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How can I show that "A center of a free group that is non-cyclic is trivial" ?

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Pick an arbitrary element other than 1. What's its first letter? Does it commute with letters other than that? –  Jyrki Lahtonen Oct 27 '11 at 8:04
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@Basil: Starting directly from the definition of the free group $F$ over a set $S$ generating $F$ (i.e., every (set-theoretic) map from $S$ to a group $G$ exists a unique group homomorphism from $F$ to $G$ that extends that map) it is maybe more work to prove your question than the answers indicate now. The "definition" given on wikipedia doesn't quite fulfill the standards of a mathematician. It's a statement about a normal form one has to prove. Be cautious of upvoted answers and comments. –  j.p. Oct 27 '11 at 16:23
    
@Basil: What's your background? How much do you know about free groups? If you already know the Nielsen-Schreier theorem (en.wikipedia.org/wiki/Nielsen%E2%80%93Schreier_theorem), you can use it to show that the center is cyclic. But also the subgroup generated by the center and any other element has to be cyclic by Nielsen-Schreier. You should be able to reach a contradiction then. –  j.p. Oct 27 '11 at 18:30
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@Basil: By Nielsen-Schreier the subgroup $H = \langle s, \mathrm{Z}(F)\rangle$ generated by an element $s \in S$ and the center of $F$ is a free group (notation like in my first comment). And it is abelian. It's quite easy to show that only the free group over $1$ element is abelian, and this group is cyclic. You can conclude that for every element $s$ of $S$ a finite power of it is contained in the center of the free group. You can lead this to a contradiction. –  j.p. Oct 28 '11 at 7:27
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@Basil: I think that ZulfiqarIII's answer can be fixed to give a proof just using the normal form, not using the deeper result of Nielsen-Schreier. –  j.p. Oct 28 '11 at 7:31
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2 Answers

up vote 3 down vote accepted

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.

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Your condition on $g_r$ is not enough: you have to ensure that $g_r \ne g_l$ (try first the case $n=2$). –  j.p. Oct 27 '11 at 16:06
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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.

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I do not like the way you start your post - "dy definition...". I much prefer the definition of free groups as, well, the free objects in the category of groups! That said, I like your proof of the result, which is a simple application of the fact that free groups are the free objects in the category of groups...(so -1 for the definition, but +2 for the v. neat proof.) –  user1729 Oct 27 '11 at 10:10
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@lhf there's something that I don't get: you said "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" why should this be true? A priori it could be clearly possible that the elements of center stay in every kernel of a homomorphism from the free group to a non abelian group. –  Giorgio Mossa Oct 27 '11 at 13:20
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@lhf: I remember your handle from some quite good answers and comments, but unfortunately this answer is not correct (see ineff's comment). So I had to downvote it. -1 –  j.p. Oct 27 '11 at 15:55
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@jug: Missing detail? Yes. Incorrect? No. Let $R$ be a word which is contained in the kernel of every mapping from $F_n$ to a group with trivial kernel. Then it must disappear in the canonical map from $F_n$ to the one-relator group with torsion $\langle x_1, \ldots, x_n; R^n\rangle$ as one-relator groups with torsion have trivial centre. However, in Magnus, Karrass and Solitar's book "combinatorial group theory" they explain the theory of one-relator groups, and one sees that $R\neq 1$ in the above group, as required. –  user1729 Oct 27 '11 at 16:59
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@lhf: How do you define a free group over $n$ elements? (IMO, wikipedia offers an engineer's definition, not a mathematician's) –  j.p. Oct 27 '11 at 18:15
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