Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ and $B$ be two groups and let $G=A*B$ be their free product. Suppose both $A$ and $B$ are nontrivial. I want now to prove that $G$ is infinite and that $Z(G)$ is trivial.

I think I can prove that $G$ is infinite using this argument: two reduced words of $A*B$ are equal if and only if they have the same length and the ith letter of the first word is the same as the ith letter of the second one. Hence one can find a reduced word of $G$ of length arbitrarily big (since we are sure it won't be equal to a shorter word). Do you think it works? Has anyone another suggestion?

On the other hand, I don't really have a clue for the second part except trying to analyze a long list of cases of reduced words and prove for each one that the only element with which they commute is the identity. Does anyone have a smarter idea?

Thanks in advance!

share|improve this question

1 Answer 1

up vote 6 down vote accepted

Well, you would have a hard time proving that a reduced word only commutes with the identity, because every word commutes with every power of itself, and of course, any element $a\in A$ will commute with all elements in $C_A(a)$, and any $b\in B$ with all elements in $C_B(b)$. So no element of $A*B$ has trivial centralizer! But you don't need a nontrivial element that doesn't commute with anything for the center to be trivial (good thing, since there is no such thing in any nontrivial group); you just need every nontrivial element to not commute with something.

You don't really need to consider a lot of cases: the only two things you have to worry a bit about are what to do if both of your groups have only two elements, and what the reduced word starts and ends with. For example, if $\mathbf{w}$ is a reduced word that starts and ends with an element of $A$, then just take $b\in B$, $b\neq e$ to get something it does not commute with, so $\mathbf{w}$ cannot be central. If it starts with an element of $A$ and ends with an element of $B$, then take a word $\mathbf{w}'$ that starts with an element of $A$ and ends with an element of $B$ (so that $\mathbf{w}\mathbf{w}'$ and $\mathbf{w}'\mathbf{w}$ are already reduced), and make $\mathbf{w}'$ either begin with something different from what $\mathbf{w}$ begins with, or end with something different from what $\mathbf{w}$ ends with; etc. You should have just 4 cases, and only two that can give you trouble if your groups are too small, in which case you can just stare at them directly.

Your argument for why $A*B$ is infinite is fine. You can even produce, explicitly, words of arbitrary length: just take $a\in A$, $b\in B$ with $a\neq e\neq b$, and consider $(ab)^n$, $n\in\mathbb{Z}$.

share|improve this answer
1  
I've always wondered about 'categorical' proofs of facts like these - that is, using the universal property of the free product to show properties of it (trivial center, what its torsion looks like, etc.). –  user641 Feb 18 '11 at 21:02
1  
@Steve D: The problem with trying to use the universal property to prove that the center is trivial is that centers don't behave very well relative to homomorphisms; the only general property you have is that if $f\colon G\to H$ is onto, then $f(Z(G))\subseteq Z(H)$. So even finding a centerless group which is generated by homomorphic images of $A$ and $B$ would only tell you that the center is contained in the kernel of the universal map, not that the center is trivial. Describing the torsion is more doable because the torsion subgroup has categorical properties that the center does not. –  Arturo Magidin Feb 18 '11 at 21:09
    
@Steve D: I take that back: you're going to have problems proving the torsion results using a categorical proof. The problem in both cases is that the maps "go the wrong way". The only reasonable to way to prove an element is central categorically is to show that you a surjection onto your group in which the element lies in the image of the center; to show an element is torsion, you show it is the image of a torsion element (or alternatively, you show it is not torsion by showing it has a nontorsion image). But categorically you get maps from the free product, not into it. (cont) –  Arturo Magidin Feb 19 '11 at 21:35
    
@Steve: So that creates a serious problem. You can characterize the center as the kernel of the natural map into the automorphism group, but that leads to computation, not categorical issues. And in a free product of two finite groups, the torsion subgroup is the entire group (because the free product is generated by the copies of $A$ and $B$, which are contained in the subgroup generated by the torsion elements). So, I don't really see a way of proving these properties in a purely categorical manner, without eventually getting down into the groups themselves. –  Arturo Magidin Feb 19 '11 at 21:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.