Consider the Dihedral group $D_{12}=\left\langle a,b: a^6=e, \ b^2=e, \ ba=a^5b \right\rangle$ of order $12$ of symmetries of regular hexagon. Every element of $D_{12}$ can be written as $a^ib^j, \ 0 \leq i \leq 5, \ 0 \leq j \leq 1$.
Find $ C_{D_{12}}(a)$=centraliser of $a$ and $ C_{D_{12}}(b)$=centraliser of $b$ .
Answer:
We know that,
$ |cl_{D_{12}}(a)|$=cardinality of the conjugacy class of the element $a$ is $2$,
$ |cl_{D_{12}}(b)|$=cardinality of the conjugacy class of the element $b$ is $3$.
Now,
cardinality of $C_{D_{12}}(a)=\frac{|D_{12}|}{|cl_{D_{12}}(a)|}=\frac{12}{2}=6,$
cardinality of $C_{D_{12}}(b)=\frac{|D_{12}|}{|cl_{D_{12}}(b)|}=\frac{12}{3}=4$.
But I can't find the group of centraliser as above
Help me
You have calculated cardinality correctly
If you want explicit centraliser of a and b then
$C_{D_{12}}(a)=<a>=${$e,a,a^2...,a^5$}
$C_{D_{12}}(b)=<a^3,b>=${$e,a^3,b,a^3b$}
Edit:
$C_{D_{12}}(b)=${$x\in D_{12} |xbx^{-1}=b $}
$<a,b>=D_{12}$
So every $x=a^ib^j$ where $0\leq i< 6 , j=0$ or $ 1$
So check for which i satisfies an equation
$a^i(b)a^{-i}=b$
$b=a^{2i}b$
implies either $i=3 or 6$
SO $a^3,e\in C_{D_{12}}$
Similarly check for which i satifyies below equation
$a^ib(b)a^ib=b$
implies $a^{2i}b=b$
implies $i=3 or 6$
That is $a^3b,b\in C_{D_{12}} $
Note:
Try to work with generating set It will reduces calculation to much more extent as we have done above rather than doing manually each term