I want to prove that there exists a non-trivial semidirect product $\mathbb{Z}/41\mathbb{Z} \rtimes \mathbb{Z}/20\mathbb{Z} $. I know this must be given by a non-trivial group homomorphism $\phi: \mathbb{Z}/20\mathbb{Z} \rightarrow \operatorname{Aut}(\mathbb{Z}/41\mathbb{Z},+) \cong (\mathbb{Z}/41\mathbb{Z})^* \cong \mathbb{Z}/40\mathbb{Z}$.
My lecture notes already give the answer:
$\phi([1]_{20})=([2]_{40})$ gives an automorphism of $\mathbb{Z}/41\mathbb{Z}$ of order $20$ because $2^{20} = 1 \mod 41$. But let’s say I don’t have a calculator or I just want to show that such a non-trivial semidirect product exists without necessarily calculating its explicit form, is that possible?
I know this is similar to my previous question about the semidirect product $\mathbb{Z}_{31} \rtimes \mathbb{Z}_5$, but in that case it was fairly easy to find an automorphism of order $5$ since the calculus was much more immediate. Also, I know that for $p$ and $q$ odd primes, a non-trivial semi-direct product $\mathbb{Z}_{q} \rtimes \mathbb{Z}_p$ exists iff $p$ divides $q-1$, but I’m not sure if that result holds in general.
Any help would be appreciated.
Best Answer
For any prime $p$, the multiplicative group of units, $\mathbb Z_p^\times$ , is a cyclic group.
I'm afraid I don't know of a completely elementary proof of this statement. I do however know that this statement is an easy consequence of the structure theorem for finite abelian groups.
The structure theorem for finite abelian groups tells us that the group $\mathbb Z_p^\times$ must be isomorphic to $\mathbb Z_{d_1} \times \mathbb Z_{d_2} \times \dots \times \mathbb Z_{d_k}$, for some choice of $d_1, d_2 \dots, d_k$ such that $1 < d_1 \ | \ d_2 \ | \ \dots | \ d_k$ and $d_1\times d_2 \times \dots \times d_k = p - 1$.
Therefore, $x^{d_k} \equiv 1 \bmod p$ for all $x \in \mathbb Z_p$. Thus the polynomial $f(x) := x^{d_k} - 1 \in \mathbb Z_p[x]$ has $p - 1$ roots. Hence the degree of $f(x)$ must be at least $p - 1$. In other words, $d_k \geq p - 1$.
The only way this can be the case is if $k = 1$ and $d_1 = p - 1$. Thus $\mathbb Z_p^\times$ is isomorphic to $\mathbb Z_{p-1}$, i.e. $\mathbb Z_p^\times$ is cyclic.
Having proved that $\text{Aut}(\mathbb Z_{41}) \cong \mathbb Z_{41}^\times$ is a cyclic group, we can immediately infer that there exists a non-trivial homomorphism from $\mathbb Z_{20}$ to $\text{Aut}(\mathbb Z_{41})$. Indeed, suppose that $g$ is a generator of the cyclic group $\mathbb Z_{41}^\times$. Then the map $\phi : \mathbb Z_{20} \to \mathbb Z_{41}^\times$ given by $\phi(n) = g^{2n}$ is a non-trivial homomorphism.
This non-trivial homomorphism $\phi : \mathbb Z_{20} \to \mathbb Z_{41}^\times$ gives rise to a non-trivial semi-direct product $\mathbb Z_{41} \rtimes_\phi \mathbb Z_{20}$.
Edit: As Daniel points out in the comment below, there are other homomorphisms from $\mathbb Z_{20}$ to $\mathbb Z_{41}^\times$. Indeed, the function $\phi: \mathbb Z_{20}$ to $\mathbb Z_{41}^\times$ defined by $\phi(n) = g^{2nk}$ is a homomorphism for any $k \in \mathbb Z$. This homomorphism is injective iff $\text{gcd}(k, 20) = 1$, and it is non-trivial iff $20 \nmid k$.
The approach I used to construct a non-trivial semi-direct product $\mathbb Z_{41} \rtimes \mathbb Z_{20}$ is pretty much the same approach that you would use to construct a non-trivial semi-direct product $\mathbb Z_q \rtimes \mathbb Z_p$ when $p$ and $q$ are prime and $p \ | \ q - 1$.
Indeed, $\text{Aut}(\mathbb Z_q) \cong \mathbb Z_q^\times$ is a cyclic group of order $q - 1$. Let $g$ be any generator of $\mathbb Z_q^\times$. Writing $q - 1 = pl$, the map $\phi : \mathbb Z_p \to \mathbb Z_q^\times$ given by $\phi(n) = g^{ln}$ is a non-trivial homomorphism, giving rise to a non-trivial semi-direct product $\mathbb Z_q \rtimes_\phi \mathbb Z_p$.
Notice that in my argument, I did not construct an explicit generator for the cyclic group $\mathbb Z_{41}^\times$. As far as I'm aware, there is no efficient way to construct generators for $\mathbb Z_{p}^\times$ in general, besides trial and error with a calculator. See this wikipedia article.