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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?

J. W. Tanner
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    This is wrong, not all groups of order $pq$ are cyclic. You need more conditions, like $p<q$ and $q\not\equiv 1(\text{mod} \ p)$. This is not the case with $3$ and $19$. – Mark Jul 09 '24 at 13:15
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    "The integers less than $57$ and coprime to $57$" is an infinite set. – Derek Holt Jul 09 '24 at 13:18
  • In the case $q\equiv 1\mod p$ as here , there is exactly one other group , a non-abelian one. – Peter Jul 09 '24 at 13:53
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    Browse for "nonabelian group of order $pq$". – Kan't Jul 09 '24 at 18:01
  • There are 2 nonisomorphic groups: https://people.maths.bris.ac.uk/~matyd/GroupNames/ – Lucenaposition Jul 10 '24 at 07:21
  • 3 downvotes, why? – Kan't Jul 10 '24 at 18:59
  • @Kan't: Not sure. It's not a bad question. But a tip to the right answer is at the link that Lucenaposition provided. – Brian Tung Jul 11 '24 at 23:11
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    The number of groups of order $pq$ with $p < q$ being distinct primes is $2$ if $p \mid q-1$ and $1$ otherwise. So, since $3 \mid 19-1=18$, there are actually $2$ groups of order $57$ (the other nonabelian one being the semidirect product $\mathbb{Z}{19} \rtimes \mathbb{Z}{3}$). – Geoffrey Trang Jul 11 '24 at 23:21
  • The edit in this answer shows how to write a presentation of the non-Abelian group. – Steve D Jul 11 '24 at 23:58
  • This video explains semi-direct product very delicately with motivation. Watch it and you will be able to construct $\mathbb Z_{19}\rtimes \mathbb Z_{3}$ which is non-abelian. Note: We don't mention the homomorphism under the $\rtimes$ symbol like in this video because there is only one non-trivial homomorphism $\mathbb Z_{3}\to\text{Aut}(\mathbb Z_{19})$. – Nothing special Jul 12 '24 at 00:35
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    @Nothingspecial: no there are $2$ such homomorphisms, they just give the same group up to isomorphism. – Steve D Jul 12 '24 at 01:00
  • @SteveD You're right... $\text{Aut}(\mathbb Z_{19})\cong \mathbb Z_{18}$ so it has two elements of order two. – Nothing special Jul 12 '24 at 01:02

1 Answers1

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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 $3$.)

arkeet
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