# A question on product of two or more elements in a group.

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

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.