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Consider the group $S_8$. What is the order of $\sigma = (4,5)(2,3,7)$ and $\tau = (1,4)(3,5,7,8)$?

My book says I should just use a trick by *the order of a permutation expressed as a product of disjoint cycles is the least common multiple of the lengths of the cycles*.

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I don’t understand the trick very well and would like to see how the counting process here actually works. I can count for non disjoint ones like say $(1,2,3,8)$ has order 4

I don’t know how to count $\sigma = \begin{pmatrix}

1 &2 &3 &4 &5 &6 &7 &8 \\

1 &3 &7 &5 &4 &6 &2 &8

\end{pmatrix}$

**EDIT** Just so I am clear, the answer for the order $\sigma$ is 6 and $\tau$ is 4. I am still confused as to how they got this

I have an extremely poor understanding of this subject so please help me! Thank you

- If $G$ acts such that $\mbox{fix}(g) \in \{0,p\}$ for $g \ne 1$, $M$ maximal with $|G : M| = p$, then $|M / L| = p$ for semiregular $L \unlhd G$.
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So take $\sigma = (4,5)(2,3,7)$. The order, by definition, is the the smallest natural number $n$ such that $\sigma^n = (1)$ (i.e. the identity element in the group, i.e. the element that sends every number to itself). Since the cycles $(4,5)$ and $(2,3,7)$ are disjoint you have

$$\begin{align}

\sigma^n &= (4,5)(2,3,7)(4,5)(2,3,7)\dots (4,5)(2,3,7)(4,5)(2,3,7)\\

&= (4,5)(4,5)\dots (4,5)(4,5)(2,3,7)(2,3,7)\dots (2,3,7)(2,3,7)\\

&= (4,5)^n(2,3,7)^n

\end{align}$$

(note that the two elements $(4,5)$ and $(2,3,7)$ commute).

So the order of $\sigma$ is exactly the smallest natural number $n$ such that $(4,5)^n = (1)$ and $(2,3,7)^n = (1)$ (think about this fact for a moment).

But what is the order of a each of $(4,5)$ and $(2,3,7)$?

Well, the order of $(4,5)$ is two exactly because $(4,5)^2 = (4,5)(4,5) = (1)$. The order of $(2,3,7)$ is $3$ because

$$\begin{align}

(2,3,7)^1 &= (2,3,7) \\

(2,3,7)^2 &= (1,7,3) \\

(2,3,7)^3 &= (1)

\end{align}

$$

Now it is not to hard to see that the order of $\sigma$ is exactly the least common multiple of $2$ and $3$ (since we need both $(4,5)^m = (1)$ and $(2,3,7)^m = (1)$ and the smallest $m$ where this happens is exactly the least common multiple). Hence the final answer is $6$.

Addendum: I just wanted to add a bit about orders of these elements. First note that for example the element $(4,5)$ is just the element

$$

(4,5) = \begin{pmatrix}

1 & 2 &3 & 4 & 5 & 6 & 7 & 8 \\ 1 & 2 & 3 &5 & 4& 6 & 7 & 8

\end{pmatrix}.

$$

(Hence $4$ maps to $5$ and $5$ to $4$). So when you compose (multiply) the element with itself, then you get

$$

\begin{align}

(4,5)(4,5) &= \begin{pmatrix}

1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 2 & 3 &5 & 4& 6 & 7 & 8

\end{pmatrix}\begin{pmatrix}

1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 2 & 3 &5 & 4& 6 & 7 & 8

\end{pmatrix} \\ &= \begin{pmatrix}

1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 2 & 3 &4 & 5& 6 & 7 & 8

\end{pmatrix} = (1)

\end{align}

$$

(I usually write the identity as $(1)$).

This means that the order of $(4,5)$ is $2$. Likewise you find that

$$

(2,3,7) = \begin{pmatrix}

1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 3 & 7 &4 & 5& 6 & 2 & 8

\end{pmatrix}.

$$

Also we find order of permutation by the formula ……

If p,q be the permutation then

o(pq) = L.C.M (length of p , length of q )

For exam:- (1 4) (3 5 7 8 )

Here length of (1 4 ) is 2 and length of (3 5 7 8) is 4

So , order of (1 4) (3 5 7 8 )= L.C.M (2 ,4 )= 4

Prove that easy lemma: the order of a permutation expressed as a product of disjoint cycles is the minimal common multiple of the lengths of the cycles.

To prove the above you may want to use the fact (also easily provable) that two disjoint cycles commute…

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