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Let $G$ be a group and $1$ is the identity in $G$. Suppose $a$, $b$ in $G$ and $ab=1$, how could one simply show that $ba=1$?

Thanks!

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    @Anonymous. Now can you prove this: if $z$ is an element of the center $Z(G)$ of $G$, and let $a, b$ in $G$ with $ab=z$. Then $ba=z$.2011-10-09

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By the axioms for a group, there exists some element $c\in G$ such that $ac=ca=1.$ Thus, if $ab=1$, then $cab=c1=c$ but also $cab=1b=b$ hence $c=b$, and therefore $ac=ca=1$ implies that $ab=ba=1$. (Incidentally, this also shows that the inverse of an element in a group is unique.)

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An alternative proof which may or may not be easier:

Multiply $ab = 1$ on the left by $a^{-1}$ to yield:

$\begin{eqnarray} a^{-1}ab & = & a^{-1}1 \\ 1b & = & a^{-1} \\ b & = & a^{-1} \end{eqnarray}$

Now multiply on the right by $a$ to obtain: $\begin{eqnarray} ba & = & a^{-1}a \\ ba & = & 1 \end{eqnarray}$

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    @ArturoMagidin: I did not know that; thanks!2011-10-09