- #1
mr.tea
- 102
- 12
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.
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.