16
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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

14
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ This is exceedingly cool. It's essentially a completely abstract definition of abelian groups! $\endgroup$
    – user1729
    Sep 26, 2019 at 16:45
  • 3
    $\begingroup$ @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
8
$\begingroup$

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.

$\endgroup$
1
7
$\begingroup$

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)

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

(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.

$\endgroup$
1
  • 2
    $\begingroup$ Also $G$ is abelian if it has order square of a prime. $\endgroup$ Dec 28, 2019 at 9:49
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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 $G'$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.