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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 $g\in G$ 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.

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  • $\begingroup$ 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, 2019 at 6:58

2 Answers 2

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

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

For 2) I think that $S_3\times \mathbb{Z}_8$ is a counterexample since $((1,2,3),\bar{0})$ has conjugacy class size $ =3$ and every element of the form $(Id,\bar{x})$ is in the center of this group.

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  • $\begingroup$ 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, 2019 at 19:23
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    $\begingroup$ @Rivaldo yes !! $\endgroup$
    – 1123581321
    Mar 23, 2019 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, 2019 at 19:28
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    $\begingroup$ @Rivaldo I don't think so, $|cl(g)| | 2^4\cdot3$ and $|cl(g)|\not=3$ so it follows immediately that $2| |cl(g)|$ $\endgroup$
    – 1123581321
    Mar 23, 2019 at 19:30
  • $\begingroup$ any hints on 2)? $\endgroup$
    – Rivaldo
    Mar 26, 2019 at 0:39
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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 $Z(G)=1$.

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  • $\begingroup$ 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, 2019 at 0:58
  • $\begingroup$ Hey @Rivaldo, did you change you original question? I was answering to a different (2) last night. $\endgroup$ Mar 26, 2019 at 7:55

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