Let $G=P\rtimes Q,P=\Bbb Z_7=\langle b\rangle,Q=\Bbb Z_3=\langle a\rangle.$ Let $\theta:Q\to{\rm Aut}P$ be $\theta(a)(b)=b^4$. Classify $G/G'$.

Question

Let $G=P\rtimes Q$ with $P=\mathbb{Z}_7=\langle b\rangle$ and $Q=\mathbb{Z}_3=\langle a\rangle.$ Take $\theta: Q\rightarrow {\rm Aut}P$ to be $\theta(a)(b)=b^4$. Classify up to isomorphism $ G/G'$.

I was able to solve this using basic techniques: $G$ is not abelian, because conjugation is not trivial. Therefore $G'\not=\{1_G\}$. Fair enough. Let us calculate a comutator:

$$[(p,q),(\tilde{p},\tilde{q})]=(\text{something},(3-q)+(3-\tilde{q})+q+\tilde{q})=(\text{something},0)$$

But this means that $G'\leq (\mathbb{Z}_7,0)$. By Lagrange's Theorem $|G'|=1$ (absurd!) or $|G'|=7$. This means $|G'|=7$ and therefore $|G/G'|=3$ and $G/G'\approx \mathbb{Z}_3$.

Fair enough. The trouble is that I have looked at the solution and I am not able to understand anything there:

$$G/G'=\langle a,b\mid a^3=1=b^7,b=baba^{-1}=b^4\rangle=$$ $$\langle a,b\mid a^3=1=b^7,b=baba^{-1}, 1=b^3\rangle=$$ $$\langle a,b\mid a^3=1=b^7,b=baba^{-1},b=1\rangle=\langle a\mid a^3=1\rangle=\mathbb{Z}_3$$

Shouldn't we consider elements of the form $gG'$? We didn't compute $G'$. What does this equality $G/G'=\langle a,b\mid a^3=1=b^7,b=baba^{-1}=b^4\rangle$ mean?

Answer 1

Let me show you a concrete realization of the group $G$: all the mod 7 invertible matrices $$ \begin{pmatrix}x&y\\0&1\end{pmatrix} $$ where $y$ is arbitrary and $x \equiv 1, 2, 4 \bmod 7$ (the subgroup of order $3$ in $(\mathbf Z/(7))^\times$). Inside this group we have the normal subgroup $$ P = \left\{\begin{pmatrix}1&y\\0&1\end{pmatrix}\right\} = \left\langle \begin{pmatrix}1&1\\0&1\end{pmatrix} \right\rangle $$ and the subgroup $$ Q = \left\{\begin{pmatrix}x&0\\0&1\end{pmatrix} : x = 1, 2, 4\right\} = \left\langle \begin{pmatrix}2&0\\0&1\end{pmatrix} \right\rangle = \left\langle \begin{pmatrix}4&0\\0&1\end{pmatrix} \right\rangle. $$ Since $$ \begin{pmatrix}x&0\\0&1\end{pmatrix}\begin{pmatrix}1&y\\0&1\end{pmatrix} \begin{pmatrix}x&0\\0&1\end{pmatrix}^{-1} = \begin{pmatrix}1&xy\\0&1\end{pmatrix}, $$ you can take $a = (\begin{smallmatrix}4&0\\0&1\end{smallmatrix})$.

The group $G/G'$ is abelian and is the "biggest" abelian quotient of $G$: if there is a homomorphism $G \to A$ for some abelian group $A$, then it is trivial on $G'$, so we get an induced homomorphism $G/G' \to A$. Thus all abelian quotients of $G$ are quotients of $G/G'$. Using the above matrix model for $G$, since $$ \begin{pmatrix}x&y\\0&1\end{pmatrix}\begin{pmatrix}x'&y'\\0&1\end{pmatrix}= \begin{pmatrix}xx'&y + xy'\\0&1\end{pmatrix} $$ one obvious homomorphism from $G$ onto an abelian group is $G \to \{1,2,4 \bmod 7\}$ by $$ \begin{pmatrix}x&y\\0&1\end{pmatrix} \mapsto x. $$ In terms of the notation you use, this is the map $G \to Q$ by $pq \mapsto q$. The kernel is $P$, so $G' \subset P$. Since $|P| = 7$, either $G'$ is trivial or $G' = P$. In this way you can figure out what $G'$ is without computing any commutators directly or having to use presentations. (After writing this up, I see now that it partly duplicates what you wrote, but my main purpose here is to show that the abstrac semidirect product in your problem does have a concrete description as a matrix group.)