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Suppose a pizza has 4 slices, and each slice can be topped with either peppers, onions, or both. How many different pizzas can be made?

I am using Burnside's counting theorem with the group $G=left {(1),r,r^{2},r^{3} right }$ acting on the set $left {P,O,B right }$. I understand why $|X_{(1)}|=81$ and $|X_{r}|=|X_{r^{3}}|=3$, but why does $|X_{r^{2}}|=9$?

I drew a diagram:

Wouldn't it still be $3$ options since none of the slices are the same as the original after two rotations?

There are $3^4$ ways to top the four slices. $|X_g|$ are the number of those ways that will still be the same four slices in the same position after we rotate the pizza rotation $g$.

Let the original slices of the pizza be $a,b,c,d$ and $g_a, g_b, g_c, g_d$. be slices after rotation.

For $1$ (don't rotate the pizza) we have $1_a = a; 1_b=b; 1_c =c; 1_d = d$.

The $X_1$ are the pizzas that don't change when we don't do anything. ... Well, that's every one of the $81$ pizzas. Or to be thorough. $X_1$ are the pizzas where $1_a = a = a; 1_b=b=b; 1_c=c=c$ and $1_d =d=d$. And $1_a = a = a$ can be any of the $3$ options and ... so on.

Fr $r$ (rotate the pizza $90^circ$) we have $r_a = d; r_b=a; r_c=d; r_d = a$. (Because the slice now in the $a$ slot was origianlly in the $d$ slot and so on. If your intuition is the exact opposite and you think $r_a$ should equal $b$ because the original $a$ slice is now in the $b$ slot... when that works too; we just have to be consistant. For me, is seems to me $g_k$ means "the slice that is now in the $k$ spot was originally in what slot".)

So For the $X_r$ are the pizzas in which $r_a = d= a; r_b= a = b; r_c = b=c; r_d = a = d$. (In other words the ones where slice $a$ once rotated to position $b$ still has the same topping that the original slice in position $b$ had.) This means $a = b = c =d$. I.e.all the slices are the same. And $|X_r| = 3$ because there are $3$ options that this topping can be.

And $r^2$ (rotating $180^circ$) means $r^2_a = c; r^2_b = d; r^2_c = a; r^2_d = b$. So $X_{r^2}$ are the pizzas where $r^2_a = c = a; r^2_b = d = b; r^2_c = a= c; $ and $r^2_d=a = d$. Or in other words where $c=a; b=d$. Slice $a = c$ can be anly of $3$ options as can $b=d$. So $|X_{r^2}| = 3^2 = 9$.

Similarly $r^3$ (rotating $270^{circ}$) mean $r^3_a = b; r^3_b=c; r^3_c = d; r^3_d =a$ and $X_{r^3}$ are where $r^3_a = b =a; r^3_b=c=b;r^3_c=d=c; r^3_d =a =d$ or $a=b=c=d$ and $|X_{r^3}| = 3$.

So by Burnside's Counting Theorem there are:

$frac 1{|G|}(|X_1| + |X_r| + |X_{r^2}| + |X_{r^3}| ) =frac 14(81 + 3+ 9 + 3) = 24$.

Which.. seemms to work:

All the same: 1111, 22222, 3333,(three choices for the 1, one way to arrange it.)

Three of one, one of another: 1222,1333,2111,2333,3111,3222 (three choices for three slices, two for the remaining one)

Two of one, two of another: 1122,1212,1133,1313,2233,2323 (three pairs, two ways to arrange)

Two of one, one of each of the others: 1123,1132,1213, 2213,2231, 2123, 3312, 3321, 3132 (Three choices for the two slices, three ways to arrange them)

Denote peppers by $1$, onions by $2$ and both by $3$.

Since there are $3$ different choices for each slice, there are $3$ ways to make a pizza with every slice the same.

Suppose $3$ of the slices are the same. Then there are $2$ choices for the fourth slice so there are $3cdot 2=6$ ways to make a pizza with $3$ slices the same.

Suppose $2$ of the slices are the same. Note that there are only $3$ choices for each of $4$ slices so at least two of the slices must be the same.

There are $3$ ways to make two of the slices the same and the remaining two slices can be either the same or different. Given a topping for the two slices that are the same, there is only one way to make the remaining two slices different for a total of $3$ ways. For each topping on the first two slices, there are $2$ ways to make the other two slices the same for a total of $6$ ways to get pairs of slices with identical toppings. But each of these gets counted twice, e.g. $(1122)$ and $(2211)$, so there are only $3$ ways to create pairs of identical slices for a total of $6$ ways to have two identical slices.

Therefore, we can make $3+6+6=15$ different pizzas.