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Can you please share any "fast" tricks you know based on the order of a group? I'm looking for something like:

  1. If the order of a group is prime then the group is cyclic
  2. If the order of the group is a prime squared $p^2$ then the group is abelian and isomorphic to $C_{p^2}$ or $C_p\times C_p$
  3. If the order of the group is the product $pq$ of two different primes $q>p$ and $p$ divides $q-1$ then there are exactly $2$ groups of that order

Do you know some similar things to look for when trying to 1) find how many groups are there of order $n$ up to isomorphism 2) know if a group of a given order is simple 3) find automorphisms 4) know if a group is solvable, etc?

user1729
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augustoperez
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    All groups of order $pqr$ (all primes, not necessary distinct) are solvable and not simple. All groups of order $1-59$ and $61-119$ are solvable. – Mark May 09 '19 at 22:16
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    all groups of odd order are solvable – Rylee Lyman May 09 '19 at 22:21
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    @Rylee Lyman True, but this is extremely hard to prove. – Mark May 09 '19 at 22:46
  • If the order of a group is $p^n$ and $n$ is large, then you should look for more information. – Servaes May 09 '19 at 23:30
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    These are no "tricks" but basic results of group theory (or hard results, concerning solvability of odd order groups etc,). – Dietrich Burde May 10 '19 at 08:54
  • @Mark Hard to prove it may be, but it addresses "2) know if a group of a given order is simple". In particular, people who study finite simple groups are somewhat obsessed with elements of order two... – user1729 May 10 '19 at 10:52
  • You better quickly take a look at the answer of Pete Clarke here https://math.stackexchange.com/questions/67407/group-of-order-15-is-abelian. I totally agree with Dietrich using the word "trick". By the way, end of the 19th century/beginning of the 20th century group theory was pretty much focussed on the classification of groups according to (prime factorization of) their order. – Nicky Hekster May 10 '19 at 18:37

2 Answers2

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Let $G$ be a group of order $2p$, where $p$ is a prime greater than $2$. Then $G$ is isomorphic to $Z_{2p}$ or $D_{p}$.

For proof you can refer this.

Rkb
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To address your question "2) know if a group of a given order is simple":

The Feit–Thompson Theorem, also called the odd order theorem, is the result quoted in the comments to the question. It is significant for its implication to simple groups, and is notorious for the length of its proof (255 pages!):

Theorem. Every group of odd order is soluble*.

The following then addresses your question 2). Recall that abelian simple groups are cyclic of order $p$ for some prime $p$.

Corollary. Non-abelian simple groups have even order.

Proof. Let $G$ be non-abelian and simple. If $G$ has odd order then it is soluble, by the theorem, so then $G'\lneq G$. As $G'$ is a normal subgroup of the simple group $G$, we have that $G'$ is trivial. Hence, $G/G'=G$. However, $G/G'$ is abelian, contracting the assumption that $G$ is non-abelian.

*American English: solvable.

user1729
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