Semidirect product of order 12

Question

I am following Milne's notes on group theory and I am trying to follow one of his examples on semidirect product. Let $C_{3}$, $C_{4}$ denote the cyclic groups of order $3$ and $4$. Also fix generators $a$ and $b$ of $C_{3}$ and $C_{4}$, respectively. Also let $e$ denote the identity in both groups.

Let $\theta : C_{4} \rightarrow \text{Aut}(C_{3})$ be the nontrivial homomorphism defined by mapping the generator $b$ to the automorphism $x \mapsto x^2$.

Then $C_{3} \rtimes_{\theta} C_{4}$ is a noncommutative group of order 12 generated by $(a,e)$ and $(e,b)$ and we have the defining relations $(a,e)^3 = (e,e)$, $(e,b)^4 = (e,e)$, and $(e,b)(a,e)(e,b^{-1}) = (a^2,e)$.

I am having some trouble seeing some of this information about $C_{3} \rtimes_{\theta} C_{4}$, or at least quickly, on my own. The two things that I am having the hardest time with is that $(a,e)$ and $(e,b)$ generate $C_{3} \rtimes_{\theta} C_{4}$ and the relation $(e,b)(a,e)(e,b^{-1}) = (a^2,e)$.

When I read through the example, I did not find it surprising that $(a,e)$ and $(e,b)$ generate $C_{3} \rtimes_{\theta} C_{4}$, but I had verify this exhaustively.

Q1: Is there a quicker way to see this other than doing a ton of calculations?

Also, it is easy to verify $(e,b)(a,e)(e,b^{-1}) = (a^2,e)$, but I feel like it would have taken some time for me to see this relation had I not read it.

Q2: Is there a quick way to notice this relation?

Last, if I had been presented with this semidirect product I likely would have determined the relations $(a,e)^3 = (e,e)$ and $(e,b)^4 = (e,e)$. However if I managed to see that $(e,b)(a,e)(e,b^{-1}) = (a^2,e)$,

Q3: how could I conclude that these are the defining relations of $C_{3} \rtimes_{\theta} C_{4}$?

Ordinarily I am not concerned with how long these things take, but I am trying to prepare for an exam later this week.

Answer 1

I'm going to write the group as $\mathbb{Z}_4\rtimes_\theta\mathbb{Z}_3$. Then the obvious generators are $(1,0)$ and $(0,1)$: for any $(a,b)\in\mathbb{Z}_4\rtimes\mathbb{Z}_3$, $a\cdot(1,0)\times b\cdot(0,1)=(a,0)(0,b)=(a+\theta(0)(0),0+b)=(a,b)$.

Noticing the relation you stated is exactly the same as saying $(0,1)(1,0)=(\theta(1),0)=(2,0)$. This lines up with what we need for the semidirect product: the internal structure of each group alongside the way that the elements relate with each other, and the automorphism structure means that we can just see how the generators relate to each other.

Answer 2

Q1 & Q2:
The idea is that the basic set of the semi-direct product is a cartesian product of $C_3$ and $C_4$ so every element of $G = C_3 \rtimes_{\theta} C_4$ can be written as $(a^i,b^j)$ for some integers $i,j$. It is clear that this is equivalent to simply write $a^ib^i$ for this since we can write $a$ for $(a,e)$ and $b$ for $(e,b)$. The problem is how do we write $ba$ and this is where the relations come in:
Q3:
Clearly $a^3 = e$ and $b^4 = e$. Using the notation from above we can write the relation coming from the automorphism as $bab^{-1} = a^2$. So what happens to $ba$? Well $$ ba = bab^{-1}b = a^2b $$ so $ba$ in its "awkward" notation is $(a^2,b)$. It should be clear that any word in $G$ can be written as $a^ib^j$ by working every $a$ to the left side until all $a$'s and $b$'s are separated. As an exercise calculate $(ab)^2$.