How to visualize permutations?

Question

I'm getting a warning that this is a subjective question, and it very well probably is. But nevertheless, it is still a valid question that helps in the studying of mathematics from my point of view. Visualizing math is very important.

I'm studying abstract algebra, reading about permutations, and I'm having a hard time visualizing the group of permutations.

For instance, a permutation is defined as a function on a set, $f:A\to A$, that is 1-1 and onto.

The binary operation on this set is composition of functions, which is associative, has an identity and has an inverse.

Usually, when I hear of functions, I picture $x^2 + 2x + 1$ or $2x + 2$. Is this what I should be thinking of when I think of a collection of permutations? Because I can't help but confuse my mind by just thinking of all the different ways you can arrange the set. How do y'all picture this set?

Answer 1

You can use matrix-like notation. For example, the bijection

$$f:\{1,2,3\}\to\{1,2,3\}\;,\;\;\begin{cases}f(1)=2\\f(2)=1\\f(3)=3\end{cases}\;\longrightarrow\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix}$$

so you can visualize the action of $\;f\;$ on each element by looking at the different columns above. In some text books that's how permutations are introduced first.

Answer 2

also we can use square matrix $P$ of size $n$ t that is made my commuting rows of $I_n$ to Define permutation group as follow :

let $X$ is $\begin{pmatrix}1 \\\vdots\\ n\\\end{pmatrix}$ and $P$ is a elementary matrix then $f_p $ is permutaion associated to $P$ $$f_{P}:\begin{pmatrix}1 \\\vdots\\ n\\\end{pmatrix}\to\left\{\begin{pmatrix}x_1 \\x_2\\\vdots\\ x_n\\\end{pmatrix}: x_i \in \{1,2,...,n\}\right\}$$ $$f_{P}\left(\begin{pmatrix}1 \\\vdots\\ n\\\end{pmatrix}\right)=P\begin{pmatrix}1 \\\vdots\\ n\\\end{pmatrix}$$ or $f_P(X)=PX$ consequencely $$S_n=\{f_P \mid P\in E(I_n)\}$$ such that $E(I_n)$ all matrices with above property. Its good to say that I didn't visualize the permutation I just said a linear algebra view of that ..

Answer 3

Saying that $f$ is a bijective function $A\to A$ is very vague and amorphous - it doesn't spell out in any detail what $f$ can look like, and what the set of all such $f$s look like. There are two main issues with thinking of graphs of polynomials functions when one thinks of functions. First, the graph of a function (on an arbitrary set) is not always the best way to think about a function; and second, not all functions have nice explicit formulas defining them (in a sense, "most" of them on an infinite set do not have any specification by symbols). Indeed the set $A$ doesn't necessarily have any algebraic structure at all, it could very easily be a barren set.

Two major ways of thinking about what a permutation looks like are inherent in the main two types of notation: one-line notation and cycle notation. The precursor to one-line notation is two-line notation. The array of values given in Timbuc's answer is an example of two-line notation: the top row is just the numbers $1$ through $n$, and the bottom row tells us where the entries in the above row go to, where they are sent by the given permutation. One-line notation compactifies this by not bothering with the top row, as it is ultimately superfluous.

In order to understand cycle notation one must first know what a cycle is. A cycle is exactly what it sounds like: when a set of numbers (not necessarily the whole set) is cycled through in some order. The notation $(a_1~a_2~\cdots~a_k)$ means $a_1\mapsto a_2$ and $a_2\mapsto a_3$ and so on up to $a_k\mapsto a_1$ (so it wraps around). Two cycles are disjoint if the sets of values they cycle are disjoint - that is, there is no value which is moved by both of them. Every permutation is a product of disjoint cycles (uniquely), and writing it as such a product constitutes what we call cycle notation.

The group of permutations of a set $A$ can be represented by a group of $n\times n$ permutation matrices, where $n=|A|$. A matrix is a permutation matrix if and only if every row and column has precisely one instance of the entry $1$ and the rest of the entries are $0$. Multiplying permutation matrices yields another permutation matrix - namely, the one you'd get if you composed the two permutations and then determined the corresponding permutation matrix.

This is a special case of a group representation. (Note that one can represent other types of algebraic objects, not just groups, and one can have the actions be more than just linear transformations on vector spaces, hence the occasional term "linear" representation.) In my philosophy, any action or representation of a group is not an intrinsic way of viewing what the group is; rather, it gives perspective on what the group can do to other things.

There is a strong colloquial understanding of what a permutation is, which suggests if you have trouble getting a feel for what a permutation intuitively is, as a student of math, you might be getting too snagged on the term "function" or otherwise overthinking things. Indeed, if you ask a lay, non-mathematician what a permutation is, they'd be able to tell you: the reordering of some arrangement, the scrambling of some sequence, the shuffling of some list, or a variant on that theme. Just imagine a shell game but with any number of shells. (The only potential issue with this lazy description is that sets aren't necessarily ordered in a sequence.)

Finally, I want to mention a generalization of permutations, braids. They are represented by (or could be defined as) so-called braid diagrams, which look like this:

$\hskip 2in$

(Source is this webpage which gives further nontechnical explanation of braids. Sometimes braids are written horizontally instead of vertically.) The binary operation with these braids is simply concatenating them, i.e. putting one on top of the other (it is not immediately obvious braids have inverses, but they do). Notice the diagrams keep track of over/under data: it depicts at each crossing which string goes over top or underneath which other string. If you lose this information, and just draw lines crossing plainly, and allow ourselves to "cancel" consecutive crossings of the same two strings, then our diagrams in fact represent permutations. We can think of the ends of the strings down below as a set, and the braid diagrams shuffle them around until they end up in some new arrangement depicted by the ends up top.

Answer 4

I sympathize with your question because this confused me too at first--what confused me was more so cycle notation, but I'll get to that in just a second.

When $S$ is the set $\{1,2,\ldots,n\}$, consisting of the first $n$ positive integers, the group $\mathrm{Sym}(S)$ is commonly denoted $S_n$, and an element $\alpha$ of $S_n$ can be conveniently represented in two-row form as follows (same as Timbuc's answer, but I'll elaborate). First write $1,2,\ldots,n$, and then below each number $k$ write its image $\alpha(k)$. Thus $$ \begin{pmatrix}1&2&3&4\\2&4&3&1\end{pmatrix}\tag{1} $$ represents the permutation in $S_4$ defined by $1\mapsto 2,2\mapsto 4,3\mapsto 3$, and $4\mapsto 1$. It may help for you to think about or visualize $(1)$ more in this manner: $$ \begin{pmatrix}1&2&3&4\\\downarrow&\downarrow&\downarrow&\downarrow\\2&4&3&1\end{pmatrix}.\tag{2} $$ Now, elements of $S_n$ are frequently written using cycle notation. If $S$ is a set, and $a_1,a_2,\ldots,a_k\in S$, then $(a_1a_2\ldots a_k)$ denotes the permutation of $S$ for which $$ a_1\mapsto a_2,\;a_2\mapsto a_3,\;\ldots,\;a_{k-1}\mapsto a_k,\; a_k\mapsto a_1 $$ and $$ x\mapsto x\quad\text{for all other $x\in S$}. $$ Such a permutation is called a cycle or a $k$-cycle. If $a$ is any element of $S$, then the $1$-cycle $(a)$ is the identity permutation of $S$.

Try to write $(1)$ using cycle notation:

• Begin with $(1$
• Next, $1\mapsto 2$, so write $(1\;2$
• Next, $2\mapsto 4$, so write $(1\;2\;4$
• Next, $4\mapsto 1$, so close the cycle, giving $(1\;2\;4)$

Now begin a new cycle with $3$, the next and only possible choice outside of $(1\;2\;4)$.

• This gives $(1\;2\;4)(3$
• But $3\mapsto 3$, so close the second cycle, giving $(1\;2\;4)(3)$

Because $(3)$ represents the identity permutation, it may be omitted. This gives the following: $$ \begin{pmatrix}1&2&3&4\\2&4&3&1\end{pmatrix}=(1\;2\;4). $$ Does that make it all a bit clearer? Of course, if you want, you can simply "reverse your steps" to turn a cycle into two-row form (this is rarely done though because cycle notation is much more compact than two-row form). Thus, if you saw $(1\;2\;4)$, for instance, then you should think the following: "$1$ maps to $2$ which maps to $4$ which maps to $1$ and $3$ maps to itself." If you were to "think symbolically," then you would think like this: $1\mapsto 2, 2\mapsto 4, 4\mapsto 1, 3\mapsto 3$. You would then represent this thinking as done in $(2)$, but you would write it down on paper as it appears in $(1)$.

Does that help? Please ask questions if something doesn't make sense.