# When will a group be Abelian?

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?

• 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. Oct 1, 2019 at 18:29

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

• This is exceedingly cool. It's essentially a completely abstract definition of abelian groups! Sep 26, 2019 at 16:45
• @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 Sep 26, 2019 at 17:05

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.

• – lhf
Sep 26, 2019 at 11:05

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 $$1 \leq k \leq a_i$$

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.

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 $$\frac{G}{Z(G)} \cong \operatorname{Inn}(G)$$.

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.

(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 , $$$$ divides $$|G|$$. Since the order of G is prime, the order of $$$$ is either 1 or p. Since $$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.

• Also $G$ is abelian if it has order square of a prime. Dec 28, 2019 at 9:49

A finite group $$G$$ is Abelian if and only if every irreducible complex representation of $$G$$ is one dimensional.

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 $$G'$$.