About free group and kernel of homomorphism I'm now reading textbook in group theory but couldn't understand its briefy explanation below
"Let $G=<a,b\mid a^4=e,b^2=e,bab^{-1}=a^{-1}>, S=\{a,b\},F(S)$ be a free group and $N$ be the smallest normal subgroup including $a^4,b^2,bab^{-1}a$ as its elements so if we define homomorphism $f:F(S)\to D_4$ such that $f(a)=\sigma,f(b)=\tau$ then $f$ is surjective and $ker(f)=N$ therefore $G=F(S)/N\cong D_4$ by homomorphism theorem"
1.why $G=F(S)/N$? I know that F(S) is like the set of words fromed  by the letter $a,b$ and quotient group defined by $F(S)/N=\{aN\mid a\in F(S)\}$ what exactly the notation of G meaning or this is just a definition?
2.why $ker(f)=N$ ? for $N\subset ker(f)$ I know that $f(a^4)=\sigma^4=1,f(b^2)=\tau^2=e,f(bab^{-1}a)=e$ what's about other elements? for $ker(f) \subset N$ I still have no idea. How to prove it? Does the fact that $ker(f)$ is always normal has something to do with this?
thanks  
 A: (1) Yes, this is exactly the definition of $$\langle S | R \rangle := \left. F(S) \middle/ \langle r^w : r \in R, w \in F(S) \rangle \right.$$
The smallest $F(S)$-normal subgroup containing $R$ is $\langle r^w : r \in R, w \in F(S) \rangle$.
(2) For $N \subseteq \ker(f)$, you have shown $r \in \ker(f)$ for all $r \in R$. Since $\ker(f)$ is a normal subgroup containing $R$, and $N$ is the smallest normal subgroup containing $R$, $N \leq \ker(f)$.
The other direction is much harder to give a rigorous proof at this stage. The gist is that in $G$, you can alphabetize any expression in $a$s and $b$s using $ba=a^{-1}b = a^3b$. You can make sure the powers on $a$ are between 0 and 3 using $a^4=e$, and you can make sure the powers on $b$ are between 0 and 1 using $b^2=e$. Hence every element of $G$ can be expressed as $a^i b^j$ for $(i,j) \in \{0,1,2,3\} \times \{0,1\}$. Hence $|G| \leq 8$, but $f:G\to D_4$ hasimage of size 8, so $|G|=8$ exactly.
A: Jack Schmidt has given a correct answer to this question, but it might be useful to point out an alternative proof that $|G|\leq8$.  It's not easier than Jack's proof in this case, but I think in more complicated cases an analogous argument may well be easier than an explicit enumeration of the elements of a group.
First, consider the subgroup $A$ of $G$ generated by $a$.  Because $a^4=e$, we know that $|A|\leq 4$.  Furthermore, $A$ is a normal subgroup of $G$, because it's closed under conjugation by $b$ (as $bab^{-1}=a^{-1}\in A$), trivially closed under conjugation by $a$, and therefore closed under conjugation by any product of $a$'s and $b$'s.  Now consider the quotient group $G/A$.  It has a presentation obtained from the presentation of $G$ by adding the relation $a=e$.  But when you add that relation, two of the original relations become redundant, and the generator $a$ becomes redundant, and so the presentation reduces to $\langle b\mid b^2=e\rangle$, which is a presentation of the $2$-element group.  So we have $|G/A|=2$.  Together with $|A|\leq 4$, this gives $|G|\leq 8$.
