I am taking a course on abstract algebra, and the lector defined $T$ to be a set, and defined $G$ to be the set of all bijections from $T$ to itself:
$$ G=\{\text{all bijections }g\colon T\rightarrow T\}=\mathrm{Sym}(T).\tag{1} $$
That's what the lector wrote on the blackboard.
Can anyone translate $(1)$ into either English or simpler math, and/or tell me what a set and a map is in simple words?
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$\begingroup$Do you know what a bijection is? It is a way to map one set ($A$) onto another set ($B$) such that every element in $B$ has something in $A$ being mapped to it (i.e., the map is onto), and no two elements of $A$ are mapped to the same element in $B$ (i.e., the map is one-to-one).
As an example, think about finite sets:
Let $A = \{ b, c, d \}$ and $B = \{a, f, g \}$.
A example of a bijection $f : A \to B$ could be the map that sends $b$ to $a$, $c$ to $f$, and $d$ to $g$. Why is this a bijection? Well, does every element of $B$ have something in $A$ being mapped to it? Yes, you can see this by looking at each element in $B$. Are any two elements in $A$ mapped to the same element in $B$? No -- you can see this again by direct observation of the elements of $A$ and where they are sent under $f$.
Well, what if I asked you how many different bijections you can have between $A$ and $B$? This is an easy exercise to solve: how many possible choices can $b$ be sent to? 3. And for each of these choices, how many possible choices can $c$ be sent to (while maintaining the bijection properties)? Only $2$ since if $b$ is sent to $a$, for example, we can't have $c$ also sent to $a$ or else the map wouldn't be one-to-one and thus a bijection.
Ok, well if $b$ can be sent to $3$ possible choices, and $c$ can be sent to $2$ possible choices after that, then how many possible choices can $d$ be sent to while still maintaining the bijective property? Only $1$. To see this, here is an example: Let $b$ be sent to any element, say, $a$. Then $c$ could only be sent to $f$ or $g$. Let's say it is sent to $f$. Then $d$ can only be sent to $g$ to have a bijection.
Ok, so if $b$ has $3$ possibilities to be sent to, and $c$ has $2$ possibilities, and $d$ has $1$ possibility, then the total number of bijections from $A$ to $B$ are $3 \times 2 \times 1 = 6$ (we can map the elements of $A$ bijectively onto $B$ in $6$ different ways).
From the above example, hopefully you can see that a bijection is just matching elements of one set up with another in a one-to-one correspondence.
Now, if we look at the bijections between a set $T$ and itself, we call each bijection a symmetry of $T$. In some sense, we are talking about the different ways we can bijectively pair up the elements in $T$ with the elements in $T$.
One last note: for two finite sets, you can only have a bijection between them if they have the same number of elements. That means any set $A$ can't have a bijection with a proper subset of itself since the proper subset has less elements. But every infinite set can be put in bijection with a proper subset of itself, so this is a big difference between finite sets and infinite sets which plenty of people have trouble understanding at first.
$\endgroup$ 2 $\begingroup$My guess is that you are learning about permutation groups (at least, what you have written fits perfectly when learning about such groups).
A permutation of a nonempty set $T$ is a one-to-one mapping from $T$ onto $T$. [As in $(1)$]
Now, consider this:
The set of all permutations of a nonempty set $T$ is a group with respect to composition. This group is called the symmetric group on $T$ and is often denoted $\operatorname{Sym}(T)$.
Suppose $T$ is the set $\{1,2,\ldots,n\}$, consisting of the first $n$ positive integers. An element of $\operatorname{Sym}(T)$ may be represented in two-row form as follows: First write $1,2,\ldots,n$, and then below each number $k$ write its image. Thus, something like \begin{pmatrix} 1&2&3&4\\ 2&4&3&1 \end{pmatrix} where $T=\{1,2,3,4\}$, represents the permutation in $\operatorname{Sym}(T)$ defined by $1\mapsto 2, 2\mapsto 4, 3\mapsto 3, 4\mapsto 1$.
Since a permutation mapping is bijective, an element of $\operatorname{Sym}(T)$ will always have the form \begin{pmatrix} 1&2&\cdots&n\\ - & - &\cdots & - \end{pmatrix} where the integers $1,2,\ldots,n$ can be placed in the $n$ blanks. There are $n$ choices for the first blank, $n-1$ choices for the second blank, and so on--this means there are $n!$ possible bijective mappings. In basic group theory parlance, the "order" of the group $\operatorname{Sym}(T)$ is $n!$.
Thus, for $T=\{1,2,\ldots,n\}$, the elements of $G=\operatorname{Sym}(T)$ will all look like \begin{pmatrix} 1&2&\cdots&n\\ - & - &\cdots & - \end{pmatrix} in some form or fashion.
$\endgroup$ 1 $\begingroup$Well, these are very basic notions in Mathematics, but I will try to explain them to you as simply as I can. A set is a collection of elements of some kind, with no order among them, and a map is a correspondence between two sets that "sends" one element of a set to just one element of the other set.
For example, Let the set $S$ be made of the numbers $1,2,3$, and the set $T$ be made of the letters $a,b$. Then you could define a correspondence $f$ between $S$ and $T$, which maps $1$ to $a$, $2$ to $a$, and $3$ to $b$. Note that $f$ can map two numbers to the same letter, but it cannot map one number to two different letters of $T$. By the way, $f$ is not the only map that can be defined between $S$ and $T$. Try to construct another one.
If the map is a one-to-one correspondence (i.e. it maps distinct elements of $S$ to distinct elements of $T$, and it turns out that all elements of $S$ and all elements of $T$ are used, then the map is called bijective. Note that in order to be able to construct a bijection between $S$ and $T$, these sets must have the same number of elements. Therefore, it is not possible to construct a bijection between the sets $S$ and $T$ given in the example above.
Nevertheless, you can construct a bijection between a set $S$ and itself. This amounts to a permutation or rearrangement of the elements of $S$. The set of all such bijections, or permutations, defined on $S$, is denoted $Sym(S)$. Hope this helps.
$\endgroup$ 0 $\begingroup$Let $A,B$ be two sets and $f:A\rightarrow B$.
We call $f$ is injective iff "For each $x,y\in A$, $f(x)=f(y)\Rightarrow x=y$."
We call $f$ is surjective iff "For each $b\in B$, there exists $a\in A$ such that $f(a)=b$".
Finally, we call $f$ is bijective if both injective and surjective.
Beyond all, why don't you study "(very) elementary set theory" before you get through abstract algebra? It will take only few weeks or a month and it would be priceless.
$\endgroup$ 1 $\begingroup$The word "map" or "mapping" is often used synonymously with "function".
A bijection on $T$ is a one-to-one function from $T$ onto $T$.
Suppose $T=\{a,b,c\}$. Then there are six bijections on $T$: $$ \begin{array}{c|c|c|c|c|c} \begin{array}{ccc} a & \mapsto & a \\ b & \mapsto & b \\ c & \mapsto & c \end{array} & \begin{array}{ccc} a & \mapsto & a \\ b & \mapsto & c \\ c & \mapsto & b \end{array} &\begin{array}{ccc} a & \mapsto & b \\ b & \mapsto & a \\ c & \mapsto & c \end{array} &\begin{array}{ccc} a & \mapsto & b \\ b & \mapsto & c \\ c & \mapsto & a \end{array} &\begin{array}{ccc} a & \mapsto & c \\ b & \mapsto & a \\ c & \mapsto & b \end{array} &\begin{array}{ccc} a & \mapsto & c \\ b & \mapsto & b \\ c & \mapsto & a \end{array} \end{array} $$ These are the six members of the set of all bijections on $T$.
Notice that I said from $T$ onto $T$ and not from $T$ into $T$.
Consider this mapping: $$\begin{array}{ccc} a & \mapsto & p \\ b & \mapsto & q \\ c & \mapsto & r \end{array} $$ This is onto the set $\{p,q,r\}$ and is into, but not onto, the set $\{p,q,r,s\}$.
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