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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?2011-10-27
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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.2011-10-27
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    @Basil: What's your background? How much do you know about free groups? If you already know the Nielsen-Schreier theorem (http://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.2011-10-27
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    @jug: Thanks so much for your advices. I know that well-known Theorem. What is clear here is that; we have a generating set with more than element(rank $F$>1). I see the generated subgroup by an element of center and any other element of $F$ is an abelian free group. And this story is true for all elemets of $F$. Do you think that I am on a right way of proof? :-)2011-10-27
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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.2011-10-28
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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.2011-10-28
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    @jug: Yes. Zulfigar's answer seems an easy proof at the first time, but, deeper result is better.Thanks much for the help. :)2011-10-28
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    @Basil: Why do you accept the **wrong** solution of lhf instead of the partially correct solution by ZulfiqarIII?2011-10-28
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    @jug: That's done.2011-10-29

2 Answers 2

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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 $n=2$).2011-10-27
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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.)2011-10-27
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    @user1729, well, I read *free* as *free from non-trivial relations*, but you're right, it depends on how you define free groups. On the other hand, *free from non-trivial relations* is a difficult concept to pin down, hence the categorical definition.2011-10-27
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    @user1729, btw, did you manage to choose your login name to be the [taxicab number](http://en.wikipedia.org/wiki/Taxicab_number) or was it just luck? "No, Hardy, it is a very interesting number".2011-10-27
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    @ lhf, that was sort of my point - presentations are nasty but people forget this. So when you can define something without using presentations it is often better to do so (although it did take me a long time to realise that homomorphisms corresponded to adding relations in a presentation because my lecturer was trying to make the point I am making now and so wasn't telling us stuff!).2011-10-27
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    Also, user1729 is purely made-up. My actual name is...user10513 (although I did call myself "Swlabr" for a while!).2011-10-27
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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.2011-10-27
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    @ineff, ok, there are probably details to be filled.2011-10-27
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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. -12011-10-27
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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.2011-10-27
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    (I should probably say that my above comment is overkill, but the word "incorrect" was used and so a point had to be made! The fact that one-relator groups with torsion have trivial center is non-trivial, but there is a paper of Baumslag and Taylor which proves it.)2011-10-27
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    @user1729: The missing detail is the correct proof. I do not see any step done into the correct direction. But I have to say that I misread ineff's comment, which is indeed not correct (but I cannot undo my upvoting of this comment). The last sentence in ineff's comment should be: A group with the same number of generators could have the center of the free group in the kernel of any onto homomorphism.2011-10-27
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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)2011-10-27
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    @user1729: yes you're right I wanted to write "onto" instead of "to": in this case what I mean is that a hypothetical element in the center of a free group should belong to every kernel from the said free group onto non abelian groups (that's include also groups which had less generators than the free group considered). Thanks for the notation.2011-10-27
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    @ineff and jug: I do not understand your point about "onto" begin different from "to". Every free group maps onto a one-relator group. You just take an arbitrary word in your free group and quotient it out...2011-10-28
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    @user1729: We both criticize the sentence "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.", but IMO ineff's argument is not correct rsp. not correctly phrased.2011-10-28
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    @user1729: The image of a group with non-trivial center can very well have a trivial center.2011-10-28
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    @user1729: In practice what jug mean is that it is possible that a free group is sent via an homomorphism onto an abelian group seen inside an non abelian group in this case its center doesn't need to be contained into the kernel of said homomorphism.2011-10-28
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    @ineff: No, I mean what I wrote.2011-10-28
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    @ jug: Yes, the image of a group with non-trivial centre can very will have trivial centre, but my point was that if $F_n$ had a non-trivial centre then there is an element in this centre, $R$, and then quotienting out $R^2$ you end up with a group which has $R$ in the center. But it is a one-relator group with torsion so it cannot have non-trivial center, and this is a contradiction. Basically, lhf's proof works...it just needs detail filled in.2011-10-28
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    @user1729: I did't doubt your proof using one-relator groups with torsions (even if it seems likely that you need the fact that free groups with more than one generator have a trivial center to get the result you cited). But your proof doesn't just fill in details. It only takes over the assumption needed for a proof by contradiction. Then lhf's and your proofs diverge completely ("then every group with the same number..." vs. a concrete quotient of $F_n$). By the way, how difficult is the proof about one-relator groups with torsion?2011-10-29
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    @lhf: Thanks for leaving your answer and adding the warning!2011-10-29
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    @jug: While I was writing my comment you hadn't wrote yours second comment, so "you mean what you said" but at the time I was writing I saw just a partial comment that's the reason for the misunderstanding.2011-10-29
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    @lhf: you're welcome and by the way as you said, your answer was correct, existence of non trivial center is not implied by the axiom for free groups so for every free group the center is trivial, by the way this isn't a trivial fact, it required a study of the presentation of groups.2011-10-29
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    @jug, ineff, user1729: thanks for putting me right.2011-10-29
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    @ jug: Yes, I suspect you are right about needing the result about free groups having trivial centre, and in retrospect it does dispense with lhf's observation. However, my point (which my example may or may not have been a good way of making) was that it is possible to fill in the detail in lhf's proof. I think one could use the fact that (finitely generated) free groups are hopfian, and then prove that $F_n/Z(F_n)$ also satisfies the universal properties...this would fill in the holes without making any big assumptions, but I haven't given much thought to proving this...2011-10-31
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    @user1729: I still don't understand what part of lhf's proof you are intending to reuse except "Assume the center is non-trivial".2011-10-31
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    The universal property of free groups. As in, if $F_n$ has centre $Z$ then prove that $F_n/Z$ is universal on $n$-generators...2011-10-31
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    @user1729: OK. Have fun!2011-10-31