Prove that for the semidirect product $\mathbb{Z}_5 \rtimes \mathbb{Z}_3$, the homomorphism $\alpha$ is trivial

Question

Suppose $G \cong \mathbb{Z}_5 \rtimes_\alpha \mathbb{Z}_3$ with respect to a homomorphism $\alpha:\mathbb{Z}_3 \to \mathrm{Aut}(\mathbb{Z}_5)$. Show that $\alpha$ is trivial and that $G \cong \mathbb{Z}_5 \times \mathbb{Z}_3$. Is $G$ a cyclic group?

My proof:

Let $([a]_5, [b]_3), ([c]_5, [d]_3) \in \mathbb{Z}_5 \rtimes_\alpha \mathbb{Z}_3$

The semidirect product gives

$$([a]_5, [b]_3) ([c]_5, [d]_3) = ([a]_5 \alpha_{[b]_3}([c]_5), [b]_3[d]_3)$$ $$= ([a]_5 ([b]_3 [c]_5 [b]_3^{-1}), [b]_3 [d]_3)$$ $$= ([a]_5[c]_5[b]_3[b]_3^{-1}, [b]_3 [d]_3)$$ $$= ([a]_5[c]_5, [b]_3[d]_3)$$ $$= \mathbb{Z}_5 \times \mathbb{Z}_3$$

the definition of the regular direct product.

In the middle of the calculations we see that the inner automorphism is trivial always since $\mathbb{Z}_n$ is abelian.

Therefore $G \cong \mathbb{Z}_5 \times \mathbb{Z}_3$.

Finally it is cyclic since $\mathrm{gcd}(5,3) = 1$ so $5$ and $3$ are coprime. Therefore the direct product is cyclic

Answer 1

The problem in your reasoning is that you assume $\alpha$ must be $$\alpha: \mathbb{Z}_3 \to \operatorname{Aut}(\mathbb{Z}_5): [b]_3 \mapsto \iota_{[b]_3},$$ with $$\iota_{[b]_3}: \mathbb{Z}_5 \to \mathbb{Z}_5: [c]_5 \mapsto [b]_3[c]_5 [b]_3^{-1}.$$

This "inner automorphism" doesn't make any sense, because you are trying to multiply elements that belong to different groups!

Let us first try to calculate what "known group" $\operatorname{Aut}(\mathbb{Z}_5)$ is isomorphic to, and then try to construct a morphism $\alpha: \mathbb{Z}_3 \to \operatorname{Aut}(\mathbb{Z}_5)$.

Since $\mathbb{Z}_5$ is cyclic and generated by $[1]_5$, any automorphism $\varphi \in \operatorname{Aut}(\mathbb{Z}_5)$ is determined completely by the value of $\varphi([1]_5)$. There are $5$ possible values of $\varphi([1]_5)$. I'll leave it to you to prove that $\varphi([1]_5) = [0]_5$ will not produce an automorphism, and the other four possibilities will form a group isomorphic to $\mathbb{Z}_4$. Thus $\operatorname{Aut}(\mathbb{Z}_5) \cong \mathbb{Z}_4$.

So we can see $\alpha$ as a morphism $$\alpha: \mathbb{Z}_3 \to \mathbb{Z}_4.$$ For the same reason as mentioned above, $\alpha$ is completely determined by the value of $\alpha([1]_3)$. Again, I'll leave it up to you to show the only possible $\alpha$ is $$\alpha: \mathbb{Z}_3 \to \mathbb{Z}_4: [b]_3 \mapsto [0]_4,$$ which translates to $$\alpha: \mathbb{Z}_3 \to \operatorname{Aut}(\mathbb{Z}_5): [b]_3 \mapsto \operatorname{id}.$$

Answer 2

It needn't to know that $\operatorname{Aut}(\Bbb Z_p)\cong\Bbb Z_{p-1}$ (not even that $\left|\operatorname{Aut}(\Bbb Z_p)\right|=p-1$).

If a group $G$ acts on another group $H$ (by automorphisms), then: $$\operatorname{Fix}(g)\le H, \forall g\in G \tag 1$$ where $\operatorname{Fix}(g):=\{h\in H\mid \phi_g(h)=h\}$, being $\phi\colon G\to \operatorname{Aut}(H)$ the action. In fact, for every $g\in G$, $\phi_g(1_H)=1_H$, whence $1_H\in\operatorname{Fix}(g)\ne\emptyset$; moreover, by definition, $\operatorname{Fix}(g)\subseteq H$; finally, for every $g\in G$ and $h_1,h_2\in\operatorname{Fix}(g)$, $\phi_g(h_1h_2^{-1})=$ $\phi_g(h_1)\phi_g(h_2^{-1})=$ $\phi_g(h_1)\phi_g(h_2)^{-1}=$ $h_1h_2^{-1}$, whence $h_1h_2^{-1}\in\operatorname{Fix}(g)$.

If $G$ and $H$ are both finite, say $G=\{g_1,\dots,g_{|G|}\}$ and $H=\{h_1,\dots,h_{|H|}\}$, then: $$\sum_{i=1}^{|H|}\left|\operatorname{Stab}(h_i)\right|=\sum_{j=1}^{|G|}\left|\operatorname{Fix}(g_j)\right| \tag 2$$ and (from Burnside's Lemma): $$|G|\text{ divides } \sum_{j=1}^{|G|}\left|\operatorname{Fix}(g_j)\right| \tag 3$$ where now the terms in the RHS of $(2)$ and in $(3)$ all divide $|H|$ (Lagrange).

In particular, take $G=\Bbb Z_p$ and $H=\Bbb Z_q$, where $p$ and $q$ are distinct primes. If a nontrivial homomorphism $\phi$ exists, then $\left|\operatorname{Fix}(g_{\bar j})\right|=1$ and $\left|\operatorname{Stab}(h_{\bar i})\right|=1$, for some $\bar j\in \{1,\dots,p\}$, $\bar i\in\{1,\dots,q\}$. Then $(2)$ and $(3)$ yield: \begin{cases} k+(q-k)p=l+(p-l)q \\ p\mid pq-l(q-1) \\ \end{cases} and rearranging: \begin{cases} k(p-1)=l(q-1) \\ p\mid pq-l(q-1) \\ \tag 4 \end{cases} for some $1\le k\le q$ and $1\le l\le p$. Now, if $\bf{p\nmid q-1}$ (in particular if $p>q$), then from $(4.2)$: \begin{alignat}{1} &p\mid l \Longrightarrow\\ &\bf{l=p}\stackrel{(4.1)}{\Longrightarrow}\\ &k(p-1)=p(q-1)\Longrightarrow\\ &k=\frac{p}{p-1}(q-1)>q-1\Longrightarrow\\ &\bf{k=q}\\ \end{alignat} But $(k,l)=(q,p)$ is not a solution of $(4.1)$, and hence there isn't any nontrivial homomorphism $\phi\colon \Bbb Z_p\to\operatorname{Aut}(\Bbb Z_q)$, namely $\Bbb Z_p\ltimes\Bbb Z_q=$ $\Bbb Z_p\times\Bbb Z_q$. Your case is just the particular one for $p=3$, $q=5$, as $3\nmid 5-1$.