Find all group of order $20$ which is a semidirect product of a cyclic group of order $4$ by a cyclic group of order $5$

Question

Problem: Find all group of order $20$ that are a semidirect product of a cyclic group of order $4$ by a cyclic group of order $5$.

My attempt: We knew that a cyclic group of order $n$ is isomorphic to $\mathbb{Z}_n$, so we have $C_4 \cong \mathbb{Z}_4$, $C_5 \cong \mathbb{Z}_5$ and $|C_4| = |\mathbb{Z}_4| = 4$, $|C_5| = |\mathbb{Z}_5| = 5$. Then we find a semidirect product of $\mathbb{Z}_4$ by $\mathbb{Z}_5$. To find this, we define an automorphism $\theta \colon \mathbb{Z}_5 \rightarrow \mathrm{Aut}(\mathbb{Z}_4)$, while $\mathrm{Aut}(\mathbb{Z}_n) = \{\sigma_k \mid k \in \mathbb{Z}, (k,n) = 1\}$, so $\mathrm{Aut}(\mathbb{Z}_4) = \{\sigma_1, \sigma_3\}$ and $|\mathrm{Aut}(\mathbb{Z}_4)| = 2$.

My professor told that there is only the trivial homomorphism $\mathbb{Z}_5 \rightarrow \mathrm{Aut}(\mathbb{Z}_4)$ since $(5,2) = 1$.

So my question:

Why from $(5,2) = 1$ can we conclude that there is only one trivial homomorphism, and for the trivial homomorphism, how do I define it?

Answer 1

Why from $(5,2)$ we can conclude that there is only one trivial homomorphism

Let $f:\Bbb Z_5 \to \text{Aut}(\Bbb Z_4) \sim \Bbb Z_2$ be a homomorphism with $f(1)=a$. Now order of $f(1)$ divides both $5$ and $2$. But their gcd is one, so $a$ must be identity and hence $f$ is trivial!

Answer 2

Since $20=2^25$, by Sylow theorem, there is only one Sylow-5 subgroup which is normal. So $G\cong C_5\rtimes C_4$ and the homomorphism is $\,\theta : \mathbb{Z}_4 \rightarrow \mathrm{Aut}(\mathbb{Z}_5)\cong \mathbb{Z}_4$. First, let's define the automorphism and homomorphism rigorously.

First, automorphisms in $\operatorname{Aut}(C_p)$ are: $$ \varphi_i: C_p \to C_p,\quad \varphi_i(a) = a^{i}, \, a\in C_p,\,(i,p)=1 $$

Suppose $C_p=\langle a\rangle$ and $C_q=\langle b\rangle$. Then a homomorphism from $\theta:C_q\to\operatorname{Aut}(C_p)$ is: $$ \theta(b^k)=\sigma^k_i(a)=b^kab^{-k}=a^{i^k},\:1\leqslant k\leqslant q\tag1 $$ Where $\sigma^k_i(a)$ is an automorphism in $\operatorname{Aut}(C_p)$. $\sigma^k_i(a)$ depends on both $i$ and $k$ because it is an automorphism based on $\varphi_i$ by conjugation of $b^k$. For any $\:1\leqslant k,j\leqslant q$, we have $$ \theta(b^kb^j)=\theta(b^{k+j})=b^{k+j}ab^{-(k+j)}=a^{i^{(k+j)}}=(a^{i^j})^{i^k}=\theta(b^k)\circ\theta(b^j) $$ Hence, $\theta(b^k)$ is a homomorphism.

If $p$ is prime, $\,i^{p-1}\equiv 1\, \operatorname{mod} p\,$ and $\operatorname{Aut}(C_p)=C_{p-1}$. Also, from (1), $\,i^{q}\equiv 1\, \operatorname{mod} p\,(b^q=1)$. So if $ (p-1,q)=1$, $\,i=1$ and $\theta$ is trivial. If $(p-1,q)\neq1$, each $s=i^{\frac{p-1}{q}d}\,$ decides a nontrivial homomorphism $\theta$ ($d$ is a common proper factor of $p-1$ and $q$) because $$\ s^{p-1}=(i^{p-1})^l\equiv 1\, \operatorname{mod} p\quad\text{and}\quad s^{q}=(i^{p-1})^d\equiv 1\, \operatorname{mod} p $$

In your case, $\,\theta : \mathbb{Z}_4 \rightarrow \mathrm{Aut}(\mathbb{Z}_5)\cong \mathbb{Z}_4$. So $\varphi_1(a)=a^i$ and $\varphi_2(a)=a^{i^2}$ are two homomorphisms, each of which decides a non-isomorphic model, $G\cong C_5\rtimes_{\varphi_1} C_4$ and $G\cong C_5\rtimes_{\varphi_2} C_4$ respectively.

Answer 3

$$ \DeclareMathOperator{\Ker}{Ker} \DeclareMathOperator{\Im}{Im} \DeclareMathOperator{\Aut}{Aut} $$

Also, there's another way to prove $f$ is trivial. Consider corollary from homomorphism theorem: $$f: G → H, \quad\textrm{where $f$ is a homomorphism}$$ $$|G| = |\Im f|\cdot|\Ker f| $$ So in our case $5 = |\Im f| |\Ker f|$. We know that $|\Im f| <= |\Aut(Z_4)| = |Z_2| = 2$. Thus, the only possible way is that $|\Ker f| = 5 $ and $|\Im f| = 1$, which concludes $f$ is trivial homomorphism.

Answer 4

A way to prove that the homomorphism $f\colon \Bbb Z_5\to\operatorname{Aut}(\Bbb Z_4)$ is only the trivial one without any prior knowledge of the structure of $\operatorname{Aut}(\Bbb Z_4)$ (not even its order), is the following.

Since $\Bbb Z_5$ is of prime order, action's pointwise stabilizers are trivial or the whole $\Bbb Z_5$. Therefore: $$\sum_{i=0}^3\left|\operatorname{Stab}(i)\right|=k+5(4-k)\tag1$$ for some integer $k$, $0\le k\le 4$. By Burnside's (counting) lemma, $5$ must divide the RHS of $(1)$, whence $k=0$: all the stabilizers do coincide with the whole $\Bbb Z_5$, and so does the kernel of the action. Therefore, the action is trivial.

---

This argument can be developed further to prove that, for $p$ and $q$ distinct primes such that $p\nmid q-1$, the only homomorphism $f\colon \Bbb Z_p\to\operatorname{Aut}(\Bbb Z_q)$ is the trivial one (so, again without any prior knowledge of the structure of $\operatorname{Aut}(\Bbb Z_q)$).

With reference to the action $f$, we get:

$$\sum_{i=0}^{q-1}\left|\operatorname{Stab}(i)\right|=\sum_{j=0}^{p-1}\left|\operatorname{Fix}(j)\right| \tag2$$ where in this case $\operatorname{Fix}(j)$ is a subgroup of $\Bbb Z_q$, for every $j$. By the primality of the orders of $\Bbb Z_p$ and $\Bbb Z_q$, $(2)$ and Lagrange yield: $$k+p(q-k)=l+q(p-l) \tag3$$ for some integers $k,l$ such that $0\le k\le q-1$ and $0\le l\le p-1$. From $(3)$: $$k(p-1)=l(q-1) \tag4$$ By Burnside's (counting) lemma, $p\mid k$ (see the LHS of $(3)$); but since by assumption $p\nmid q-1$, from $(4)$ follows $p\mid l$: contradiction, because $p>l$. So, if $p\nmid q-1$, there isn't any nontrivial action (by automorphisms) of $\Bbb Z_p$ on $\Bbb Z_q$.