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My Attempt:

$1.$ $G$ is abelian if and only if the mapping $g\mapsto g^{-1}$ is an isomorphism on the group $G$.

$2.$If $G$ is finite and every irreducible character is linear then $G$ is abelian.

$3.$If $\operatorname{Aut}(G)$ acts on the set $G-\{e\}$ transitively then $G$ is abelian.

$4.$If $\mathbb Z_2$ acts by automorphism on a finite group $G$ fixed point freely then $G$ is abelian.

$5.$ If $\forall a,b\in G$ $ ab=ba$ then $G$ is Abelian.

My Question:

The above are the things which I already use to show a group will be Abelian.

Is/are there any other way(s) to show a group $G$ to be Abelian?

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  • $\begingroup$ It took me way too long to realize that by “$g\mapsto g^{-1}$ is an isomorphism” you actually mean “$g\mapsto g^{-1}$ is a homomorphism”. I was grinding my gears how it could fail to be bijective. $\endgroup$ Oct 1, 2019 at 18:29

9 Answers 9

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A group $G$ is abelian if and only if the multiplication map $\circ:G\times G\to G$ is a homomorphism.

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  • $\begingroup$ This is exceedingly cool. It's essentially a completely abstract definition of abelian groups! $\endgroup$
    – user1729
    Sep 26, 2019 at 16:45
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    $\begingroup$ @user1729 Related facts are that an abelian group is a group object internal to the category of groups, and that the 2-morphisms of a 2-category with one object and only the identity 1-morphism form an abelian group. ncatlab.org/nlab/show/Eckmann-Hilton+argument $\endgroup$ Sep 26, 2019 at 17:05
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If $G/Z(G)$ is cyclic, then $G$ is abelian.

and its corollary for finite groups:

If $|Z(G)| > \frac {1}{4} |G|$, then $G$ is abelian.

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If $G$ is finite of order $n$ and $n$ is an abelian number, then $G$ is abelian.

$n$ is an abelian number when $n$ is a cubefree nilpotent number, that is, if $n = p_1^{a_1} \cdots p_r^{a_r}$, then

  • $a_i < 3$
  • $p_i^k \not \equiv 1 \bmod{p_j}$ for all $1 \leq k \leq a_i$

(adapted from this answer)

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If you know a generating set for the group $G$, so a set $S$ such that $G=\langle S\rangle$, then $G$ is abelian if and only if $xy=yx$ for every $x, y\in S$.

That is, you just need to check commutativity of the generators rather than of all elements.

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How about if $G = Z(G)$, or if $\operatorname{Inn}(G) \cong 1$, ($\operatorname{Inn}(G)$ is the group of inner automorphisms).

These two criteria are related to eachother quite directly by the isomorphism $\frac{G}{Z(G)} \cong \operatorname{Inn}(G)$.

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If we know (e.g. by Sylow's theorems) that there exist an abelian normal subgroup $H \unlhd G$ and another subgroup $U$ such that $HU = G$ and $H\cap U = \{e\}$, then $G = U \rtimes H$. If the only group action of $U$ on $H$ (i.e. the only homomorphism $U \rightarrow \operatorname{Aut}(H)$) is trivial, the group is abelian.

This worked for me in practice in a few undergrad problems on finite groups. Not sure if it's helpful elsewhere.

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(1) A group G is abelian iff G is equal to its center

(2) A group G is abelian iff G is isomorphic to an abelian group.

(3) If G is a group and every element in G has order 1 then G is abelian.

(4) If G is a group and every element in G has order 2 then G is abelian.

(5) Every cyclic group is abelian.

(6) If a group G has prime order then it is abelian.

Proof of (6): Let G be a group of prime order, p. Let $g\in G$ be such that $g$ is not the identity, by lagranges theorem , $<g>$ divides $|G|$. Since the order of G is prime, the order of $<g>$ is either 1 or p. Since $g$ is not the the identity, it follows that G is cyclic, and thus abelian by (5).

(7) A subgroup H of G is abelian if G is abelian.

(8) Let G be a finite group and H an abelian subgroup. If the order of H is the order of G then G is abelian.

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    $\begingroup$ Also $G$ is abelian if it has order square of a prime. $\endgroup$
    – J. De Ro
    Dec 28, 2019 at 9:49
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A finite group $G$ is Abelian if and only if every irreducible complex representation of $G$ is one dimensional.

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You could show the first derived subgroup, the commutator, is trivial. It's the group generated by all $[x,y]=xyx^{-1}y^{-1}$ for $x,y\in G$. Denoted $[G,G]$ or $G'$.

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