Work out all the automorphisms of $C_7$. Now using a semidirect product, construct a nonabelian group of order 21

Question

I know that ${\rm Aut}(C_7$) is $C_6$, and ${\rm Aut}(C_6$)= $C_4 \times C_2$ but I do not know if that will help.

Any hints or tips welcome!

Answer 1

$\operatorname{Aut}C_7\cong C_7^×\cong C_6$. So there are $6$, each of the form $\phi(g)=g^k$ for $k$ relatively prime to $7$.

$C_7\rtimes_\varphi C_3$ where $\varphi:C_3\to C_6$ is a nontrivial homomorphism, will be a nonabelian group of order $21$.

Answer 2

You are almost there. To construct a semidirect product of 2 groups $A$ and $B$ you need a homomorphism $\psi:A\to Aut(B)$. In your case $A=C_{3}$ and $B=C_{7}$. So, you can pick an homomorphism $\psi: C_{3} \to Aut(C_{7})\cong C_{6}$ that maps $x$ to $x^{2}$. If you are ok with presentations, then you can define it as $G = C_{7}\rtimes C_{3} = \langle x, y \mid x^3=1, y^7=1, yx=xy^2\rangle$.

Actually, it does not matter what (nontrivial) homomorphism you pick (because you will get an isomorphic group). For further information you can read this.