Counting elements of a given order in $S_5$ using the Burnside formula

Question

Let $G=S_7$, we want to count the number of elements of a given order in $G$.

Reminder of the Burnside formula:

$\forall X$ finite set, $\forall H$ finite group, et for all action of $H$ on $X$, we denote $Orb_H(X)$ the set of H-orbite in $X$ and $\text{Fix}_X(h)=\{x \in X, h\cdot x=x\}$, then $\forall h \in H$:

$$\text{Card}(Orb_H(X)) = \frac{1}{\text{Card}(H)}\sum_{h\in H}\text{Card}(\text{Fix}_X(h))$$

We denote $A=\{a,b,c,d,e,f,g\}$ and $Y=\{(u_i)_{i=0}^6 \in A^7, u_i \neq u_j \forall 0 \leq i \neq j \leq 6\}$

1. Show that $\Bbb Z/7\Bbb Z$ acts on $Y$ using: $g\cdot((u_i)_{i=0}^6)=(u_{i+g (\mod 7)})_{i=0}^6$ $ \forall g \in \Bbb Z/7\Bbb Z$
2. Using the Burnside formula find the total number of orbits for this action.
3. Deduct the number of 7-cycles in $G=S_7$.
4. Adapt to find the number of elements of order 6 in $G$.

What I have so far:

1. Let us denote $\phi: \rightarrow Y$ such that

\begin{align*} \phi\ \colon \Bbb Z/7\Bbb Z \times Y &\to Y\\ (g, u_i) &\mapsto u_{i+g \mod 7} \end{align*} Then $\phi$ is just the map that shifts an element of $Y$ by $g$ positions. How to properly show that it makes $\Bbb Z/7\Bbb Z $ act on $Y$?

2. We see that $\forall i \in \{0,...6\}$ we have $\bar{7} \cdot u_i = u_i$ (as $i+ \bar{7} \mod 7 \equiv i \mod 7$) and that $\forall g \neq \bar{7}$ we have $ g\cdot u_i \neq u_i$.

So that $\forall h \neq \bar{7}$ we have $\text{Card}(\text{Fix}_{Y}(h))=0$ and we have $\text{Card}(\text{Fix}_{Y}(\bar{7}))=7$

Finally as $\text{Card}(\Bbb Z/7\Bbb Z )=7$ the Burnside formula yields $\text{Card}(Orb_{}(Y))=\frac{7}{7}=1$

3. How does the number of orbits help finding the number of 7-cycles (who are the elements of order 7 in $G=S_7$?
4. Applying the same thing as previously but using $\Bbb Z/6\Bbb Z$, we have that for $h=\bar{6}$, $h\cdot u_i=u_i$ only if $i \in \{0,...,5\}$ because for $i=6$ we have $\bar{6}\cdot u_6=u_{6+6 \mod 6} = u_0 \neq u_6$

So that $\forall h \neq \bar{6}$ we have $\text{Card}(\text{Fix}_{Y}(h))=0$ and we have $\text{Card}(\text{Fix}_{Y}(\bar{6}))=6$

And finally applying the Burnside formula we get $\text{Card}(Orb_{}(Y))=\frac{6}{6}=1$ as well.

Answer 1

1. Think of $Y$ as a subset of the functions from $\mathbb{Z}/7$ to $A$.

2. Yep

3. In $S_7$, being an element of order $7$ is equivalent to being a $7$-cycle but I don't think it is relevant here. To answer 3 in light of the previous question, you should ask yourself what is the link between $Y$ and the set of $7$-cycles. More precisely:

How do you associate a $7$-cycle $c(y)$ to an element of $y\in Y$? Once you found the map from $Y$ to the set of $7$-cycles, find under which conditions $c(y_1)=c(y_2)$.

4. I will let you adapt it.