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? 

 A: 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.
A: 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)
A: 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 $\frac{G}{Z(G)} \cong \operatorname{Inn}(G)$.
A: 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.
A: 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.
A: (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.
A: A finite group $G$ is Abelian if and only if every irreducible complex representation of $G$ is one dimensional. 
A: 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 $G'$.
A: A group $G$ is abelian if and only if the multiplication map $\circ:G\times G\to G$ is a homomorphism.
