Let $S_n$ be the symmetric group on $n$-symbols say $\{1,2, \cdots , n \}.$ Let $\sigma \in S_n,$ Let $\sigma_1, \sigma_2, \cdots , \sigma_r$ be $r$ disjoint cycles in $S_n$ such that $\sigma = \sigma_1 \sigma_2 \cdots \sigma_r.$ Then show that $$\text {Ord}\ (\sigma) = \text {lcm}\ \{\text {Ord}\ (\sigma_1), \text {Ord}\ (\sigma_2), \cdots , \text {Ord}\ (\sigma_r) \}.$$
Here I observe that any two disjoint cycles will commute with each other. Tempted by this observation and a theorem I read in my first course in group theory (which states that Let $G$ be a finite group and $a,b \in G$ be two elements of $G$ which commute with each other. Let $\text {Ord}\ (a) = m$ and $\text {Ord}\ (b) = n.$ If $m$ and $n$ are relatively prime to each other then $\text {Ord}\ (ab) = mn$) I have tried the whole long day to prove a lemma which is as follows.
Lemma $:$ Let $G$ be a finite group. Let $a,b \in G$ be two elements commute with each other. Then $\text {Ord}\ (ab) = \text {lcm}\ \left (\text {Ord}\ (a), \text {Ord}\ (b) \right ).$
I saw that if we can prove this lemma the required result follows. Just now I observed that it is false for every finite group of cardinality $\geq 2.$ Because if $G$ is a group with $|G| \geq 2$ then it contains a non-identity element say $a.$ Then I noticed $aa^{-1} = a^{-1}a = e.$ So $a$ always commutes with $a^{-1}.$ Also we have $\text {Ord}\ (a) = \text {Ord}\ (a^{-1}).$ So if the above lemma would hold then $$1 = \text {Ord}\ (e) = \text {Ord}\ (aa^{-1}) = \text {lcm}\ \{\text {Ord}\ (a), \text {Ord}\ (a^{-1}) \} = \text {Ord}\ (a).$$
This shows that $\text {Ord}\ (a) = 1 \implies a = e,$ which is a contradiction to our assumption. Hence the lemma is false.
Now I don't find any other approach to prove the required result. Can anybody please give me any suggestion regarding this?
Thanks for your valuable time.
EDIT $:$ I have managed to prove that $\text {Ord}\ (\sigma)\ \bigg |\ \text {lcm}\ \{\text {Ord}\ (\sigma_1), \text {Ord}\ (\sigma_2), \cdots , \text {Ord}\ (\sigma_r) \}.$ How do I prove the other way round?
Best Answer
I have started from the stage where I got stuck in proving the above lemma. It is easy to show what I just mentioned in the edit is that $\text {Ord}\ (ab)\ \big |\ \text {lcm}\ \left (\text {Ord}\ (a), \text {Ord}\ (b) \right ).$ To prove equality we need to prove the other way round which is not true for arbitrary finite groups even if $a$ and $b$ commute. We are so lucky that the other part is true for our case. Why? Lets discuss.
Before proving the required result I noticed that if we can prove the following lemma we are through.
Lemma $:$ Let $\sigma, \tau \in S_n$ be two disjoint cycles. Then $\text {Ord}\ (\sigma \tau ) = \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$
For proving the equality in the lemma let us first introduce the following definition.
Let $\rho = (a_1,a_2, \cdots , a_r) \in S_n$ be an $r$-cycle. Then the support of $\rho$ is denoted by $\text {Supp}\ (\rho)$ and it is defined as $\text {Supp}\ (\rho) = \{a_1,a_2, \cdots , a_r \}.$ So $\text {Supp}\ (\rho)$ consists of those points in $\{1,2, \cdots, n \}$ which are disturbed by the operation of $\rho.$
Observation $:$ If $\rho,\rho' \in S_n$ are two cycles inverses of each other then $\text {Supp}\ (\rho) = \text {Supp}\ (\rho').$ (Because inverse cycles fix same points).
Now let us take two disjoint cycles $\sigma , \tau \in S_n.$ On contrary let us assume that $\text {Ord}\ (\sigma \tau) = m < \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$ Then it is easy to see that $m\ \bigg |\ \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$ Let us assume that $\sigma^m \neq \text {id}$ and $\tau^m \neq \text {id}$ for otherwise $m = \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ),$ a contradiction to our assumption. Since fixed points of $\sigma$ and $\tau$ are respectively fixed points of $\sigma^m$ and $\tau^m$ respectively it follows that $\text {Supp}\ (\sigma^m) \subseteq \text {Supp}\ (\sigma)$ and $\text {Supp}\ (\tau^m) \subseteq \text {Supp}\ (\tau).$ Since $\sigma$ and $\tau$ are disjoint cycles so we have $\text {Supp}\ (\sigma) \cap \text {Supp}\ (\tau) = \varnothing.$ Hence $\text {Supp}\ (\sigma^m) \cap \text {Supp}\ (\tau^m) = \varnothing.\ \ \ \ (*)$
Now since $\text {Ord}\ (\sigma \tau) = m$ so we have $$\begin{align*} (\sigma \tau)^m & = \text {id} \implies \sigma^m \tau^m = \text {id} \implies \sigma^m = (\tau^m)^{-1} \end{align*}$$
So $\sigma^m$ is the inverse of $\tau^m.$ So from our Observation it follows that $\text {Supp}\ (\sigma^m) = \text {Supp}\ (\tau^m).$ Since $\sigma^m \neq \text {id}$ and $\tau^m \neq \text {id}$ it follows that $\text {Supp}\ (\sigma^m) = \text {Supp}\ (\tau^m) \neq \varnothing$ and hence $\text {Supp}\ (\sigma^m) \cap \text {Supp}\ (\tau^m) \neq \varnothing,$ which contradicts $(*).$ That implies either $\sigma^m = \text {id}$ or $\tau^m = \text {id}.$ But if one of $\sigma^m$ or $\tau^m$ is identity then by using the equation $\sigma^m \tau^m = \text {id}$ we find that the other is also an identity. So we must have $\sigma^m = \tau^m = \text {id}.$ This implies $\text {Ord}\ (\sigma)\ \big |\ m$ and $\text {Ord}\ (\tau)\ \big |\ m.$ But it means that $\text {lcm}\ \left ( \text {Ord}\ (\sigma),\text {Ord}\ (\tau) \right )\ \bigg |\ m,$ which is a contradiction to our assumption that $m < \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$ Hence our assumption is false. So $m \geq \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$ But since $m\ \bigg |\ \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right )$ it follows that $m \leq \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$ Hence combining these two inequalities it follows that $m = \text {lcm}\ \left (\text {Ord}\ (\sigma), \text {Ord}\ (\tau) \right ).$
QED