The free group $Z*Z$ is isomorphic to which group?

Question

The free group $\mathbb{Z}*\mathbb{Z}$ is basically an infinite non abelian gorup with each of its elements having infinite order except identity. I can't think of any other group which can be thought as an isomorphic group to this free group. And can someone please define any homomorphism from $\mathbb{Z}*\mathbb{Z}$ to $\mathbb{Z}$?

Answer 1

Up to isomorphism there are many, many different infinite nonabelian groups with each of its elements having infinite order except the identity.

But free groups are special. Perhaps the most special property of the free group $$\mathbb{Z} * \mathbb{Z} = \langle a \rangle * \langle b \rangle $$ is that it is the only 2-generator group (up to isomorphism) having the property that for any group $G$ and any function $$h : \{a,b\} \to G $$ there is a unique homomorphism $$H : \mathbb{Z} * \mathbb{Z} \to G $$ such that $H(a)=h(a)$ and $H(b)=h(b)$. In other words, a homomorphism with domain $\mathbb{Z} * \mathbb{Z}$ is uniquely and freely characterized by where it maps each of the two elements $a,b$.

So, for example, to define a homomorphism $H : \mathbb{Z} * \mathbb{Z} \to \mathbb{Z}$, pick what you want for $H(a)$, for example $$H(a) = 42 $$ Then pick what you want for $H(b)$, for example $$H(b)=-34890567 $$ There is a unique homomorphism $H : \mathbb{Z} * \mathbb{Z} \to \mathbb{Z}$ with those two values. You can compute any value of $H$ that you want very easily, for example $$H(aba)= H(a) + H(b) + H(a) = 42 -34890567 +42 = -34890483 $$

As for your title question, perhaps the first time you meet a free group, it will seem unfamiliar to you. But free groups are so special that mathematicians have given them their own name and their own notation. So the only real answer to your title question is: free groups are what they are, just as say infinite cyclic groups are what they are, or other very important named groups.

Answer 2

Here is an explicit construction of the free group.

The underlying set is all* 'words' with letters $a,b,c,d$.

How do we multiply two words? For example $aca$ and $bda$?

Just put one after another and apply the rule

$ab=ba=cd=dc=$ the empty string.

Let's have a go:

$aca *bda = acabda = ac(ab)da = acda = a(cd)a = aa$

Exercise: Prove this is a group.

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*To be absolutely rigorous we have to forbid any strings that can be reduced by applying the rule. In other words only allow the strings that have been reduced as far as they will go. So $dbaa$ is not allowed because it has a $ba$ in the middle, but the reduced form $da$ is allowed.

Answer 3

An example is the commutator subgroup of the modular group is a free group of rank 2 isomorphic to $F:=\mathbb{Z}*\mathbb{Z}$. Given two generators $x$ and $y$ of $F$, any integers $n$ and $m$ and the generator $z$ of $\mathbb{Z}$, then the general homomorphism $f:F \to \mathbb{Z}$ given by $f(x)=z^n,\;f(y)=z^m$ is uniquely well defined since $x$ and $y$ are generators of the free group $F$.