I am trying to work out this exercise, in which I have to prove there are only 2 groups of order 231 (up to isomorphism), one of those is of course $C_{231}$, and the other one will be a non-abelian group.
I already showed in a previous exercise that in the case $|G|=3\cdot 7\cdot 11$ we have that $G = H\rtimes C_3$, in which $H$ is a group of order 77. As a consequence of Sylow's theorems we have that $H = C_{77}$, thus we need to determine the possible semi-direct products $C_{77}\rtimes C_3$, this means we need to analyze the possible homomorphisms
$$\sigma:C_3 \rightarrow \text{Aut}(C_{77})$$
but we have $\text{Aut}(C_{77})\cong (C_{77})^* \cong C_{59}$, so in the end we need to see what are the homomorphisms
$$ \rho:C_3\rightarrow C_{59} $$
But the only possibility is the trivial homomorphism, which results in the direct product $C_{77}\times C_3 \cong C_{231}$.
I don't know what I'm doing wrong, because according to the book there are two possibilities, and I showed that there is only one, so there has to be something wrong in my solution, can anyone help me with that?
The mistake is writing $(C_{77})^* \cong C_{59}$. Notice $\phi(77) = \phi(7) \times \phi(11) = 6 \times 10 = 60$, not $59$.
Instead, we have that $(C_{77})^* \cong (C_7)^* \times (C_{11})^* \cong C_6 \times C_{10}$.