Free Group Proofs I am completely new to free groups and I have no idea how to solve this, so if anyone could help that would be amazing. Thanks! The problem goes as follows:
Considering the diagram:

Let $S=\{x,y,z\}$ and let $G$ be the free group $G=F\{a,b\}$. Let $\phi$ be the map $x\mapsto a^2, y\mapsto b^2, z\mapsto ab$. Then the universal property says $\Phi$ is a unique homomorphism $F\to G$ with $\Phi \circ i=\phi$. Prove:
$a)$ $\Phi:F\to G$ is injective
$b)$ the image $\Phi(F)$ is a normal subgroup of $G$
$c)$ $G/\Phi(F) \cong \Bbb Z/2$
Thanks a lot!
 A: Though there may be slick proofs involving just the universal properties themselves, it's often helpful to see what these constructions concretely are.


*

*$F$ is the free group on 3 elements; it consists of all possible words made of $x$, $y$, $z$, their inverses, and a unit. For example, $xy^{-1}zyz^{-1}z^{-1}$ is a word in $F$.

*$i : 3 \hookrightarrow F$ sends the letters $x, y, z$ to their corresponding "one-letter words" in $F$.

*$G$ is the free group on 2 elements; it consists of all possible words made of $a$ and $b$, their inverses, and a unit.

*$\varphi : 3\rightarrow G$ sends the letters $x,y,z$ to the words $aa$, $bb$, and $ab$ respectively.

*The unique function $\Phi$ must exist because of $F$'s universal property. Hence $\Phi: F\xrightarrow{!}G$ sends words made out of $x$, $y$, $z$ (etc.) into words made out of $a$ and $b$ (etc.) by systematically replacing all instances of $x$ with $aa$, $y$ with $bb$, $z$ with $ab$, and similarly for their inverses.

*(In general, the free group property means that if you have a function $\varphi$ sending letters in $L$ to a group $G$, there's a unique function $\Phi$ which extends $\varphi$ to words made of letters in $L$ (i.e. the free group on $L$). That unique function $\Phi$ applies $\varphi$ to each of the letters in the word, then multiplies the results together.)
Proving that $\Phi$ is injective:


*

*Intuitively, you can tell that the translation process mentioned above is uniquely reversible: given a $\Phi$-made word like $aab^{-1}b^{-1}ab$ in $G$, you can find the unique string $xy^{-1}z$ in $F$ that produced it.

*You could formalize this process to show that $\Phi$ is injective. Alternatively, another approach is to show that $\Phi$ sends nothing to the identity of $G$ except for the identity of $F$ itself. ("$\Phi$ has a trivial kernel"). 
Proving that $\Phi(F)$ is a normal subgroup of $G$:


*

*A normal subgroup is one that is closed under conjugation by members of $G$. 

*Because $G$ is a free group, it is made up of words of $a$ and $b$ and their inverses and unit. Hence we only need to show that it's closed under conjugation by these one-letter words and their inverses. 


Proving that $G/\Phi(F)$ is the cyclic group of order 2:


*

*You know that $\Phi(F)$ contains, for example, $aa$, $bb$, and $ab$. When you quotient by a normal subgroup, you cause every member of that subgroup to become equal to the identity. 

*Note that $\Phi(F)$ also contains $ba$, because $\Phi(yz^{-1}x)=\Phi(y)\Phi(z^{-1})\Phi(x) = bb(ab)^{-1}aa = bbb^{-1}a^{-1}aa = ba$.

*Hence in $G/\Phi(F)$, we have that $aa = 1$, $bb = 1$, $ab = 1$, $ba=1$, among other things. Hence all higher powers of $a$ and $b$ are equal to the identity, and $a$ and $b$ are inverses of each other. The group consists of just $\{1, a\}$, and has the same multiplication structure as $\mathbb{Z}/2$.
