# Free product of non-trivial groups is non-abelian

1. Dec 15, 2011

### James4

Hello

I have to show that the free product of a collection of more than one non-trivial group is non-abelian.

But doesn't this just follow from the definition of the free product?
Or how would you tackle this question?

2. Dec 15, 2011

### Office_Shredder

Staff Emeritus
Well... what definition of free product are you using, and how does non-abelian follow from it?

3. Dec 15, 2011

### James4

You mean the definition of the free product?

As a set, the free product  G consists of all
words g_1g_2.. .g_m of arbitrary finite length m > 0, where each letter g_i belongs to
a group G_ i and is not the identity element of G_ i , and adjacent letters g_i and g_(i+1)
belong to different groups G .

For a counterexample to being abelian i thought about:
$\pi_1(S^1 \vee S^1)$ which corresponds to the free product of the fundamental groups of S^1 and which is not abbelian

4. Dec 15, 2011

### Office_Shredder

Staff Emeritus
So if you have two groups G and H, and you look at G*H, can you identify two elements in G*H which do not commute?

5. Dec 15, 2011

### James4

Hi Office_Shredder

>So if you have two groups G and H, and you look at G*H, can you identify two elements in G*H which do not commute?

Thats what I meant when I wrote in the beginning that it almost follows form the definition, because it holds for any two non trivial items g,h from G,H respectively that g*h is not the same as h*g.

6. Dec 15, 2011

### Dick

Yes, it's almost trivial. g*h is not equal to h*g if g and h are not the identities of their respective groups.

7. Dec 15, 2011

### James4

So is there actually something to prove here?

8. Dec 15, 2011

### Dick

Not much. gh is not equal to hg because g isn't equal h and neither is the identity.