3

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.

Gokimo
  • 355
  • 1
    Could you please elaborate on how the approach in the lecture notes requires a calculator? Also, how comfortable are you with the relationship between semi-direct products and automorphisms? – Kenny Wong May 26 '23 at 22:41
  • Here is a hint in two sentences. Square and reduce modulo $41$ repeatedly. Write powers in binary. – Shaun May 26 '23 at 22:57
  • @KennyWong Well, I suppose there’s a way of calculating $2^{20} \mod 41$ without using a calculator (I don’t know how to do it right now, but I didn’t intend that to be the question). I was thinking it wouldn’t be that easy to find an element of a given order in a general cyclic group, and was wondering if it can be proven that such an element exists (and then that the non-trivial semi-direct product exists) without resorting to explicit calculations. For instance, as I said, I know that there is such a condition when $p$ and $q$ are odd primes. – Gokimo May 26 '23 at 23:19
  • @KennyWong Regarding your second question: it that’s what you mean, I know that a semi-direct product $G=H \rtimes N$ corresponds to a group homomorphism $\phi: H \rightarrow \operatorname{Aut}(N)$ and that the semi-direct product will be different from the direct product $H \times N$ iff the homomorphism $\phi$ isn’t trivial… right? – Gokimo May 26 '23 at 23:29
  • note that $20$ divides $41-1$ and $5$ divides $31-1$. That's because for $p$ prime the automorphism group of $p\mathbb{Z}$ has order $p-1$ in these cases. It's isomorphic to the nonzero elements of $p \mathbb{Z}$ under multiplication which must be cyclic (why?) and that motivates the homomorphisms. For example $2 \times 20 = 40$ is where the map $[1] \rightarrow [2]$ comes from. – CyclotomicField May 27 '23 at 00:15
  • For the record, there are five distinct (non-isomorphic) nontrivial semidirect products of this form. – Derek Holt May 27 '23 at 07:49
  • One can construct such a group concretely: Consider the affine group $\operatorname{Aff}(\Bbb F_{41}) \cong \Bbb F_{41} \rightthreetimes \Bbb F_{41}^\times$ of affine transformations $t \mapsto a t + b$ of invertible affine transformations of the affine line over $\Bbb F_{41}$. Then, the subgroup consisting of the transformations with square $a$, that is, $\Bbb F_{41} \rightthreetimes \Bbb (\Bbb F_{41}^\times)^2$, is a semidirect product isomorphic to $\Bbb Z_{41} \rightthreetimes \Bbb Z_{20}$. – Travis Willse May 27 '23 at 14:38
  • We can realize that example as the matrix group $\left{\pmatrix{1&b\\cdot&a} : a \in (\Bbb F_{41}^\times)^2, b \in \Bbb F_{41}\right} \subset \operatorname{GL}(2, \Bbb F_{41})$. – Travis Willse May 27 '23 at 14:41
  • @CyclotomicField Why $2 \times 20 = 40$ is where the map $[1] \mapsto [2]$ come from? – Gokimo May 27 '23 at 15:48
  • 1
    @Gokimo if I have two cyclic groups $\mathbb{Z}_m$ and $\mathbb{Z}_n$ with $m$ dividing $n$ then $n=mk$ for some integer $k$ and the map $x \rightarrow kx$ will be a homomorphism. Basically you're embedding a smaller cycle into a larger one. So like $\mathbb{Z}_2$ is in $\mathbb{Z}_6$ you can find a map $x \rightarrow 3x$ because $2 \times 3 = 6$ so $k$ is $3$. It's important the groups be cyclic for this sort of reasoning which happens because the automorphism group of a cyclic group of order $p$ for some prime $p$ then the automorphism group is of order $p-1$ and is cyclic. – CyclotomicField May 27 '23 at 16:55

1 Answers1

1

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.

Kenny Wong
  • 32,192
  • Once you know that $\mathbb{Z}{41}^{\times}$ is a cyclic group, it follows that there is a homomorphism $\mathbb{Z}{20} \to \mathbb{Z}_{41}^{\times}$ with $\phi(n) = g^n$ exactly when $g$ is a nonzero quadratic residue $\pmod{41}$. And for example, 2 is a quadratic residue since $41 \equiv \pm 1 \pmod{8}$. – Daniel Schepler May 27 '23 at 00:31
  • Yes, I agree. $\ $ – Kenny Wong May 27 '23 at 05:08
  • @KennyWong Thank you! Just to check that I understood correctly: in the case $\mathbb{Z}{31} \rtimes \mathbb{Z}_5$, since $30=6*5$, the map $\phi: \mathbb{Z}_5 \rightarrow \mathbb{Z}{31}^{\times}$ such that $\phi(n)=g^{5n}$ (where $g$ is a generator of $\mathbb{Z}_{31}^{\times}$) would be a non-trivial homomorphism giving rise to a non-trivial semi-direct product? – Gokimo May 27 '23 at 12:08
  • I think you mean $\phi(n) = g^{6n}$? If so, then yes, that's correct. And as Daniel points out, $\phi(n) = g^{12n}$, $\phi(n) = g^{18n}$ and $\phi(n) = g^{24n}$ will also work. – Kenny Wong May 27 '23 at 12:45
  • You really don't need to know that $\frac{\mathbb Z}{41}^\times$ is cyclic to solve this problem. You can define a nontrivial homomorphism from $({\mathbb Z}/20,+)$ to $\frac{\mathbb Z}{41}^\times$ by mapping a generator to $-1$. – Derek Holt May 27 '23 at 14:24
  • That's true, good point! – Kenny Wong May 27 '23 at 15:12
  • @KennyWong Yes sorry, that’s what I meant. I think all is clear now, thank you. – Gokimo May 27 '23 at 15:48
  • @KennyWong Just a last question, if you don’t mind: if I knew beforehand that there exists $\mathbb{Z}{41} \rtimes \mathbb{Z}_2$ and $\mathbb{Z}{41} \rtimes \mathbb{Z}5$ both non-trivial, could I conclude directly that there exists at least one non-trivial semi-direct product $\mathbb{Z}{41} \rtimes \mathbb{Z}_{20}$ as well? I mean, If $\mathbb{Z}_n \rtimes \mathbb{Z}_p$ and $\mathbb{Z}_n \rtimes \mathbb{Z}_q$ are two non-trivial semi-direct products (where $p$ and $q$ are prime), then $\mathbb{Z}_n \rtimes \mathbb{Z}_p \times \mathbb{Z}_q$ is also a non-trivial semi-direct product, right? – Gokimo May 27 '23 at 19:04
  • Note that $\mathbb Z_{20} \cong \mathbb Z_4 \times \mathbb Z_5$. If you have homomorphisms $\phi_4: \mathbb Z_4 \to \mathbb Z_{41}^\times$ and $\phi_5: \mathbb Z_5 \to \mathbb Z_{41}^\times$, then the function $\phi : \mathbb Z_4 \times \mathbb Z_5 \to \mathbb Z_{41}^\times$ defined by $\phi(n_1, n_2) = \phi_4(n_1)^{k_1} \phi_5(n_2)^{k_2}$ is a homomorphism, for any choice of $k_1, k_2 \in \mathbb Z$. And for a lot of choices of $k_1$ and $k_2$, $\phi$ will be non-trivial. You can even take $k_1 = 1$ and $k_2 = 0$ or $k_1 = 0$ and $k_2 = 1$ (assuming $\phi_4$ and $\phi_5$ are non-trivial). – Kenny Wong May 27 '23 at 19:13