How to find the order of a symmetric group S4?

Question

Let $\omega = \bigl(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 3 & 2 & 4 & 1 \end{smallmatrix}\bigr)$ and $\tau = \bigl(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{smallmatrix}\bigr)$, both elements of $S_4$.

How exactly do I compute the orders of $\omega$ and $\tau$ ?

My professor said $\ |\omega|= 3$ and$\ |\tau|= 4$. But he did not explain how to get these answers, also why are they both not just 4? I thought order was the number of elements in G, in this case they both have 4. Does $\ |\omega|= 3$ because $\omega(2)=2$, so it sends itself to itself?

Sorry, if this question seems dumb I just don't fully understand where the answer is coming from.

Answer 1

The order of a group is not the same thing as the order of an element. There is a connection, but don't worry about that yet. The order of a group, written $|G|$, is the number of elements it has. The order of an element in a group - say $g \in G$, is the smallest $n$ such $g^n = e$ where $e$ is the identity element of the group $G$. In this case we also write $|g| = n$.

You are not being asked to find the order of a group. $\tau$ and $\omega$ are elements of a group and you are being asked to find the order of $\tau$ and $\omega$ as elements of $S_4$. Elements of $S_4$, such as $\tau$ and $\omega$, are bijections (functions) from the set $\{1,2,3,4\}$ to the set $\{1,2,3,4\}$. To find their order you need to investigate how many times you need to repeatedly apply the function in order to get the identity function. It doesn't really make sense to take about the order of $\tau$ and $\omega$ based on how many elements they have as you did in your original post, however we do have this.

It does turn out that if the order of an element is $n$, say $|g| = n$, then the cyclic subgroup generated by $g$ will have $n$ elements. So I guess if you wanted to find the orders of $\tau$ and $\omega$ you could go count the elements of their cyclic subgroups but this seems superfluous.

Answer 2

We first need to know that when talking about $S_4$, we are talking about the set of all bijections from the set $\{1,2,3,4\}$ to itself.

So, the elements of the set $S_4$ are functions. An example of a function in $S_4$ is the function $f: \{1,2,3,4\} \to \{1,2,3,4\}$ given by: $$1 \mapsto 2, 2\mapsto 3, 3\mapsto 4, 4 \mapsto 1 $$

However, this is incredibly tedious to write out. Instead, we will use matrices to represent functions in $S_4$. In other words, we will let:

$$f := \bigl(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{smallmatrix}\bigr)$$ where an element in row 1 of the matrix is mapped to the element below it in row 2 of the matrix by $f$.

Now, when we are talking about order, we can talk about 2 things: the order of a group, and the order of an element of the group.

The order of a group is the number of elements in that set, but we can also pick out an element from that group (in this case the group is $S_4$ and we're picking out an element from $S_4$) and we can ask what is the order of the element. The order of an element is the smallest positive number $n$ such that $x^n = 1$ where $x$ is the element of the group we are considering, and $1$ is the identity element in the group.

In your case, when your professor says that $|\omega| = 3$, what he means is that if you compose the function $\omega \in S_4$ with itself 3 times, you will get back the identity element of $S_4$. The identity element of $S_4$ is the identity function on $\{1,2,3,4\}$, i.e. the function:

$$i = \bigl(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{smallmatrix}\bigr)$$

For instance, although $\omega = \bigl(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 3 & 2 & 4 & 1 \end{smallmatrix}\bigr)$, we see that $\omega^2 = \bigl(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 4 & 2 & 1 & 3 \end{smallmatrix}\bigr)$, and $\omega^3 = \bigl(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{smallmatrix}\bigr) = i$

Answer 3

In "cycle notation ", $\omega=(134)$ is a $3$-cycle, and $\tau=(1234)$ is a $4$-cycle.

It happens that $n$-cycles always have order $n$.

Answer 4

The common way how determine the order of an element of a finite symmetric group is to subdivide this element — permutation — into one or more cyclic permutations, because (obviously):

$$ \bbox[lightyellow,5px,border:0px solid yellow]{\text {The order of a cyclic permutation is equal to the number of permuted elements.}}$$

Then the order of such subdivided element is calculated as the least common multiple of orders of these cyclic permutations.

I will show you in detail how to do it.

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For $\omega = \bigl(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 3 & 2 & 4 & 1 \end{smallmatrix}\bigr)$: Select a number from the top row — for example $1$ - and trace the path of its gradual change until you meet the same element. You will see this path (traced numbers are on the left of the green line): or $$\color{red}1 \mapsto 3 \mapsto 4 \mapsto \color{red}1$$

It is a cyclic permutation of 3 elements:

Then select one of the remaining numbers — in our example it remains only one, namely number $2$ — and do the same with it. Tracing the remaining number $2$ we see that

$$\color{red}2 \mapsto \color{red}2$$

which is the identity of 1-element set, considered as a cyclic permutation, too.

Using the notation for cyclic permutations, the first one is

$$(1\ 3\ 4)$$

and the second one is simply $$(2)$$

As the result, we may write

$$\bbox[lightcyan,5px,border:2px solid blue]{\omega = \bigl(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 3 & 2 & 4 & 1 \end{smallmatrix}\bigr)= (1\ 3\ 4)\ (2)}$$

Now, the first, 3-element cyclic permutation has the order $\mathbb 3$ (its "cycle" must turn 3-times $120^o$ to return to its initial position)

and the order of the second one is $\mathbb 1$ (order of identity). Their least common multiple $lcm(3, 1)=3$, so

$$\bbox[yellow,5px,border:2px solid red]{ord(\omega) = 3}$$

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For $\tau = \left(\begin{matrix}1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{matrix}\right)$ it is even easier, because it is a cyclic permutation itself:

$$\color{red}1 \mapsto 2 \mapsto 3 \mapsto 4 \mapsto \color{red}1$$

so we may write

$$\bbox[lightcyan,5px,border:2px solid blue]{\tau = \left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{smallmatrix}\right) = (1\ 2\ 3\ 4)}$$

and

$$\bbox[yellow,5px,border:2px solid red]{ord(\tau) = 4}$$