Show that G has a conjugacy class of size 3. 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.
 A: 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.
A: 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 $Z(G)=1$.
