Isn't it the same to say that a group is abelian, and that the center of the group is all the group?
I have an exercise to prove that this is true, and it's exactly one stroke for each direction of the proof, correct me if I wrong:
First direction: The group $G$ is abelian, therefore for each $a,b\in G$: $$ab\:=\:ba$$
Therefore, in other words:
$$a\in G\::\:Z(G)=\{ab\:=\:ba\::\:b\in G\}$$
and by definition of center (which here I symbolized as $Z(x)$), the center here is every element of $G$, therefore it is whole $G$ itself.
Second direction:
That is true: $$a\in G\::\:Z\left(G\right)=\left\{ab\:=\:ba\::\:b\in G\right\}$$
and in other words, for each for each $a,b\in G$: $$ab\:=\:ba.$$
What should I add in this proof, does it seems that i missed something? I mean it seemed pretty trivial, both directions.
Best Answer
You've definitely got all the ideas. Your notation may need to be changed up a bit, though: $Z(G) = \{a \mid a \in G \land (\forall b \in G)[ab = ba]\}$. Once this is done, you can ground your proof in it:
($\Rightarrow$): Suppose $G$ is Abelian. Let $a \in G$. Then for each $b \in G$, $ab = ba$, and so $a \in Z(G)$. Thus $G \subseteq Z(G)$. Clearly $Z(G) \subseteq G$, so $G = Z(G)$.
($\Leftarrow$): Suppose $G = Z(G)$. Let $a,b \in G$. Then $a \in Z(G)$ and so $ab = ba$. Thus, $G$ is Abelian.
No new ideas here, to be clear; this just slightly more directly grounds the proof using sets.