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I'm trying to prove that the free group $A=A_1*A_2$, where $A_1, A_2\neq 1$ is not abelian. Following the hints below:

Let $x,y\in A_1*A_2$, where $x\neq y$.

Suppose now $A_1=F(S)$ and $A_2=F(T)$, where $S=\{\alpha_1,\ldots,\alpha_n\}$ and $T=\{\beta_1,\ldots\beta_m\}$

Let $x,y\in A_1*A_2$, where $x\neq y$, then we have the words

$x=\alpha_1^{n_1}\ldots\alpha_k^{n_k}$ and $y=\beta_1^{m_1}\ldots\beta_l^{m_l}$

Thus using the definition of the operation of the free products, we have

$x\cdot y=\alpha_1^{n_1}\ldots\alpha_k^{n_k}\beta_1^{m_1}\ldots\beta_l^{m_l}$

$y\cdot x=\beta_1^{m_1}\ldots\beta_l^{m_l}\alpha_1^{n_1}\ldots\alpha_k^{n_k}$

Am I correct so far?

I can't continue from that point, since $k$ and $l$ can be different.

Thanks

  • 0
    What do you know about the definition of a free product?2012-11-28
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    Your claim as stated is false: $1 * 1 \cong 1$, and $1$ is certainly abelian.2012-11-28
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    @YACP I'm talking about free groups2012-11-28
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    @ZhenLin I'm talking about free groups2012-11-28
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    @ZhenLin yes, sorry, you're right, I will edit my question2012-11-28
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    Well, actually $1$ is free. And with the assumption that neither $A_1$ nor $A_2$ is trivial, your claim holds (not only for free groups!), plainly by the definition of free product.2012-11-28
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    @Berci yes, I realized that, I've already edited the question2012-11-28
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    I don't see why this question should deserve a down vote. Hence I will up vote to cancel the down vote.2012-11-28
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    The claim is obvious for free groups: the free product of free groups is again a free group, and any free group on more than one generator is non-abelian. Otherwise some work is needed: the easiest way to proceed is to find a pair of homomorphisms $A_1 \to S$, $A_2 \to S$, $S$ non-abelian, such that the subgroup generated by the union of the images of $A_1$ and $A_2$ is non-abelian.2012-11-28
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    Do you perhaps mean free product, not free group? $A$ is only a free group if $A_1$ and $A_2$ are both free groups...2012-11-28
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    @user1729 yes, $A_1$ and $A_2$ are free groups.2012-11-28
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    I still can't understand what's going on here: If $\,A\,$ is a free group, then expressing it as a *free product* $\,A=A_1*A_2\,$ (as this is the meaning of the operation * in group theory) means that both $\,A_1\,,\,A_2\,$ are free groups themselves, but then why in the hell use this to talk about a *free group*?!2012-11-28
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    The existence of free products is frequently shown using words of finite length with elements of the factors as letters. This construction obviously yields a non-abelian group, because by construction the words $a_1a_2$ and $a_2a_1$ for elements $a_1\in A_1\setminus 1$ and $a_2\in A_2\setminus 1$ are different.2012-11-28
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    I edited your question to change your statement to "$A_1, A_2$ free groups", and then I realised you had added in the line "I'm trying to prove first for finitely presented free groups $A_1$, $A_2$." So, again, are you trying to do it for all groups $A_1$, $A_2$ or just for free groups?2012-11-28
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    @user1729 all free groups, sorry I'm really a beginner in this subject, maybe I'm using some wrong definition.2012-11-28
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    There are two (related) concepts - free groups and free products. The result holds for all non-trivial groups, not just free groups (a free product of non-trivial groups is always non-abelian). I am just confused because in your question you say you want to do the case of free groups first, which implies you will do the other cases after.2012-11-28
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    @user1729 ok I'm going to edit the question, thank you2012-11-28
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    Is there a reason for the algebraic topology tag?2012-11-28
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    @JasonDeVito yes, good question, usually we can study this also in algebraic topology courses, in fact I'm studying this in my algebraic topology course right now. Thanks for the remark.2012-11-28

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I could find a clue for this question as follows which should be verified independently.

If $a\in A$ and $b\in B$ are nontrivial elements in $A*B$, then $aba^{-1}b^{-1}$ has infinite order and so the above group is an infinite centerless group$^1$.

$1$. An introduction to the Theory of Groups by J.J.Rotman.

If this clue is useful for paving the way of any answer, I will delete it. :)

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    [While it is of course of infinite order, we only need the much much weaker fact that $[a,b]$ is nontrivial. Moreoever, this is why the free product $A*B$ is nonabelian when $A,B$ are *any* nontrivial groups. Why would we bother restricting ourselves to $A,B$ free groups, or bother restricting ourselves to finitely presented groups, OP?]2012-11-28
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    @anon: Exactly! we just need to see that $[a,b]\neq 1$. I saw a problem at this book before in chapter of free products. I just added this strongest fact, maybe it helps the OP and others to get out of this speech so soon. Sorry if I did nothing for the OP. Sorry.2012-11-28