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Show that G has a conjugacy class of size 3.

3

1

$begingroup$

1)Let $G$ be a group of size $48$ with centre consisting of the identity element only. Show that $G$ has a conjugacy class of size $3$ .

My attempt:
I know $C(g)$ centraliser of $g$ is a subgroup of $G$, so its size must divide that of $G$.
And that $|cl(g)|=|G|/|c(g)|$ , but how am I sure there exists a $gin G$ such that$|c(g)|=16$?

2)Also say I had a group of size 48 and at least one conjugacy class of size 3 does that imply the centre only consists on the identity element?

Any help on 2) would be appreciated thanks.

group-theory

share|cite|improve this question

edited Mar 26 at 0:34

Rivaldo

asked Mar 23 at 18:41

RivaldoRivaldo

236210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Elements with conjugacy class size $3$ have a centralizer of order $48/3 = 16$ and subgroups of order $16$ of $G$ are Sylow $2$-subgroups. So you are looking at central elements of a Sylow $2$-subgroup that are not in the center of the whole group. As a (finite) $2$-group always has a non-trivial center, (1) follows and it should be now easy to find a group showing that the implication in (2) does not hold.
  $endgroup$
  – j.p.
  Mar 26 at 6:58
  

  
  
  
  

add a comment | 

3

1

$begingroup$

1)Let $G$ be a group of size $48$ with centre consisting of the identity element only. Show that $G$ has a conjugacy class of size $3$ .

My attempt:
I know $C(g)$ centraliser of $g$ is a subgroup of $G$, so its size must divide that of $G$.
And that $|cl(g)|=|G|/|c(g)|$ , but how am I sure there exists a $gin G$ such that$|c(g)|=16$?

2)Also say I had a group of size 48 and at least one conjugacy class of size 3 does that imply the centre only consists on the identity element?

Any help on 2) would be appreciated thanks.

group-theory

share|cite|improve this question

edited Mar 26 at 0:34

Rivaldo

asked Mar 23 at 18:41

RivaldoRivaldo

236210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Elements with conjugacy class size $3$ have a centralizer of order $48/3 = 16$ and subgroups of order $16$ of $G$ are Sylow $2$-subgroups. So you are looking at central elements of a Sylow $2$-subgroup that are not in the center of the whole group. As a (finite) $2$-group always has a non-trivial center, (1) follows and it should be now easy to find a group showing that the implication in (2) does not hold.
  $endgroup$
  – j.p.
  Mar 26 at 6:58
  

  
  
  
  

add a comment | 

$begingroup$

1)Let $G$ be a group of size $48$ with centre consisting of the identity element only. Show that $G$ has a conjugacy class of size $3$ .

My attempt:
I know $C(g)$ centraliser of $g$ is a subgroup of $G$, so its size must divide that of $G$.
And that $|cl(g)|=|G|/|c(g)|$ , but how am I sure there exists a $gin G$ such that$|c(g)|=16$?

2)Also say I had a group of size 48 and at least one conjugacy class of size 3 does that imply the centre only consists on the identity element?

Any help on 2) would be appreciated thanks.

group-theory

share|cite|improve this question

edited Mar 26 at 0:34

Rivaldo

asked Mar 23 at 18:41

RivaldoRivaldo

236210

$endgroup$

share|cite|improve this question

edited Mar 26 at 0:34

Rivaldo

asked Mar 23 at 18:41

RivaldoRivaldo

236210

share|cite|improve this question

edited Mar 26 at 0:34

Rivaldo

asked Mar 23 at 18:41

RivaldoRivaldo

236210

asked Mar 23 at 18:41

RivaldoRivaldo

236210

asked Mar 23 at 18:41

RivaldoRivaldo

236210

2 Answers
2

active

oldest

votes

1

$begingroup$

1)$|G | = | Z(G) | + ∑_i [G : C(g_i)]Rightarrow 48=2^4cdot3=1+∑_i [G : C(g_i)]$

2) the elements in the conjugacy class of $g$ are in one-to-one correspondence with cosets of the centralizer $C(g)$

So 1) gives $48=2^4cdot3=1+∑_i [G : C(g_i)]=1+∑_i |cl(g_i)|$ so if for all $i$ it was $|cl(g_i)|not=3$ then $2| |cl(g_i)| quad forall iRightarrow 2| ∑_i |cl(g_i)|Rightarrow 2|2^4cdot3-1$, contradiction!

For 2) I think that $S_3times mathbbZ_8$ is a counterexample since $((1,2,3),bar0)$ has conjugacy class size $ =3$ and every element of the form $(Id,barx)$ is in the center of this group.

share|cite|improve this answer

edited Mar 26 at 6:27

answered Mar 23 at 18:52

giannispapavgiannispapav

1,968325

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so you're saying that possible values of size of cl(g) are 2,3,4,8,12,16,24 so if its not 3 then it must be divisible by 2 for all i , so 2/sum ,but 2 doesn't divide 47 which leads to a contradiction ?
  $endgroup$
  – Rivaldo
  Mar 23 at 19:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Rivaldo yes !!
  $endgroup$
  – giannispapav
  Mar 23 at 19:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks a lot , also in exam do you think I have to say all the possible values of |cl(g)|?
  $endgroup$
  – Rivaldo
  Mar 23 at 19:28
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Rivaldo I don't think so, $|cl(g)| | 2^4cdot3$ and $|cl(g)|not=3$ so it follows immediately that $2| |cl(g)|$
  $endgroup$
  – giannispapav
  Mar 23 at 19:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  any hints on 2)?
  $endgroup$
  – Rivaldo
  Mar 26 at 0:39
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

Hint on (2): define a map $f: Cl_G(g) rightarrow tC_G(g): t in G$ by $f(xgx^-1)=xC_G(g)$. Show that $f$ is well-defined, injective and onto.

Now let us assume that there is a unique conjugacy class of order $3$. Let $g in G$ with $|Cl_G(g)|=3$. The class formula (taken mod $2$) implies that $|Z(G)|$ must be odd. Since $Z(G) subset C_G(g)$ and $|C_G(g)|=16$, this implies that $Z(G)=1$.

share|cite|improve this answer

edited Mar 26 at 7:59

answered Mar 26 at 0:53

Nicky HeksterNicky Hekster

29.1k63456

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So from that we could deduce domain =codomain in size right? which leads to this observation $|cl(g)|=|G|/|c(g)|$ so we have |c(g)|=16 for that particular g so centre must be less than or equal to 16 , and odd from observations in (1) , so I get centre could be of order 1 or 3
  $endgroup$
  – Rivaldo
  Mar 26 at 0:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hey @Rivaldo, did you change you original question? I was answering to a different (2) last night.
  $endgroup$
  – Nicky Hekster
  Mar 26 at 7:55
  

  
  
  
  

add a comment | 

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1

$begingroup$

1)$|G | = | Z(G) | + ∑_i [G : C(g_i)]Rightarrow 48=2^4cdot3=1+∑_i [G : C(g_i)]$

2) the elements in the conjugacy class of $g$ are in one-to-one correspondence with cosets of the centralizer $C(g)$

So 1) gives $48=2^4cdot3=1+∑_i [G : C(g_i)]=1+∑_i |cl(g_i)|$ so if for all $i$ it was $|cl(g_i)|not=3$ then $2| |cl(g_i)| quad forall iRightarrow 2| ∑_i |cl(g_i)|Rightarrow 2|2^4cdot3-1$, contradiction!

For 2) I think that $S_3times mathbbZ_8$ is a counterexample since $((1,2,3),bar0)$ has conjugacy class size $ =3$ and every element of the form $(Id,barx)$ is in the center of this group.

share|cite|improve this answer

edited Mar 26 at 6:27

answered Mar 23 at 18:52

giannispapavgiannispapav

1,968325

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so you're saying that possible values of size of cl(g) are 2,3,4,8,12,16,24 so if its not 3 then it must be divisible by 2 for all i , so 2/sum ,but 2 doesn't divide 47 which leads to a contradiction ?
  $endgroup$
  – Rivaldo
  Mar 23 at 19:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Rivaldo yes !!
  $endgroup$
  – giannispapav
  Mar 23 at 19:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks a lot , also in exam do you think I have to say all the possible values of |cl(g)|?
  $endgroup$
  – Rivaldo
  Mar 23 at 19:28
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Rivaldo I don't think so, $|cl(g)| | 2^4cdot3$ and $|cl(g)|not=3$ so it follows immediately that $2| |cl(g)|$
  $endgroup$
  – giannispapav
  Mar 23 at 19:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  any hints on 2)?
  $endgroup$
  – Rivaldo
  Mar 26 at 0:39
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

1)$|G | = | Z(G) | + ∑_i [G : C(g_i)]Rightarrow 48=2^4cdot3=1+∑_i [G : C(g_i)]$

2) the elements in the conjugacy class of $g$ are in one-to-one correspondence with cosets of the centralizer $C(g)$

So 1) gives $48=2^4cdot3=1+∑_i [G : C(g_i)]=1+∑_i |cl(g_i)|$ so if for all $i$ it was $|cl(g_i)|not=3$ then $2| |cl(g_i)| quad forall iRightarrow 2| ∑_i |cl(g_i)|Rightarrow 2|2^4cdot3-1$, contradiction!

For 2) I think that $S_3times mathbbZ_8$ is a counterexample since $((1,2,3),bar0)$ has conjugacy class size $ =3$ and every element of the form $(Id,barx)$ is in the center of this group.

share|cite|improve this answer

edited Mar 26 at 6:27

answered Mar 23 at 18:52

giannispapavgiannispapav

1,968325

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so you're saying that possible values of size of cl(g) are 2,3,4,8,12,16,24 so if its not 3 then it must be divisible by 2 for all i , so 2/sum ,but 2 doesn't divide 47 which leads to a contradiction ?
  $endgroup$
  – Rivaldo
  Mar 23 at 19:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Rivaldo yes !!
  $endgroup$
  – giannispapav
  Mar 23 at 19:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks a lot , also in exam do you think I have to say all the possible values of |cl(g)|?
  $endgroup$
  – Rivaldo
  Mar 23 at 19:28
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Rivaldo I don't think so, $|cl(g)| | 2^4cdot3$ and $|cl(g)|not=3$ so it follows immediately that $2| |cl(g)|$
  $endgroup$
  – giannispapav
  Mar 23 at 19:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  any hints on 2)?
  $endgroup$
  – Rivaldo
  Mar 26 at 0:39
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

1)$|G | = | Z(G) | + ∑_i [G : C(g_i)]Rightarrow 48=2^4cdot3=1+∑_i [G : C(g_i)]$

2) the elements in the conjugacy class of $g$ are in one-to-one correspondence with cosets of the centralizer $C(g)$

So 1) gives $48=2^4cdot3=1+∑_i [G : C(g_i)]=1+∑_i |cl(g_i)|$ so if for all $i$ it was $|cl(g_i)|not=3$ then $2| |cl(g_i)| quad forall iRightarrow 2| ∑_i |cl(g_i)|Rightarrow 2|2^4cdot3-1$, contradiction!

For 2) I think that $S_3times mathbbZ_8$ is a counterexample since $((1,2,3),bar0)$ has conjugacy class size $ =3$ and every element of the form $(Id,barx)$ is in the center of this group.

share|cite|improve this answer

edited Mar 26 at 6:27

answered Mar 23 at 18:52

giannispapavgiannispapav

1,968325

$endgroup$

share|cite|improve this answer

edited Mar 26 at 6:27

answered Mar 23 at 18:52

giannispapavgiannispapav

1,968325

answered Mar 23 at 18:52

giannispapavgiannispapav

1,968325

answered Mar 23 at 18:52

giannispapavgiannispapav

1,968325

0

$begingroup$

Hint on (2): define a map $f: Cl_G(g) rightarrow tC_G(g): t in G$ by $f(xgx^-1)=xC_G(g)$. Show that $f$ is well-defined, injective and onto.

Now let us assume that there is a unique conjugacy class of order $3$. Let $g in G$ with $|Cl_G(g)|=3$. The class formula (taken mod $2$) implies that $|Z(G)|$ must be odd. Since $Z(G) subset C_G(g)$ and $|C_G(g)|=16$, this implies that $Z(G)=1$.

share|cite|improve this answer

edited Mar 26 at 7:59

answered Mar 26 at 0:53

Nicky HeksterNicky Hekster

29.1k63456

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So from that we could deduce domain =codomain in size right? which leads to this observation $|cl(g)|=|G|/|c(g)|$ so we have |c(g)|=16 for that particular g so centre must be less than or equal to 16 , and odd from observations in (1) , so I get centre could be of order 1 or 3
  $endgroup$
  – Rivaldo
  Mar 26 at 0:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hey @Rivaldo, did you change you original question? I was answering to a different (2) last night.
  $endgroup$
  – Nicky Hekster
  Mar 26 at 7:55
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint on (2): define a map $f: Cl_G(g) rightarrow tC_G(g): t in G$ by $f(xgx^-1)=xC_G(g)$. Show that $f$ is well-defined, injective and onto.

Now let us assume that there is a unique conjugacy class of order $3$. Let $g in G$ with $|Cl_G(g)|=3$. The class formula (taken mod $2$) implies that $|Z(G)|$ must be odd. Since $Z(G) subset C_G(g)$ and $|C_G(g)|=16$, this implies that $Z(G)=1$.

share|cite|improve this answer

edited Mar 26 at 7:59

answered Mar 26 at 0:53

Nicky HeksterNicky Hekster

29.1k63456

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So from that we could deduce domain =codomain in size right? which leads to this observation $|cl(g)|=|G|/|c(g)|$ so we have |c(g)|=16 for that particular g so centre must be less than or equal to 16 , and odd from observations in (1) , so I get centre could be of order 1 or 3
  $endgroup$
  – Rivaldo
  Mar 26 at 0:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hey @Rivaldo, did you change you original question? I was answering to a different (2) last night.
  $endgroup$
  – Nicky Hekster
  Mar 26 at 7:55
  

  
  
  
  

add a comment | 

$begingroup$

Hint on (2): define a map $f: Cl_G(g) rightarrow tC_G(g): t in G$ by $f(xgx^-1)=xC_G(g)$. Show that $f$ is well-defined, injective and onto.

Now let us assume that there is a unique conjugacy class of order $3$. Let $g in G$ with $|Cl_G(g)|=3$. The class formula (taken mod $2$) implies that $|Z(G)|$ must be odd. Since $Z(G) subset C_G(g)$ and $|C_G(g)|=16$, this implies that $Z(G)=1$.

share|cite|improve this answer

edited Mar 26 at 7:59

answered Mar 26 at 0:53

Nicky HeksterNicky Hekster

29.1k63456

$endgroup$

share|cite|improve this answer

edited Mar 26 at 7:59

answered Mar 26 at 0:53

Nicky HeksterNicky Hekster

29.1k63456

answered Mar 26 at 0:53

Nicky HeksterNicky Hekster

29.1k63456

answered Mar 26 at 0:53

Nicky HeksterNicky Hekster

29.1k63456