This is from a previous exam paper, several exercises have already been completed, and their results are given here since they are often used in the subsequent exercises.
Pre-exercise:
We have $G$, which is a non-abelian group of order $2p$, where $p$ is an odd prime.
Results from exercise 1
$G$ has a normal subgroup $P$ of order $p$. $G$ consists of the neutral element $e$, plus $p$ elements of order $2$ and $p-1$ elements of order $p$.
Results from exercise 2
We have shown that $x \in G$ where $ord(x)=2$ has a centralizer $Z(x)$, where $|Z(x)|=2$. Furthermore $y \in P$ with $ord(y)=p$ has centralizer with order $|Z(y)|=p$.
The exercise I'm needing help with is as follows:
"Show that $G$ has $\frac{p+3}{2}$ different conjugation classes".
The explanatory answer says:
"From the previous exercise, we see that the number of elements in the conjugation class for an element $p$ is $\frac{|G|}{p}=2$, so that there are $\frac{p-1}{2}$ conjugation classes consisting of elements of order $p$".
I do not understand how we can use information about the centralizer in order to obtain information about the number of conjugation classes. What information from exercise 2 (or 1?) can be used to conclude the above?
It continues:
"Additionally we see that there are $\frac{|G|}{2}=p$ elements in the conjugation class for an element of order 2. Together with the conjugation class $\{e\}$ we get $\frac{p-1}{2}+1+1=\frac{p+3}{2}$ different conjugation classes."
This final step would appear to make sense, but I do not see why we "see" that there are $\frac{|G|}{2}=p$ elements in the conjugation class for elements of order 2. Does this also follow directly from something shown previously?
Thank you.
Observe that $\;G\times G\to G\;,\;\;g\cdot x:=x^g=g^{-1}xg\;$ is an action of $\;G\;$ on itself, and thus by the orbit-stabilizer theorem, for any $\;x\in G\;$:
$$|\mathcal O(x)|=[G:G_x]\;,\;\;G_x:=\{g\in G\;|\;g\cdot x:=x^g=x\}=C_G(x)=\text{the centralizer of}\;x$$
in the whole group $\;G\;$
Can you take it from here?