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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?

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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 $(C_{77})^* \cong (C_7)^* \times (C_{11})^* \cong C_6 \times C_{10}$.

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  • $\begingroup$ Thanks, I have just one more question, how can we know that $(C_{77})^*$ is cyclic? With my mistake it was trivial when the order was 59, but with order 60 there are other possibilities. $\endgroup$
    – D18938394
    Commented May 1, 2018 at 14:07
  • $\begingroup$ My mistake! We actually don't know it's cyclic. I'll edit $\endgroup$
    – B. Mehta
    Commented May 1, 2018 at 14:13
  • $\begingroup$ Ok, thanks, but is that last part true for any pair of distinct primes? If p,q are distinct primes then $(C_{pq})^* \cong (C_p)^*\times(C_q)^*$? $\endgroup$
    – D18938394
    Commented May 1, 2018 at 14:46
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    $\begingroup$ Yes, by the chinese remainder theorem $(C_{pq})^* \cong (C_p)^* \times (C_q)^* \cong C_{p-1} \times C_{q-1}$. $\endgroup$
    – B. Mehta
    Commented May 1, 2018 at 14:46
  • $\begingroup$ I see that the chinese remainder theorem gives $C_{pq} = C_p\times C_q$, but is it true in general that $(C_n\times C_m)^* \cong (C_n)^*\times (C_m)^*$, for any two primes $m,n$? $\endgroup$
    – D18938394
    Commented May 1, 2018 at 14:50

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