We have two sylow subgroups of orders 7 and 3. Let $n_3$ and $n_7$ denote the number of sylow subgroups for 3 and 7, respectively.
$n_7 \equiv 1 \mod 7$ and $n_7 | 3 \implies n_7 = 1$
$n_3 \equiv 1 \mod 3$ and $n_3 | 7 \implies n_3 = 1, 7$
Let $P_3 \cong \Bbb{Z}_3 $ and $P_7 \cong \Bbb{Z}_7$. Since $P_7$ is always normal in G, we know that $G \cong \Bbb{Z}_7 \rtimes_{\alpha} \Bbb{Z}_3$.
Case 1:
Let $n_3 = 1$. Then we know that $P_3$ is also normal, and so $G \cong \Bbb{Z}_7 \times \Bbb{Z}_3$.
Case 2:
Let $n_3=7$. We have $\alpha : \Bbb{Z}_3 \rightarrow \Bbb{Z}_7^{\times} \cong \Bbb{Z}_6$
So the generator $a$ of $\Bbb{Z}_3$ needs to be sent to an element of order 3 in $\Bbb{Z}_6$, which means that it needs to be sent to either $2$ or $4$.
Let $\alpha'(a) = 2$ and $\alpha(a)=4$.
There are two theorems from J.S. Milne's notes that says:
Theorem 1 If there exists an $\alpha \in Aut(N)$ such that $$\theta'(q) = \alpha \circ \theta(q) \circ \alpha^{-1} \forall q \in Q$$ then the map $$(n,q) \rightarrow (\alpha(n), q): N \rtimes_{\theta} Q \rightarrow N \rtimes_{\theta'} Q$$ is an isomorphism.
Theorem 2 If $\theta = \theta' \circ \alpha$ with $\alpha \in Aut(Q)$, then the map $$(n,q) \rightarrow (n, \alpha(q)) : N \rtimes_{\theta} Q \approx N \rtimes_{\theta'} Q$$
We know that there are two automorphisms of $\Bbb{Z}_3$, the identity and $f: n \rightarrow 2n$. Since $\alpha(a) = \alpha' \circ f (a)$, we know that $\Bbb{Z}_7 \rtimes_{\alpha} \Bbb{Z}_3 \cong \Bbb{Z}_7 \rtimes_{\alpha'} \Bbb{Z}_3$.
So I was able to show that they are isomorphic by using theorem 2. But I was wondering if I could use theorem 1 instead. Here is the problem that I ran into when I attempted to use the first one: We know that $Aut(\Bbb{Z}_7) \cong \Bbb{Z}_6$, and it's easier to see which elements have order three in $\Bbb{Z}_6$ than in $Aut(\Bbb{Z}_7)$. But after finding the elements 2 and 4, how can I use theorem 1? In other words, we need automorphisms in order to use theorem 1, but I only have elements of $\Bbb{Z}_6$ which are obviously not automorphisms.
Thank you in advance
Best Answer
In your example case Theorem 1 cannot be used. This is because this time $Aut(N)$ is abelian. Observe that $\theta(q)$ and $\alpha$ are both in $Aut(N)$, so from its commutativity it follows that $$ \alpha\circ \theta(q)\circ \alpha^{-1}=\theta(q) $$ for all choices of $\alpha$.