Is this definition of a cycle in symmetric groups​ correct?"

Question

An elementary text on group theory applied to the Rubik's cube defines $S_n$, the symmetric group on n letters, as the set of bijections from $\{1,2,..., n\}$ to $\{1,2,...,n\}$ with the operation of composition.

A shorthand notation for such bijections is illustrated for some $\sigma\in S_{12}$: $$\sigma = (1\ 12\ 8\ 3\ 5\ 6\ 9\ 10)\ (2\ 4)\ (7)\ (11)$$

Some excerpts from the text:

Here, $(1\ 12\ 8\ 3\ 5\ 6\ 9\ 10)$, $(2\ 4)$, $(7)$, and $(10)$ are called cycles.

Definition 5.6. The cycle $(i_1\ i_2\ \dots\ i_k)$ is the element $\tau \in S_n$ defined by $τ(i_1)=i_2$, $\tau (i_2) = i_3$, ... , $\tau (i_{k−1}) = i_k$, $\tau (i_k) = i_1$ and $\tau (j) = j$ if $j \neq i_r$ for any $r$. The length of this cycle is $k$, and the support of the cycle is the set $\{i_1, \dots , i_k\}$ of numbers which appear in the cycle. The support is denoted by $\operatorname{supp}(\tau)$. A cycle of length $k$ is also called a k-cycle.

Definition 5.7. Two cycles $\sigma$ and $\tau$ are disjoint if they have no numbers in common; that is, $\operatorname{supp}{\sigma} \cap \operatorname{supp} \tau = \varnothing$.

My question: Is Definition 5.6 correct?

I find it problematic because it appears incompatible with the provided example $\sigma \in S_{12}$​, where multiple "cycles" contain more than one number. Furthermore, the definition seems ambiguous since it refers to a cycle both as a bijection $\tau \in S_n$​ as well as part of it ("The length of this cycle is $k$").

Answer 1

You've correctly identified a very common (and useful) abuse of notation. When a small cycle is written in a "large" symmetric group, the implicit assumption is that all numbers not identified in the cycle (or in a product of disjoint cycles) are fixed by the bijection.

So $\sigma= (1 ~2)$ is an element of $S_{87}$ once you realize that all of the numbers from $3$ through $87$ are implictly assumed to remain fixed by $\sigma$. Similarly, $\tau=(3~4)(36~51)$ is likewise an element of $S_{87}$ with this same understanding that all numbers other than $3, 4, 36$, and $51$ are fixed by $\tau$.

Answer 2

Yes it is correct. The cycle (2 4) tells you that something is happening to the numbers 2 and 4. Implicitly, all other numbers do not change, so the cycle bijection keeps them fixed, and it is truly a bijection in $S_{12}$. The same representation may be used for $S_{15}$ and so on. As for what the cycle does, it is "circular", reading left-to-right, so 2 maps to the right of it: 4. On the other hand, 4 has nothing to the right of it because it is the last number in the cycle representation, and it maps all the way back to the first number, 2.