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I know this may sound a bit trivial, but I just want to check whether this is true:

[Question] Let $G$ be a group and $a,\,b\in G$, and $e$ be an identity element. Suppose $ab=e$. If $a\neq b^{-1}$, then $a=e$ and $b=e$

[My attempt] So I prove by contrapositive. If $a\neq e$ or $b\neq e$, we go by cases. Suppose $a\neq e$ but $b=e$, this cannot happen because we assume that $ab=e$. Thus $a\neq e$ and $b\neq e$. Then $a=(a^{-1}a)b=a^{-1}(ab)=a^{-1}e=a^{-1}$.

I think there is something wrong with the proposed question or my attempt because in the later part, even though I assume that both $a$ and $b$ are not the identity element, I didn't really use the assumption.

Now, if I rephrase the question in another way: Let $\{G_{\lambda}\}_{\lambda\in\Lambda}$ be a collection of disjoint groups and consider its free product $*_{\lambda}G_{\lambda}$. For each element $g=g_{1}g_{2}\ldots g_{m}\in *_{\lambda}G_{\lambda}$ which is a reduced word, $g=e$ if and only if $g_{1}=g_{2}=\ldots g_{m}=e$.

Is the question true and whether there is any proof for this?

Many thanks!

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Ehr... if $ab=e$, then $a=b^{-1}$; in fact, the case of $a=b=e$ also satisfies $a=b^{-1}$, since $e^{-1}=e$. In other words, the premise of your implication is always false! –  Arturo Magidin May 20 '12 at 20:23
    
Good point! I realised this after i have read the answer below. –  enoughsaid05 May 21 '12 at 2:30
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1 Answer

up vote 1 down vote accepted

Perhaps you should learn elementary group theory first before considering free products of groups. It is trivial and holds in a general group, that $ab=e$ implies $a=b^{-1}$. Just read the definitions.

The second question about free products follows directly from the definition of the free product as the set of reduced words. If you use another definition, you should name it and make clear what is the precise problem.

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