# How to compute the automorphism group of split metacyclic groups?

I am trying to calculate the automorphism group of an affine subgroup $$G=\mathbb{Z}_p\rtimes\mathbb{Z}_{k}\leq\text{AGL}(1,p).$$

One might guess $\text{Aut}(G)=\text{AGL}(1,p)$. And this matches what I got in GAP after checking a couple examples. However, it seems proving it through by definition is messy and not particular easy. So I have been pondering a while what will be a better way to approach? Any suggestions?

#### Solutions Collecting From Web of "How to compute the automorphism group of split metacyclic groups?"

Here is a quick outline argument. You need to assume that $k>1$ or the result isn’t true. Then $Z(G)=1$, so we can embed $G$ in $A:={\rm Aut}(G)$ and, since $G \unlhd {\rm AGL}(1,p)$, we have $G < {\rm AGL}(1,p) \le A$.

Write $G = P \rtimes Q$ with $P=C_p$, $Q=C_k$. Then $P \lhd A$ and the complements of $P$ in $G$ are conjugate so, by the Frattini Argument $A=GR$ with $R=N_A(Q)$ and, since $R \cap P=1$, $A = P \rtimes R$.

Now, $[C_R(P),Q] \le C_G(P) \cap Q = 1$, so $C_R(P) \le C_A(G) = 1$. Hence $|R| \le |{\rm Aut}(P)|=p-1$, so $G={\rm AGL}(1,p)$.

That was trickier that I expected – more detail on request!