Explanation of semidirect products

Question

I have been reading many pdfs and books about abstract algebra, covering almost completely the topic of groups. Fortunately, I have been able to understand almost everything with the exception of semidirect products. As I have seen, if one has two groups $H,K$, one can define the semidirect product as $(h_1,k_1)(h_2,k_2) = (h_1\phi_{k_1}(h_2),k_1k_2)$, where $\phi: K \rightarrow Aut(H)$. The definition seems right (however, it would be nice if someone could confirme it), but I get struck when trying to calculate, for example, all the semidirect products between $\mathbb{Z}_4$ and $\mathbb{Z}_7$. Another important point of this concept is knowing how to calculate all groups given a certain order, so I would be really grateful if someone could help me to extend my comprehension in this topic and to get all the semidirect products given the certain groups I have mentioned before, which were just an example.

Answer 1

Semidirect products of $\mathbb{Z}_7$ by $\mathbb{Z}_4$.

In $\mathbb{Z}_7$ every non-identity element has order $7$ and so is a generator. To fix notation let $s$ be one of the generators. Clearly an automorphism of $\mathbb{Z}_7$ carries a generator to a generator; and conversely any map carry a generator to a generator provides one automorphism. Then there are $6$ automorphisms, $\alpha_r: y^m\mapsto y^{mr}$ with $r=1,2,\dots,6$.

Now fix notation by letting $f$ be a generator of $\mathbb{Z}_4$. We are seeking homomorphisms $\phi:\mathbb{Z}_4\to \{\alpha_1,\dots,\alpha_6\}$. Now we have $f^4=1$ and so $\phi(f)^4=1$. That is, the element $\phi(f)$ has order dividing $4$, that is one of $1,2,4$.

It is an easy calculation to see that $\alpha_1$ has order $1$ and that $\alpha_6$ has order $2$; and that all the other $\alpha_r$ have orders $3$ or $6$.

So there are two possible homomorphisms $\phi:\mathbb{Z}_4\to \{\alpha_1,\dots,\alpha_6\}$. There is $\phi_0:f\mapsto \alpha_1$ and $\phi_1:f\mapsto \alpha_6$.

There are therefore two semidirect products.

With $\phi_0$ note that $\alpha_1$ is the identity map, and so by your definition of how the multiplication is defined we have $(y^i, f^j)(y^I,f^J)=(y^{i+I},f^{j+J})$. (The indices are to be calculated mod $7$ and mod $4$ of course.) This is just the direct product.

With $\phi_1$ note that $\alpha_6$ is the map that inverts $y$ and each of its powers. So by your definition of how the multiplication is defined we have $(y^i, f^j)(y^I,f^J)=(y^{i+(-1)^j I},f^{j+J})$. (Again the indices are to be calculated mod $7$ and mod $4$ respectively.)