Problem statement: Find four groups of order 20 not isomorphic to each other.
In the accepted answer for this question, it says "If $\varphi_4(x) = \alpha^3$, then $\mathsf{im}\ \varphi_4 = \mathsf{im}\ \varphi_2$. Since $\mathbb{Z}_4$ is cyclic, by a previous theorem, $H \rtimes_{\varphi_4} K \cong H \rtimes_{\varphi_2} K$."
I am assuming the "previous theorem" is related to rules about isomorphic semidirect product. (i.e. if we can make two homomorphisms $K \to{\rm Aut}(H)$ equal by precompsing something from ${\rm Aut}(K)$ or conjugate postcomposing something from ${\rm Aut}(H)$)
The question is, what is an efficient way to fully explain the argument without "by a previous theorem"? The only thing I can think of is:
- Given $\phi_2(x) = \alpha$ and $\phi_4(x) = \alpha^3$, we find that $\phi_2(x^3) = \alpha^3$.
- we find an automorphism $f$ from ${\rm Aut}(K)$ that maps $x$ to $x^3$.
- now since $\phi_4(x) = \phi_2 \circ f(x)$, we can claim that these two maps are equal as $x$ is a generator. Hence we are done.
Also, I do not seem to use the fact that $\Bbb Z_4$ is cyclic anywhere?
