Automorphisms, Cyclic Groups, and Order Question: Let $q=7$ and $r=11$.   Classify  groups  of  order $rq$ and compute  their automorphism groups.
Since $r>q$, $q$ cannot divided $r-1$, making $G$ cyclic.
Let $R$ and $Q$ be $p$-subgroups with order $r$ and $q$, respectively. I have found that both $Q$ and $R$ are normal and I have classify the groups ($G$ is isomorphic to $R\times Q$).  But I don't know how to find a group of automorphisms.
 A: $$Aut(\mathbb{Z_{n}})\cong U_{n}$$ .
$U_{n}$ is the group of units modulo $n$  under multiplication
And you have your $G\cong \mathbb{Z_{77}}$
So you have $Aut(G)\cong U_{77}$. Which is a group of order $\phi(77)=60$
To construct the isomorphism from $Aut(\mathbb{Z_{n}})\to U_{n}$ you need to see that any isomorphism from $\mathbb{Z_{n}}\to \mathbb{Z_{n}}$ takes a generator to a generator. And there are precisely $\phi(n)$ generators of a cyclic group of order $n$. So each such isomorphism forms the distinct elements of the Automorphism group.
Also just a word of caution. You are using the facts in the reverse order. It is because $R$ an $Q$ are normal(which is derived from that 11-1=10 is not divisible by 7) and they intersect trivially....that is why you can conclude that $G\cong R\times Q$. And because $R$ and $Q$ are cylcic and their orders are relatively prime...you can conclude that $R\times Q$ is cyclic and hence $G$ is cyclic. In the statement of your question you directly wrote that as $q$ does not divide $r-1$ , $G$ is cyclic and then you are saying that you have found unique normal subgroups. That is the reverse and incorrect way of thinking.
