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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.

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    $\begingroup$ It is so trivial that I would not call this an exercize. $\endgroup$ – Crostul Aug 10 '15 at 19:57
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    $\begingroup$ I'm absolutly agree with you, it is a kind of a joke )) $\endgroup$ – Ilya.K. Aug 10 '15 at 20:14
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    $\begingroup$ its almost harder to write proofs like this where the result is so obvious. I would informally describe the center as "the part of a group which is abelian", so to me there is nothing to prove. i think your work definitely suffices. $\endgroup$ – Elliot G Aug 10 '15 at 20:18
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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.

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It is not that hard to tell. By definition the center $Z(G)$ of a group are the elements that conmute for all the others.

i.e

$$Z(G) = \{g \in G : gh=hg , \forall h \in G\} $$ If $Z(G)=G$, is trivial to see that all the group conmute, meaning that $G$ is abelian.

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$Z(G) = $ $\{$ x $\in$ G : for every y in G , $xy=yx$ $\}$.

Claim: A group G is abelian iff $Z(G)=G$.

Suppose G is abelian. That means for any $x,y \in G$, $xy=yx$. Clearly, $Z(G)$ $\subset G$. So it suffices to show the reverse inclusion. Note that since every element commutes with every other element, it is necessarily the case that G is contained in the center, $Z(G)$.

Now suppose $Z(G)=G$. $Z(G)$ contains all the elements in G which commute so it must be abelian. Therefore by assumption it follows that G is also abelian.

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