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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.

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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 $A\rtimes_f B$.

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