I'm working through some questions on semi-direct products, and although I can work out these problems (for the most part), I usually have trouble completing them. I have identified some of the things that typically throw me off, and I thought I could share my work on an example from my textbook so that someone could clarify some of my problems.

Since there are many cases to consider, I'll only post those that have given me trouble.

Let $|G| = 12$. Let V be a Sylow $2$-subgroup of $G$ and let $T$ be a Sylow $3$-subgroup of $G$. It is either the case that $V \unlhd G$ or that $T \unlhd G$.

Suppose $T \unlhd G$. So we determine all possible homomorphisms $\varphi: V \longrightarrow \text{Aut}(T)$. First note that $\text{Aut}(T) = \langle x \rangle \cong Z_2$. Assume first $V = Z_4 = \langle y \rangle$. We have the trivial homomorphism $\varphi(y)(x) = x$, but I'm aware that we also have $\varphi(y)(x) = x^{-1}$. I know that inverting $x$ is an automorphism (since $Z_2$ is abelian), so I can see why this is true. But I'm not entirely clear how I could deduce this.

I would guess that it must be the case that $y \mapsto x$, in which case the action determines $yxy^{-1} = x \iff yx = xy$ (so this is the trivial homomorphism, giving the direct product). But I'm not sure what $y$ needs to map to give the second homomorphism.

In the second case, suppose $V \cong Z_2 \times Z_2 = \langle a \rangle \times \langle b \rangle $. Here there are three nontrivial homomorphisms $\varphi: V \longrightarrow \text{Aut}(T)$. Two of them are clearly the same, so we can only consider these two: $\varphi(a) = x, \varphi(b) = x$ and another one is given by $\varphi(a) = x, \varphi(b) = 1$. That much I can figure out, but I can't figure out what their semi-direct products are, and why they both yield the same semi-direct product.

I also am confused by my textbook's explanation that our first homomorphism here has kernel $\langle ab \rangle$ and that in the semi-direct product, both $a$ and $b$ act by inverting $T$ and that $ab$ centralizes $T$.


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