How the center of a non-cyclic free Group is trivial? How can I show that "A center of a free group that is non-cyclic is trivial" ?
 A: Say the group is generated by $g_1,\ldots,g_n$ for $n\geq 2$.
Suppose some non-identity element $h$ is central. Now $h$ can be written in a unique way in terms of the generating elements ; say $g_l$ is the leftmost element in this expansion, and say $g_r$ is some element different from the rightmost.
Set $g := g_r g_l^{-1}$, since $h$ is assumed to be central we must have $gh = hg$. But on the lhs some factors cancel, while on the rhs we just added some factors which cannot be simplified. Hence the number of factors on the lhs and rhs differ, yet they still yield the same element? This is obviously not possible in a free group.
A: By definition, a free group has no non-trivial relations, only those implied by the axioms of groups. Since being in the center is certainly not implied by the axioms, the center of a free group is trivial. In other, more fancy words, if there was a free group with non-trivial center, then every group with the same number of generators would have a non-trivial center, because free groups are universal objects. Hence the restriction  that the free group be non-cyclic, ie, not be of rank 1.
Edit: From the comments it seems that what I wrote above was wishful thinking. Perhaps having a non-trivial center cannot be captured in a relation. I could delete the answer but the comments contain an instructive discussion and so I'll leave it here.
