Describe all non-isomorphic groups of order $57$, such that for each of them you write down its generators and the connections between them.
Attempt: 57 is the product of two primes, specifically $57 = 3 \times 19$. By the classification of groups of order $pq$ where $p$ and $q$ are primes, there is only one group of this order, which is cyclic. Thus, any group of order 57 is isomorphic to the cyclic group $\mathbb{Z}_{57}$. The cyclic group $\mathbb{Z}_{57}$ has a generator, which is any element $g$ of order 57. An element $g \in \mathbb{Z}_{57}$ is a generator if and only if $\gcd(g, 57) = 1$. Since 57 has the prime factors 3 and 19, the generators of $\mathbb{Z}_{57}$ are the integers less than 57 and coprime to 57. So what is the overall answer here, as I got lost?
Best Answer
As noted in the comments, $|G| = pq$ a product of distinct primes doesn't imply $G$ is cyclic (for example, $S_3$ has order $2 \cdot 3$), so the attempt in the question fails.
If $|G| = 3 \cdot 19$, then Sylow tells us that $G$ has a subgroup $H = \langle h \rangle$ of order $3$, and a normal subgroup $K = \langle k \rangle$ of order $19$; then $hkh^{-1} = k^s$ for some $s$, and $G = HK$. This already gives a presentation $$ G = \langle h,k \mid h^3 = k^{19} = 1, hkh^{-1} = k^s \rangle $$ since the relations let you write any element in the form $h^i k^j$ (with $0 \le i < 3$ and $0 \le j < 19$), so the group given by the presentation has order at most $57$ (i.e. no more relations are needed).
Since $k = h^3 k h^{-3} = k^{s^3}$ we get $s^3 \equiv 1 \pmod{19}$, which implies $s \equiv 1, 7, 11 \pmod{19}$. If $s \equiv 1$, then $G$ is abelian (hence cyclic, so there is a simpler presentation). The other two solutions give nonabelian groups which are isomorphic (I leave this as an exercise).
We still need to show a nonabelian group of order $57$ exists. But each possible value of $s$ above gives a homomorphism $\theta \colon \mathbb{Z}_3 \to \operatorname{Aut}(\mathbb{Z}_{19})$ (where $\theta(1)$ is multiplication by $s$), and hence a semidirect product $\mathbb{Z}_3 \ltimes_\theta \mathbb{Z}_{19}$ with the above presentation.
(What happens with the presentation if $s^3 \not\equiv 1$? Then you can actually deduce $k = 1$ from the relations, since $\gcd(s^3-1,19)=1$, so you only get a group of order $3$.)