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I look for some theorems which tell us about the relation between the property of being abelian for groups and the order of the group.

I think these theorems are provided in a second course of group theory or more advanced courses. At this stage, however, I'm interested in knowing some of these theorems without the details of proofs of this theorems.

I hope that you can give me some of these theorems!


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One such theorem is if the order of the group is prime, then the group is cyclic, hence abelian. – Fredrik Meyer Mar 8 '13 at 17:46
You should also note that you cannot conclude anything about the order from the group being abelian, as there are abelian groups of any (finite) order (and probably any infinite order as well, but there are set theorists around so I'm being careful). So any result has to deduce that groups of certain orders (maybe with some other properties) are abelian, as in Fredrik's example. – Matthew Pressland Mar 8 '13 at 17:48
There's the Fundamental Theorem of Finite Abelian Groups: – Todd Wilcox Mar 8 '13 at 17:51
$G$ is abelian if $|G|=p^2$ and $p$ is prime, from here – valtron Mar 8 '13 at 17:52
Theorem: If $|G|=pq$, where $p<q$ are primes such that $q\not\equiv1\pmod p$, then $G$ is cyclic, hence abelian. – Jyrki Lahtonen Mar 8 '13 at 17:54

Here are some facts. Assume all groups below are finite.

  • All finite abelian groups look like this: $\mathbb{Z}_{p_1^{e_1}}\oplus \mathbb{Z}_{p_2^{e_2}}\oplus \mathbb{Z}_{p_3^{e_3}}\oplus \cdots \oplus \mathbb{Z}_{p_n^{e_n}}$, where the $p_i$ are not necessarily distinct.

  • $\mathbb{Z}_{pq}\cong \mathbb{Z}_p\oplus\, \mathbb{Z}_q$ if and only if $p$ and $q$ are coprime.

  • A group $G$ is cyclic if and only if it has exactly one subgroup of order $d$ for every divisor $d$ of $|G|$.

  • Lagrange's theorem states that whenever $H$ is a subgroup of $G$, the order of $H$ divides the order of $G$. The converse does not hold in general, but it does for abelian groups - that is, for every divisor $d$ of $|G|$, there is a subgroup of order $d$ in $G$.

  • If $3/4$ths or more of a group's elements have order $2$, the group is abelian. In particular, if every element in a group has order $2$, that group is abelian. (Note: the latter obviously follows from the former, but is proved much more easily by itself.)

  • The derived subgroup and inner automorphism group of an abelian group have order $1$.

  • Of all groups of order $p^e$, groups which look like $\underbrace{\mathbb{Z}_p\oplus \mathbb{Z}_p \oplus \cdots \oplus \mathbb{Z}_p}_{\text{e times}}$ have the largest number of subgroups. Groups which look like $\mathbb{Z}_{p^e}$ have the fewest. (This includes nonabelian groups.)

  • The order of the automorphism groups of the last bullet point are $\prod_{k=0}^{e-1}p^e-p^k$ and $p^{e-1}(p-1)$, respectively. In general the order of the automorphism group of $\mathbb{Z}_n$ is $\varphi(n)$.

  • If a group is simple and solvable, it is cyclic of prime order.

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$|ab|=lcm(|a|,|b|)$ is wrong (take for instance $a=b^{-1}$). – Martin Brandenburg Mar 8 '13 at 18:48
Some people are more familiar with $\times$ than $\oplus$, but if you've never seen it before, $A\oplus B$ and $A\times B$ are isomorphic. (things get more complicated if there are "infinitely many" $\oplus$ signs) – Eric Stucky Mar 9 '13 at 6:16
@EricStucky I hate to admit it but pretty much the only reason I use $\oplus$ instead of $\times$ is I think it looks cooler. – Alexander Gruber Mar 9 '13 at 6:53
@AlexanderGruber Although they do have different meanings. One is a "coproduct" and one is a "product". The fact that they coincide (they are a "biproduct") has to do with the fact that the category of abelian groups is an abelian category. – Alex Youcis Mar 9 '13 at 22:14

One can classify those integers $n$ for which every group of order $n$ is abelian, see here, the last answer by Robin Chapman. This has also appeared here and there on math.stackexchange.

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The condition you have given classifies $n$ for which every group of order $n$ is cyclic. For example, for $p$ prime every group of order $p^2$ is abelian but $p^2$ and $\phi(p^2)$ are not coprime. – Mikko Korhonen Mar 9 '13 at 11:10
Ok, I copied from one of the linked threads. – Martin Brandenburg Mar 9 '13 at 22:10

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