Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is my first time using Stack Exchange, but it looks like a good resource. I am here to ask a couple questions about my homework. We're working from the latest edition of Abstract Algebra by Herstein. This is problem #3 from p. 73 of that book, and it reads as follows:

Let $G$ be any group and $A(G)$ the set of all 1-1 mappings of $G$, as a set, onto itself. Define $L_{a} \colon G \to G$ by $L_{a}(x)=xa^{-1}$. Prove that:
(a) $L_{a} = A(G)$.
(b) $L_{a}L_{b} = L_{ab}$.
(c) The mapping $\psi:G \rightarrow A(G)$ defined by $\psi(a) = L_{a}$ is a monomorphism of G into A(G).

I believe that I have answered parts b & c correctly, but have some concerns about the rigorousness of my approaches to all three problems. I will start out here by including my proposed solution to part a. Any advice or comments would be greatly appreciated.

(a) $A(G)$, as the set of all 1-1 mappings of $G$ onto itself, can be represented as the set of all operations on an element $x \in G$ such that the result is also in $G$. Since $G$ is a group, we know that this set can be represented as the set of all group multiplications $yx \mid x \in G, y \in G$ for a given element $x$. This is because any $x$ element can be fixed as the first parameter, while the $y$ elements are taken over every element of G. We have then that $yx \in G$, due to G's closure under group multiplication. Furthermore, any element $e \in G$ has an inverse element $e^{-1} \in G$, since $G$ is a group. This means that we can consider any element in $G$ as the inverse of its inverse: $e = (e^{-1})^{-1}$. This, when plugged in for our variable $x \in G$ above, gives us that $\forall x \in G, \forall y \in G, yx^{-1} \in G$. This is exactly our given mapping $L_{a}$ above, with the labels rearranged. This shows that $L_{a}$ is a mapping from $G \rightarrow G$, as required. In order to show why this set contains every such possible mapping, we will assume that there is a mapping $M_{a}(x) : G \rightarrow G$ such that $M_{a}(x) \notin L_{a}$. This mapping, as a mapping from G to G, must take the form of a group multiplication: $M_{a}(x) = x*a, x\in G$. However, since G is a group, we have again that $a = (a^{-1})^{-1}$, or, if we let $m = a^{-1}$, that $a = m^{-1}$, and so our mapping can be rewritten as $M_{a}(x) = x*m^{-1} m \in G$. However, this is exactly the same mapping as our above $L_{a}(x)$, showing that every mapping in $A(G)$ can indeed be written as a multiplication between some element $x \in G$ and another element's inverse $a^{-1} \in G$, and thus that $L_{a} \in A(G)$. $\blacksquare$

Thanks for taking the time to check this out, even if you don't feel you can offer any help. EDIT: Thanks to those who pointed out that I had used =, not $\in$, above by mistake. Also appreciated is the edit to italicize my variables. Now I know how to as well. :)

EDIT: The responses so far have been so helpful, I'd like to put the other two parts of my solution up to solicit feedback on them as well. Hopefully they aren't as muddled as the first part's was.
changed a bit in response to feedback
(b) $L_{a}(x) = xa^{-1}$
$L_{b}(x) = xb^{-1}$
$(L_{a}L_{b})(x) = L_{a}(L_{b}(x))$
$L_{a}(L_{b}(x)) = L_{a}(xb^-1)$
$L_{a}(xb^-1) = xb^{-1}a^{-1}$
$xb^{-1}a^{-1} = x(ab)^{-1} = L_{ab}(x)$
$L_{a}(x)L_{b}(x) = L_{ab}(x) \quad\blacksquare$

(c) $\psi(a) = L_{a}(x) = xa^{-1}$. To show that this mapping is a monomorphism of $G$ into $A(G)$, we will first rely on part (a) to state that $\psi(a) = L_{a}$ is indeed a mapping from $G$ to $A(G)$. Now we must show that $\psi$ is a monomorphism of $G$ into $A(G)$. First we will show that $\psi$ is a homomorphism of $G$ into $A(G)$. To this end, we will appeal to the results of our calculations in part (b) to state that $L_{a}L_{b} = L_{ab}$, which of course implies that $\psi(a)\psi(b) = L_{a}L_{b} = L_{ab} = \psi(ab)$, which proves that $\psi$ is a homomorphism. In order to continue and show that $\psi$ is a monomorphism, we must show that it is an injective (1-1) mapping. Let $\psi(a) = Z = \psi(b)$. This can be written as: $\psi(a) = L_{a}(e) = ea^{-1} = Z$
$\psi(b) = L_{b}(e) = eb^{-1} = Z$
$Z = ea^{-1} = eb^{-1}$
$a^{-1} = b^{-1} \Rightarrow a = b$
The last line of the above follows from the uniqueness of inverse elements in $G$. This shows that the only way for two output values of $\psi$ to be equal is for their inputs to be equal as well, and thus $\psi$ is an injective homomorphism, or a monomorphism from $G$ to $A(G)$. $\blacksquare$

share|improve this question
4  
+1 for a thought-out, typeset, respectful question. You are a model new user of the site :) –  Zev Chonoles Oct 12 '11 at 19:17
1  
Do you want $L_a \in A(G)$, in part (a)? –  Dylan Moreland Oct 12 '11 at 19:23
    
It is still incorrect that $\psi(a) = L_a(e)$. Again: $\psi(a)$ is a function from $G$ to $G$ but $L_a(e)$ is an element of $G$. They cannot be equal. Don't confuse a function with one of its values. –  Arturo Magidin Oct 12 '11 at 20:51
    
OHHH, I see now what you mean. Thank you for your efforts to communicate that distinction! Hmmm... –  karmic_mishap Oct 12 '11 at 20:52
add comment

3 Answers

up vote 2 down vote accepted

I'm somewhat concerned with what you write, as I am not sure what it is you are trying to say.

You seem to be trying to show that if $f\in A(G)$, then $f=L_a$ for some $a$; first, nobody is asking you to prove that. And second, you don't seem to be doing that. And third, what you write, $L_a=A(G)$, does not even make sense! $L_a$ is a function from $G$ to itself; $A(G)$ is a set of functions from $G$ to itself. You are trying to show that a single function is equal to a set of functions. That is going to be rather hard to do...

Suppose that $f\colon G\to G$ is a (set-theoretic) 1-1 function.

It is true that for each $y\in G$, since $f(y)\in G$ by hypothesis, and since in a group we can always solve any equation of the form $ay=b$, we can find some $x$, *which depends on $y$ *, such that $f(y) = yx$. However, in general, different $y$'s will require different $x$s with the same function $f$.

So I do not see how you can simply state, as you do when you write:

Since G is a group, we know that this set can be represented as the set of all group multiplications $yx | x\in G,y\in G$ for a given element $x$.

In fact, this assertion is wrong: suppose that every bijection $f\colon G\to G$ is indeed of the form $f(y)=yx$ for some $x$. Then $f(e) = ex = x$, so for every $y$ we would have $f(y)=yf(e)$. It is very easy to see that this cannot be true for most groups, because if you have any bijection $G\to G$, you can compose it with a function that transposes $e$ and $x\neq e$ that does not have order $2$, and still get a bijection. This composition will be a bijection, but will map $e$ to $xf(e)$, but $x$ will not be mapped to $x^2f(e)$, but to $f(e)$. So not every function in $A(G)$ can be of the form $L_x$ for some $x$. The rest of the paragraph is, in my opinion, a big muddle.

In any case, you are not being asked to prove that. You are being asked to show that if $a\in G$, then the map $L_a\colon G\to G$ is an element of $A(G)$. That is, you need to show that $L_a$ is a bijection of $G$ onto itself. You are given that $L_a$ is a function from $G$ to $G$, so what you need to show is that $L_a$ is one-to-one and onto.

(As an aside, your attempt to argue about a function $M_a$ which is not one of the $L_a$ also gets off on the wrong foot; if you wanted to show by contradiction that every element of $A(G)$ is some $L_a$, you would need to start by assuming there is a function $M$ in $A(G)$ such that *for every $b\in G$ * we have $M\neq L_b$; you only assume that $M\neq L_a$ for a particular $a$, and that's no good).


Added. The penultimate line for part (b) should have $L_{ab}(x)$, not $L_{ab}$. Otherwise, it is correct.

The first line of (c) is incorrect. $L_a(x)$ is an element of $G$ (namely, $xa^{-1}$). But $\psi(a)$ is an element of $A(G)$ (namely, $L_a$). $\psi(a)$ does not equal $L_a(x)$.

You don't need to show that $\psi$ is a map from $G$ to $A(G)$: it is defined to be a map from $G$ to $A(G)$. This follows from (a), since you know, if you do (a) correctly, that $L_a\in A(G)$ for any $a\in G$.

Several times you confuse the function $L_a$, with the value of the function at $x$, $L_a(x)$. Remember that "$x$" is one of the names for elements of $G$: don't use it as if this were calculus and you call the function "$f(x)$"! The name of the function is just $L_a$, not $L_a(x)$.

You correctly show it is a homomorphism.

To prove it is 1-1, you are not really doing a proof by contradiction, you are doing a direct proof: you are showing that if $\psi(a)=\psi(b)$, then $a=b$. This is a direct proof, do it as a direct proof. But the line right after that you again commit the faux pas of confusing the function $\psi(a)$ with the particular value of $\psi(a)$ at the element $x$. $\psi(a)$ is not equal to $xa^{-1}$, and $\psi(b)$ is not equal to $xb^{-1}$.

Instead, you have to assume that $L_a=L_b$ as functions, meaning that for every $x\in G$ you have $L_a(x)=L_b(x)$. At that point you can plug in some values for $x$ and see what you can conclude. Might I suggest using the fact that $L_a(e) = L_b(e)$ to conclude that $a=b$?

share|improve this answer
add comment

To be clear, $A(G)$ is the set of all bijective set maps $G \to G$. This $A(G)$ is even a group under composition, but an element of $A(G)$ doesn't have to be one of the $L_a$ or pay much attention at all to the group structure of $G$. For example, look at the cyclic group $G$ of order $3$, which I'll write as $\{1, x, x^2\}$. I can define a bijective map of sets $G \to G$ by swapping $x$ and $x^2$, fixing $1$. You can check that this is not of the form $L_a$.

Anyway, what you want to show for (a) is that each $L_a$ is bijective, and hence a member of $A(G)$. Perhaps you can find an inverse map: what operation will undo $x \mapsto xa^{-1}$? You shouldn't have to look very far.

And your solution for (b) looks good! I'm worried about (c) only because we seem to be mixing up functions and the formulas defining them. I would write this as: If $L_a$ and $L_b$ agree as maps of sets then in particular $L_a(1) = L_b(1)$, so $a^{-1} = b^{-1}$ and hence $a = b$, using the fact that $(a^{-1})^{-1} = a$. (Or whatever suits you.)

share|improve this answer
    
Your guess is correct: if you read carefully, you'll see that it is defined as "the set of all 1-1 from $G$, as a set, onto itself." The "onto" signals surjectivity. –  Arturo Magidin Oct 12 '11 at 20:11
    
@Arturo Ah, good point! Then nothing is ambiguous. –  Dylan Moreland Oct 12 '11 at 22:25
add comment

Take a look at the problem again:

enter image description here

Part a) says that $$L_a\in A(G),$$ where the symbol in the middle denotes "is an element of" (see here), not $$L_a=A(G).$$ The latter expression doesn't make sense; $L_a$ is a function from $G$ to $G$, while $A(G)$ is a collection of functions. We want to prove that $L_a$ is a member of this collection.

share|improve this answer
    
D'oh! This is a big problem. Thanks for catching that. Does the rest of the proof lead to this as its conclusion? –  karmic_mishap Oct 12 '11 at 19:27
    
No, I'm afraid the argument you have written in your question is incorrect (or does not make sense); not every element of $A(G)$ is of the form $L_a$ for some $a\in G$ (nor is that what the statement "$L_a=A(G)$" means), and not every function from $G$ to $G$ is in $A(G)$ (only the invertible, i.e. bijective, ones are). In order to prove that $L_a\in A(G)$ for all $a\in G$, you must prove that $L_a$ is a bijection. –  Zev Chonoles Oct 12 '11 at 19:33
    
@karmic_mishap: No, the rest of the argument is, unfortunately, a muddled mess. –  Arturo Magidin Oct 12 '11 at 19:33
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.