# Cycle notation in symmetric groups

There is a con­ve­nient way to rep­re­sent the el­e­ments of a sym­met­ric group on a finite set.

# $$k$$-cycle

A $$k$$-cy­cle is a mem­ber of $$S_n$$ which moves $$k$$ el­e­ments to each other cycli­cally. That is, let­ting $$a_1, \dots, a_k$$ be dis­tinct in $$\{1,2,\dots,n\}$$, a $$k$$-cy­cle $$\sigma$$ is such that $$\sigma(a_i) = a_{i+1}$$ for $$1 \leq i < k$$, and $$\sigma(a_k) = a_1$$, and $$\sigma(x) = x$$ for any $$x \not \in \{a_1, \dots, a_k \}$$.

We have a much more com­pact no­ta­tion for $$\sigma$$ in this case: we write $$\sigma = (a_1 a_2 \dots a_k)$$. (If spac­ing is am­bigu­ous, we put in com­mas: $$\sigma = (a_1, a_2, \dots, a_k)$$.) Note that there are sev­eral ways to write this: $$(a_1 a_2 \dots a_k) = (a_2 a_3 \dots a_k a_1)$$, for ex­am­ple. It is con­ven­tional to put the small­est $$a_i$$ at the start.

Note also that a cy­cle’s in­verse is ex­tremely easy to find: the in­verse of $$(a_1 a_2 \dots a_k)$$ is $$(a_k a_{k-1} \dots a_1)$$.

For ex­am­ple, the dou­ble-row no­ta­tion $$\begin{pmatrix}1 & 2 & 3 \\ 2 & 3 & 1 \\ \end{pmatrix}$$\$ is writ­ten as $$(123)$$ or $$(231)$$ or $$(312)$$ in cy­cle no­ta­tion.

How­ever, it is un­clear with­out con­text which sym­met­ric group $$(123)$$ lies in: it could be $$S_n$$ for any $$n \geq 3$$. Similarly, $$(145)$$ could be in $$S_n$$ for any $$n \geq 5$$.

# Gen­eral el­e­ments, not just cycles

Not ev­ery el­e­ment of $$S_n$$ is a cy­cle. For ex­am­ple, the fol­low­ing el­e­ment of $$S_4$$ has or­der $$2$$ so could only be a $$2$$-cy­cle, but it moves all four el­e­ments:

$$\begin{pmatrix}1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \\ \end{pmatrix}$$

How­ever, it may be writ­ten as the com­po­si­tion of the two cy­cles $$(12)$$ and $$(34)$$: it is the re­sult of ap­ply­ing one and then the other. Note that since the cy­cles are dis­joint (hav­ing no el­e­ments in com­mon), it doesn’t mat­ter in which or­der we perform them. It is a very im­por­tant fact that ev­ery per­mu­ta­tion may be writ­ten as the product of dis­joint cy­cles. If $$\sigma$$ is a per­mu­ta­tion ob­tained by first do­ing cy­cle $$c_1 = (a_1 a_2 \dots a_k)$$, then by do­ing cy­cle $$c_2$$, then cy­cle $$c_3$$, we write $$\sigma = c_3 c_2 c_1$$; this is by anal­ogy with func­tion com­po­si­tion, in­di­cat­ing that the first per­mu­ta­tion to ap­ply is on the right­most end of the ex­pres­sion. (Be aware that some au­thors differ on this.)

## Order of an element

Firstly, a cy­cle has or­der equal to its length. In­deed, the cy­cle $$(a_1 a_2 \dots a_k)$$ has the effect of ro­tat­ing $$a_1 \mapsto a_2 \mapsto a_3 \dots \mapsto a_k \mapsto a_1$$, and if we do this $$k$$ times we get back to where we started. (And if we do it fewer times—say $$i$$ times—we can’t get back to where we started: $$a_1 \mapsto a_{i+1}$$.)

Now, sup­pose we have an el­e­ment in dis­joint cy­cle no­ta­tion: $$(a_1 a_2 a_3)(a_4 a_5)$$, say, where all the $$a_i$$ are differ­ent. Then the or­der of this el­e­ment is $$3 \times 2 = 6$$, be­cause:

• $$(a_1 a_2 a_3)$$ and $$(a_4 a_5)$$ are dis­joint and hence com­mute, so $$[(a_1 a_2 a_3)(a_4 a_5)]^n = (a_1 a_2 a_3)^n (a_4 a_5)^n$$

• $$(a_1 a_2 a_3)^n (a_4 a_5)^n$$ is the iden­tity if and only if $$(a_1 a_2 a_3)^n = (a_4 a_5)^n = e$$ the iden­tity, be­cause oth­er­wise (for in­stance, if $$(a_1 a_2 a_3)^n$$ is not the iden­tity) it would move $$a_1$$.

• $$(a_1 a_2 a_3)^n$$ is the iden­tity if and only if $$n$$ is di­visi­ble by $$3$$, since $$(a_1 a_2 a_3)$$’s or­der is $$3$$.

• $$(a_4 a_5)^n$$ is the iden­tity if and only if $$n$$ is di­visi­ble by $$2$$.

This rea­son­ing gen­er­al­ises: the or­der of an el­e­ment in dis­joint cy­cle no­ta­tion is equal to the least com­mon mul­ti­ple of the lengths of the cy­cles.

# Examples

• The el­e­ment $$\sigma$$ of $$S_5$$ given by first perform­ing $$(123)$$ and then $$(345)$$ is $$(345)(123) = (12453)$$. In­deed, the first ap­pli­ca­tion takes $$1$$ to $$2$$ and the sec­ond ap­pli­ca­tion does not af­fect the re­sult­ing $$2$$, so $$\sigma$$ takes $$1$$ to $$2$$; the first ap­pli­ca­tion takes $$2$$ to $$3$$ and the sec­ond ap­pli­ca­tion takes the re­sult­ing $$3$$ to $$4$$, so $$\sigma$$ takes $$2$$ to $$4$$; the first ap­pli­ca­tion does not af­fect $$4$$ and the sec­ond ap­pli­ca­tion takes $$4$$ to $$5$$, so $$\sigma$$ takes $$4$$ to $$5$$; and so on.

This ex­am­ple sug­gests a gen­eral pro­ce­dure for ex­press­ing a per­mu­ta­tion which is already in cy­cle form, in dis­joint cy­cle form. It turns out that this can be done in an es­sen­tially unique way.

## Cy­cle type

The cy­cle type of a per­mu­ta­tion is given by tak­ing the list of lengths of the cy­cles in the dis­joint cy­cle form.

Children:

Parents:

• Symmetric group

The sym­met­ric groups form the fun­da­men­tal link be­tween group the­ory and the no­tion of sym­me­try.

• I’m cu­ri­ous if the in­verse has any par­tic­u­lar use in this field.

• None that I’m aware of, but I’ve found it con­ve­nient to know when I was do­ing ex­er­cises in a first course in group the­ory.