Consider the group generated by $<a,b : a^4 = b^4 = e,a^2b^{-2} = e, bab^{-1}a = e>$.I would like to prove that $Q_8$ is that group. Set $X = \{a,b\}$ and consider the F(X) of X. We know that there exists a homomorphism $\phi : F(X) \rightarrow Q_8$ defined by $\phi(a) = i$ and $\phi(b) = j$.
I showed that $R = <a^4,b^4,a^2b^{-2},bab^{-1}a>$ is subgroup of the kernel. I would like now to prove that $F(X) / R$ is isomorphic to $Q_8$, but I have no idea how to do that. This example was from hungerford but he left out the details of things, so I decided to complete it myself. The book mentioned that all elements F/R can be written in the form $a^ib^jN$ for $ 0 \leq i \leq 3$ and $j = 0,1$, but why is that ?
It is the intention of the author that you show that both groups $Q_8$ and $F/X$ have the same structure. Let $C_i$ be the cyclic group of order $4$ generated by $i$. This group is a normal subgroup of $Q_8$ because $i$ acts trivially on it, and the action of $j$ is given by $j^{-1}ij = j^{-1}ji^{-1} = i^{-1}$.The quotient group of $Q_8/C_i$ is of order 2 and let us denote it by $C_\textbf{j}$, where $\textbf{j}$ is the left coset of $j$. Then $Q_8 = C_i \rtimes C_\textbf{j}$ the semidirect product of two cyclic groups and so every element can be written as $i^u\textbf{j}^v$ with $u \in \{0,\ldots,3\}$ and $v \in \{0,1\}$. The same can be said about $F/X$. Let $F_a$ be the group generated by $a$. It is a cyclic group of order $4$ (as can be seen by the first relation) that is mapped isomorphically by $\phi$ to $C_i$. The fourth relation, with a little rewriting, shows that $b^{-1}ab = a^{-1}$. If we denote the quotient $(X/F)/F_a$ by $F_\textbf{b}$ where $\textbf{b}$ denotes the coset of $b$ then $\phi$ maps $F_\textbf{b}$ to $C_\textbf{j}$ and so also the semidirect product $F_a \rtimes F_\textbf{b}$ $C_i to \rtimes C_\textbf{j}$; mapping the eight elements $a^u\textbf{b}^v$ to the eight elements $i^u\textbf{j}^v$.