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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.


share|cite|improve this question
Your claim as stated is false: $1 * 1 \cong 1$, and $1$ is certainly abelian. – Zhen Lin Nov 28 '12 at 11:00
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. – Berci Nov 28 '12 at 11:40
I don't see why this question should deserve a down vote. Hence I will up vote to cancel the down vote. – Rudy the Reindeer Nov 28 '12 at 11:42
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?! – DonAntonio Nov 28 '12 at 12:50
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. – user1729 Nov 28 '12 at 14:47
up vote 2 down vote accepted

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. :)

share|cite|improve this answer
[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?] – anon Nov 28 '12 at 13:06
@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. – Babak S. Nov 28 '12 at 13:13
+1 Nice work, once again! – amWhy Apr 8 '13 at 0:37

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