How many necklaces can be made with 6 beads of 3 different colors Good day!
Disclaimer: Not a homework question
How many necklaces can be made with 6 beads of 3 different colors?
I know i'm supposed to use the formula: $\frac{1}{ \left |G  \right |} \sum_{g \in G} \left | Fix(g) \right |$
My books gives the example of 9 beads of 3 different colors, and gives 
$$
\frac{1}{ \left |G  \right |} \sum_{g \in G} \left | Fix(g) \right |= \frac {1}{18}(3^9+2 \times 3^3+ 6 \times 3 +9 \times 3^5)= 1219
$$
Can anyone explain this? Where do the terms in the parenthesis come from? And the $ \frac {1}{18}$ ? Thanks!
 A: Basically you have to find out which necklaces are identical. Necklaces are identical if they are identical under symmetric operations just as rotate them ($r$) or turning them around ($s$). 
First of all I explain why you get the powers of 3. 
Look at the case that you do not turn the necklace (you apply the identity $\{e\}$). Because all of 9 beads can have different colors, you can have $3^9$ different necklaces. 
Now look at what happen if you consider the rotation $r$, that means you rotate the necklace $2 \pi / 9$ around. If you do so until you reach the point where you began you will see that the necklace stays only identical for all $r, r^2, r^3,..., r^8$ if all the breads have an identical colour, so there are 3 options to create such a necklace. 
Now turn the necklace upside down. There is 1 bead that is fixed and all the other 8 beads are at a new position. For the necklace to be identical under this symmetric operation (call it $s$) the 8 beads that change their position have to have the same colour as the bread that was at this position at the beginning. Because of this you have 5 breads where you can choose the colour and then the remaining four must have fixed colours. This corresponds to $3^5$. 
Now look at the case if you rotate the necklace around $2 \pi / 3$ corresponding to $r^3$. Then you can choose the colour of three breads and through rotation the other 6 breads must have colours depending on the colours of the first 3. Hence you have $3^3$ choices for the colours. 
Take a look at the conjugacy classes of the symmetric operations. 
The case $3^9$ corresponds to the identity {e} from which there is one. The rotation $r$ could aswell be $\{r, r^2, r^4, r^5, r^7, r^8\}$ because all the elements are of order 9, which means that they bring the necklace back in the starting position after 9 times applied. Notice that this are 6 elements, so its 6 times 3. 
$\{r^3, r^6\}$ form another conjugacy class, so you have $2 \cdot 3^3$. 
The last conjugacy class is $\{s, rs, r^2s, r^3s, r^4s, ..., r^8s\}$ containing 9 elements, thus $9 \cdot 3^5$.
At last, notice that $|G| = 18$ because we have 18 possible symmetric operations here: $\{e, r, r^2, ..., r^8, s, rs, r^2s, ... r^8s\}$.
Maybe you can try the case of 6 breads of 3 different colours yourself now and ask if you get stuck anywhere. 
