = ge$.

Now, we will show that it is closed. Let $a,b in H$, we know that $forall g: ag = ga$ and $gb = gb$. Thus, $gab = agb = abg$ and thus $ab in H$.

Now we only have to show that every $h in H$ has an inverse and we are done.
Let $h in H$, we know that $forall g in G: gh = hg$, thus
$$beginalign*h^-1(gh)h^-1 &= h^-1(hg)h^-1\
h^-1g(hh^-1) &= (h^-1h)gh^-1\
h^-1(ge) &= (eg)h^-1\
h^-1g &= gh^-1
endalign*$$

Which implies that $h^-1 in H$.

share|cite|improve this answer

edited Sep 20 '11 at 12:37

answered Sep 20 '11 at 12:30

sxdsxd

2,85831836

$endgroup$

add a comment | 

17

$begingroup$

As you said in a comment you already showed that it is normal. So I will only show that it is a subgroup.

Clearly it contains $e$, since $eg = ge$.

Now, we will show that it is closed. Let $a,b in H$, we know that $forall g: ag = ga$ and $gb = gb$. Thus, $gab = agb = abg$ and thus $ab in H$.

Now we only have to show that every $h in H$ has an inverse and we are done.
Let $h in H$, we know that $forall g in G: gh = hg$, thus
$$beginalign*h^-1(gh)h^-1 &= h^-1(hg)h^-1\
h^-1g(hh^-1) &= (h^-1h)gh^-1\
h^-1(ge) &= (eg)h^-1\
h^-1g &= gh^-1
endalign*$$

Which implies that $h^-1 in H$.

share|cite|improve this answer

edited Sep 20 '11 at 12:37

answered Sep 20 '11 at 12:30

sxdsxd

2,85831836

$endgroup$

add a comment | 

$begingroup$

As you said in a comment you already showed that it is normal. So I will only show that it is a subgroup.

Clearly it contains $e$, since $eg = ge$.

Now, we will show that it is closed. Let $a,b in H$, we know that $forall g: ag = ga$ and $gb = gb$. Thus, $gab = agb = abg$ and thus $ab in H$.

Now we only have to show that every $h in H$ has an inverse and we are done.
Let $h in H$, we know that $forall g in G: gh = hg$, thus
$$beginalign*h^-1(gh)h^-1 &= h^-1(hg)h^-1\
h^-1g(hh^-1) &= (h^-1h)gh^-1\
h^-1(ge) &= (eg)h^-1\
h^-1g &= gh^-1
endalign*$$

Which implies that $h^-1 in H$.

share|cite|improve this answer

edited Sep 20 '11 at 12:37

answered Sep 20 '11 at 12:30

sxdsxd

2,85831836

$endgroup$

share|cite|improve this answer

edited Sep 20 '11 at 12:37

answered Sep 20 '11 at 12:30

sxdsxd

2,85831836

answered Sep 20 '11 at 12:30

sxdsxd

2,85831836

answered Sep 20 '11 at 12:30

sxdsxd

2,85831836

5

$begingroup$

sxd has shown that it is a subgroup.

That it is normal follows from here:

Let $xin Z(G)$ (center of $G$).

Then for any $gin G$, $gxg^-1=gg^-1x=xin Z(G)$.

This proves $Z(G)$ is a normal subgroup.

share|cite|improve this answer

answered Dec 17 '16 at 15:46

yoyosteinyoyostein

8,155104073

$endgroup$

add a comment | 

5

$begingroup$

sxd has shown that it is a subgroup.

That it is normal follows from here:

Let $xin Z(G)$ (center of $G$).

Then for any $gin G$, $gxg^-1=gg^-1x=xin Z(G)$.

This proves $Z(G)$ is a normal subgroup.

share|cite|improve this answer

answered Dec 17 '16 at 15:46

yoyosteinyoyostein

8,155104073

$endgroup$

add a comment | 

$begingroup$

sxd has shown that it is a subgroup.

That it is normal follows from here:

Let $xin Z(G)$ (center of $G$).

Then for any $gin G$, $gxg^-1=gg^-1x=xin Z(G)$.

This proves $Z(G)$ is a normal subgroup.

share|cite|improve this answer

answered Dec 17 '16 at 15:46

yoyosteinyoyostein

8,155104073

$endgroup$

share|cite|improve this answer

answered Dec 17 '16 at 15:46

yoyosteinyoyostein

8,155104073

answered Dec 17 '16 at 15:46

yoyosteinyoyostein

8,155104073

answered Dec 17 '16 at 15:46

yoyosteinyoyostein

8,155104073

1

$begingroup$

i think the kernel argument (Mikko's no 1) is the most elegant approach. but sxd shows with thoroughness that the centre is a group, and it is perhaps worth a passing mention that once we know it is a group its normality is implied by the fact that it consists of complete conjugacy classes, albeit they are all singletons.

share|cite|improve this answer

answered Nov 17 '13 at 2:26

David HoldenDavid Holden

14.9k21225

$endgroup$

add a comment | 

1

$begingroup$

i think the kernel argument (Mikko's no 1) is the most elegant approach. but sxd shows with thoroughness that the centre is a group, and it is perhaps worth a passing mention that once we know it is a group its normality is implied by the fact that it consists of complete conjugacy classes, albeit they are all singletons.

share|cite|improve this answer

answered Nov 17 '13 at 2:26

David HoldenDavid Holden

14.9k21225

$endgroup$

add a comment | 

$begingroup$

i think the kernel argument (Mikko's no 1) is the most elegant approach. but sxd shows with thoroughness that the centre is a group, and it is perhaps worth a passing mention that once we know it is a group its normality is implied by the fact that it consists of complete conjugacy classes, albeit they are all singletons.

share|cite|improve this answer

answered Nov 17 '13 at 2:26

David HoldenDavid Holden

14.9k21225

$endgroup$

share|cite|improve this answer

answered Nov 17 '13 at 2:26

David HoldenDavid Holden

14.9k21225

answered Nov 17 '13 at 2:26

David HoldenDavid Holden

14.9k21225

answered Nov 17 '13 at 2:26

David HoldenDavid Holden

14.9k21225