In Pinter's A Book of Abstract Algebra, Chapter 13 Exercise Set J aims to construct the orbit-stabilizer theorem from several proofs. After concluding that the size of every orbit with respect to $G$ (…a subgroup of $S_A$ that acts on $A$) is a factor of the order of $G$, Pinter states the following:
Given the orbit-stabilizer theorem, explain why the length of each cycle of $f \in S_A$ is a factor of the order of $f$ in $S_A$
For a hypothetical case, consider $A=\{1,2,3,4,5\}$ and the permutation $f = (1 2)(345)$
The length of $(12)$ is $2$, the length of $(345)$ is $3$, and the order of $f$ is $6$.
So I see that this statement has some truth to it…but I am failing to see why this is connected to the orbit-stabilizer theorem.
The orbit-stabilizer theorem states: $|O(u)|=\frac{|G|}{|G_u|}$, where $G_u$ is Pinter's notation for stabilizer.
I see that $f^0, f^1, f^2, f^3, f^4$, and$f^5$ generate their own subgroup of $S_A$. Therefore, we could let $G$ be equal to $\langle f\rangle$.
So the orbit-stabilizer theorem could be rewritten as $|O(u)|=\frac{|\langle f\rangle|}{|\langle f\rangle_u|}$
This is where I get sort of stumped as to how to continue my thought process. It seems to me that the length of each cycle can only be spoken about meaningfully in relation to the orbit-stabilizer theorem if I know that $f$, to begin with, is composed of disjoint cycles…but Pinter makes no mention of this.
Any help regarding where to go from here would be appreciated!
Best Answer
The action of $f$ on $\{1,\ldots,n\}$ does not depend on the expression of $f$ as a product of disjoint cycles: it just depends on the fact that $f$ is a bijection from $\{1,\ldots,n\}$ to itself. This induces a partition of $\{1,\ldots,n\}$ into orbits via the equivalence relation $a\sim b$ if and only if there exists an integer $k$ such that $f^k(a)=b$.
So now that we have an action and the orbits, we can pick an arbitrary orbit $\mathcal{O}(u)$ of this action. Then, as you note, the Orbit Stabilizer Theorem tells you that $$|\mathcal{O}(u)| = \frac{|\langle f\rangle|}{|\langle f\rangle_{u}|}.$$ Since $|\langle f\rangle|$ is the order of $f$, this immediately tells you that the size of any orbit of the action is necessarily a divisor of the order of $f$.
The next step is to note that when you express $f$ as a product of disjoint cycles, what you are doing is describing the orbits of the action of $f$. That is, if we write $f$ as a product of disjoint cycles $$ f=\sigma_1\cdots\sigma_r$$ (including $1$-cycles), and we define an equivalence relation on $\{1,\ldots,n\}$ by $aRb$ if and only if there exists $j$ such that $a$ and $b$ are both in the support of $\sigma_j$, then you can show that $\sim$ and $R$ are in fact the exact same equivalence relation. So that the support of the cycles are precisely the orbits of the action, and so you get that the length of the cycles are exactly the sizes of the orbits.
I don't have your book on hand, so I don't know if they define the expression of $f$ as a product of disjoint cycles before discussing general actions (this is done, for example, in Hungerford, where the decomposition of an arbitrary element of $S_n$ into disjoint cycles precedes the discussion of group actions and the Orbit-Stabilizer Theorem); or whether the expression of $f$ as a product of disjoint cycles is done following group actions and as a consequence (by describing the cycles as the restriction of $f$ on the orbits of its actons). Either way does not affect the result on hand: the length of the cycle is the size of the orbit of any of the elements in its support, and then the Orbit-Stabilizer Theorem directly yields the desired conclusion, that this quantity is a divisor of $|\langle f\rangle|$.