I Help Needed: Understanding Hungerford's Algebra Book Proofs

[mr.tea]

I am trying to learn about free groups(as part of my Bachelor's thesis), and was assigned with Hungerford's Algebra book. Unfortunately, the book uses some aspects from category theory(which I have not learned). If someone has an access to the book and can help me, I would be grateful.

First, in theorem 9.1(2003 edition), he proves that the set of all reduced words, $F(X)$, of a given set X is a group(under the operation defined), using "Van Der Waerden trick". At some point in the proof he says:" since $1 \mapsto x_{1}^{\delta_{1}} \cdots x_{n}^{\delta_{n}}$ under the map $|x_{1}^{\delta_{1}}|\cdots |x_{n}^{\delta_{n}}|$, then it follows that $\varphi$ is injective". Can you please explain to me how did he get this conclusion?

Second, in the same proof, he found that this $\varphi$ is a bijection between a set and a group, and therefore concludes that since the group is associative then also the set is. Why is that?

Thank you.

 

[fresh_42]

I am trying to learn about free groups(as part of my Bachelor's thesis), and was assigned with Hungerford's Algebra book. Unfortunately, the book uses some aspects from category theory(which I have not learned). If someone has an access to the book and can help me, I would be grateful.

First, in theorem 9.1(2003 edition), he proves that the set of all reduced words, $F(X)$, of a given set X is a group(under the operation defined), using "Van Der Waerden trick". At some point in the proof he says:" since $1 \mapsto x_{1}^{\delta_{1}} \cdots x_{n}^{\delta_{n}}$ under the map $|x_{1}^{\delta_{1}}|\cdots |x_{n}^{\delta_{n}}|$, then it follows that $\varphi$ is injective". Can you please explain to me how did he get this conclusion?

Second, in the same proof, he found that this $\varphi$ is a bijection between a set and a group, and therefore concludes that since the group is associative then also the set is. Why is that?

Thank you.

I cannot answer your first question, because I don't have the book and thus don't know, what exactly your ingredients are: Bartel's trick, the definition of $|.|$ or $\varphi$ and what had happened in the proof until this point. So I'll fold here.

The second question is easier to answer. Let's say we have a bijection $\varphi : (G,\cdot) \longrightarrow X$ from a group into and onto a set. Then we simply can define $x_1 \circ x_2 := \varphi (\varphi^{-1}(x_1) \cdot \varphi^{-1}(x_2))$ as a binary operation on $X$ and the group properties are $1:1$ transported from $G$ to $(X,\circ )$.

Of course you'll still have to formally prove, that this definition provides associativity. First check whether it is well-defined, then that $\varphi (g) \circ \varphi (h) = \varphi (g\cdot h)$ and next $(x \circ y) \circ z = x \circ (y \circ z)$.

 

[martinbn]

First, in theorem 9.1(2003 edition), he proves that the set of all reduced words, $F(X)$, of a given set X is a group(under the operation defined), using "Van Der Waerden trick". At some point in the proof he says:" since $1 \mapsto x_{1}^{\delta_{1}} \cdots x_{n}^{\delta_{n}}$ under the map $|x_{1}^{\delta_{1}}|\cdots |x_{n}^{\delta_{n}}|$, then it follows that $\varphi$ is injective". Can you please explain to me how did he get this conclusion?

The map $\varphi$ is from the set $F$ of all reduced words to a subgroup of the group of all bijections of $F$. It maps the word $x_{1}^{\delta_{1}} \cdots x_{n}^{\delta_{n}}$ to the bijection of $F$ denoted by $|x_{1}^{\delta_{1}}|\cdots |x_{n}^{\delta_{n}}|$. If $\varphi$ is not injective, then it would take a word $x_{1}^{\delta_{1}} \cdots x_{n}^{\delta_{n}}\not = 1$ to the identity map of $F$. But the image is the map $|x_{1}^{\delta_{1}}|\cdots |x_{n}^{\delta_{n}}|$, which takes the element $1$ of $F$ to the element $x_{1}^{\delta_{1}} \cdots x_{n}^{\delta_{n}}$, and therefore is not the identity map of $F$, so $\varphi$ is injective.

 

[martinbn]

Second, in the same proof, he found that this $\varphi$ is a bijection between a set and a group, and therefore concludes that since the group is associative then also the set is. Why is that?

It is a bit more than that. The set also has an operation and the map is not just a bijection, but also respects the operation i.e. $\varphi(ab)=\varphi(a)\varphi(b)$. So you have

$\varphi(a(bc))=\varphi(a)\varphi(bc)=\varphi(a)(\varphi(b)\varphi(c))=(\varphi(a)\varphi(b))\varphi(c)=\varphi(ab)\varphi(c)=\varphi((ab)c)$

In the middle it is used that the you have associativity in the image, hence $\varphi(a(bc))=\varphi((ab)c)$. And since $\varphi$ is a bijection you have $a(bc)=(ab)c$.

 

[mr.tea]

I cannot answer your first question, because I don't have the book and thus don't know, what exactly your ingredients are: Bartel's trick, the definition of $|.|$ or $\varphi$ and what had happened in the proof until this point. So I'll fold here.

The second question is easier to answer. Let's say we have a bijection $\varphi : (G,\cdot) \longrightarrow X$ from a group into and onto a set. Then we simply can define $x_1 \circ x_2 := \varphi (\varphi^{-1}(x_1) \cdot \varphi^{-1}(x_2))$ as a binary operation on $X$ and the group properties are $1:1$ transported from $G$ to $(X,\circ )$.

Of course you'll still have to formally prove, that this definition provides associativity. First check whether it is well-defined, then that $\varphi (g) \circ \varphi (h) = \varphi (g\cdot h)$ and next $(x \circ y) \circ z = x \circ (y \circ z)$.

Thank you for your answer!

If $\varphi$ is not injective, then it would take a word $x_{1}^{\delta_{1}} \cdots x_{n}^{\delta_{n}}\not = 1$ to the identity map of $F$. But the image is the map $|x_{1}^{\delta_{1}}|\cdots |x_{n}^{\delta_{n}}|$, which takes the element $1$ of $F$ to the element $x_{1}^{\delta_{1}} \cdots x_{n}^{\delta_{n}}$, and therefore is not the identity map of $F$, so $\varphi$ is injective.

I am sorry. Can you explain a bit more about this part?

It is a bit more than that. The set also has an operation and the map is not just a bijection, but also respects the operation i.e. $\varphi(ab)=\varphi(a)\varphi(b)$. So you have

$\varphi(a(bc))=\varphi(a)\varphi(bc)=\varphi(a)(\varphi(b)\varphi(c))=(\varphi(a)\varphi(b))\varphi(c)=\varphi(ab)\varphi(c)=\varphi((ab)c)$

In the middle it is used that the you have associativity in the image, hence $\varphi(a(bc))=\varphi((ab)c)$. And since $\varphi$ is a bijection you have $a(bc)=(ab)c$.

Thank you very much, now it's understandable.

Thank you.