This is in reference to the question already posted here. Suppose $F$ is the free group on $\{a,b\}$ and $N$ is the normal subgroup generated by $\{b^3, a^7, aba^{-2}b^{-1}\}$ (not the subgroup generated by $\{b^3, a^7, aba^{-2}b^{-1}\}$). I am trying to prove that $F/N$ is isomorphic to the group $A\rtimes_f B$ where $B=\{1,x,x^2\}$ and $A=\{1,a,a^2,\dots,a^6\}$ are cyclic groups, $\phi\in Aut(A)$ is the automorphism $\phi(a^i)=a^{2i}$ and $f$ is the homomorphism $f(x^k)=\phi^k$.
Clearly elements of $F/N$ are all of the form $Nb^ia^j; 0\le i\le 2,0\le j\le 6$ due to $Nab=Nba^2, Nb^3=Na^7=N$.
My question is as follows:
How do I establish an isomorphism from $F/N$ to $A\rtimes_f B$ in the natural way, i.e. $g:F/N\to A\rtimes_f B$ given by $g(Nb^ia^j)=(a^j,x^i)$? I am not even able to show that $g$ is well defined. It is not clear to me that $F/N$ cannot have less then 21 elements either, although that part would follow if the isomorphism is established.
To establish the isomorphism, you need to use universal properties.
To avoid any confusion, I chose to write $A=\{1,\dots, y^6\}$ so that $a$ and $b$ are elements of the free group whereas $x$ and $y$ are elements of the semi-direct product.
Let's say that $\phi(y):=y^4$ instead.
By universal property of free groups, there exists a group morphism $\psi:F\to A\rtimes_f B$ such that $F(a)=y\in A$ and $F(b)=x\in B$.
Clearly, $a^7$ and $b^3$ belongs to $\ker\psi$. One verifies "by hand" $yxy^{-2}x^{-1}=y\phi(y^{-2})=y^{-7}=1$ in $ A\rtimes_f B$ is trivial. Therefore $\ker \psi$ contains $N$.
As a result, applying universal property of quotients, there exists a group morphism $\psi_N:F/N\to A\rtimes_fB$ such that $\psi_N(N\gamma):=\psi(\gamma)$.
Since $y$ and $x$ generate $A\rtimes_f B$ on one hand and that they belong to the image of $\psi$ whence of $\psi_N$, it follows that $\psi_N$ is onto.
Finally your argument shows that $F/N$ has at most $21$ elements. Therefore $\psi_N$ is a group morphism of a group $F/N$ with at most $21$ elements onto a group $A\rtimes_f B$ with exactly $21$ elements. This shows that $F/N$ is isomorphic to $A\rtimes_f B$.