2

Right now, I am reading Evan Chen's Napkin to study Abstract Algebra and other various topics. In lemma 1.2.5, he states:

Let $G$ be a group, and pick a $g\in{G}$. Then the map $G\rightarrow{G}$ given by $x\mapsto{gx}$ is a bijection.

to which he asks the reader to prove this lemma. As I am not familiar with this notation, I think the map $G\rightarrow{G}$ given by $x\mapsto{gx}$ is the same as a function $f$ such that $f:G\to{G}$ and $f(x)=gx$, though I am not sure (I used Is there any difference between mapping and function? and Different arrows in set theory: $\rightarrow$ and $\mapsto$ as points of reference). Given that my assumption is right, then surjection would be proven by:

Let $y=f(x)$. Then $y=gx$ and $x=\frac{y}{g}$, demonstrating $\forall{y}\in\mathbb{R},\exists{x}\in\mathbb{R}$ such that $y=f(x)$ and proving surjectivity.

Then for injectivity, suppose $f(x_1)=f(x_2)$. Then $gx_1=gx_2$ and thus $x_1=x_2$, proving injectivity.

Thus, since $f$ is both injective and surjective, then $f$ is bijective. Similarly, this implies the map $G\rightarrow{G}$ given by $x\mapsto{gx}$ is bijective.

My main question is whether the map $G\rightarrow{G}$ given by $x\mapsto{gx}$ is the same as a function $f$ such that $f:G\to{G}$ and $f(x)=gx$. Could you also check my proof to make sure I haven't missed any considerations (such as whether I need to consider the binary operator of G). If my assumption in the main question is correct, this is just the same as proving the function $f(x)=gx$ is bijective (something very easy to do).

JC12
  • 1,040
  • 3
    Your interpretation of the symbols is correct. The $\rightarrow$ is used for representing a function between the sets. The symbol $\mapsto$ is used to represent an element of the domain maps to the corresponding element of the co-domain. Regarding the proofs, the idea is okay but you need to use the inverse element $g^{-1}$ as opposed to $\frac{1}{g}$. So for surjectivity you are claiming that for $y \in G$, there exists $g^{-1}y$ that will map to $y$. Also keep in mind that $g^{-1}y$ need not be the same as $yg^{-1}$. – Anurag A Jul 08 '20 at 03:16
  • 2
    There is no such thing as $\frac{y}{g}$ unless the group is Abelan. You can use $g^{-1}y$, though. – markvs Jul 08 '20 at 03:17
  • 2
    @AnuragA, thanks for the clarification. If I change $\frac{y}{g}$ to $g^{-1}y$ and rewrite that $gx_1=gx_2$ implies $g^-1gx_1=g^-1gx_2$, leading to $1_Gx_1=1_Gx_2$, would the proof be correct. – JC12 Jul 08 '20 at 03:19
  • 1
    @JC12 Yes that would fix it. – Anurag A Jul 08 '20 at 03:20
  • Btw another way to show that a function $f\colon X\to Y$ is bijective is to show that there is a function $g\colon Y\to X$ such that $f\circ g$ is the identity on $Y$ and $g\circ f$ is the identity on $X$. Such a $g$ exists if and only if $f$ is bijective and in this case, $g$ is the inverse of $f$. I think this is often the quicker way to check bijectivity. – Filippo Mar 06 '22 at 22:25

1 Answers1

1

In the surjectivity part, "$\Bbb R$" is out of place. Group definition entails that every element $g\in G$ has an inverse, namely some $g^{-1}\in G$ such that $gg^{-1}=g^{-1}g=1_G$ (here $1_G$ is the identity element of $G$, whose existence is another group axiom, and the omitted operation sign in between elements is group's operation and has nothing to do with the multiplication in $\Bbb R$). Therefore, for every $y\in G$, there is $x:=g^{-1}y$, such that $f(x)=$ $gx=$ $g(g^{-1}y)=$ $(gg^{-1})y=$ $1_Gy=$ $y$ (note that another axiom of group's definition is used in the chain, namely the associativity of group's operation). The injectivity is okay, though it may worth mentioning that the result holds for the cancellation law, which is among the very first lemmata on groups in general.