# Groups: showing $ab=1$ implies $ba=1$.

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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Multiply $ab=1$ on the left by $a^{-1}$. – Arturo Magidin Oct 8 '11 at 20:27
@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 $ba=z$. – Nicky Hekster Oct 9 '11 at 21:40

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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MathJax gets a little confused when you use eqnarray outside of mathmode. The two ways to fix the issue are: (i) enclose it in ; then it works as usual. And (ii) use a "triple backslash", \\, for the end-of-line; the first backslash informs the parser that the next characters should be treated literally, and that solves the issue as well. I used the former to fix it; then I rolled back your edit because the eqnarray displays are better than a sequence of displays. – Arturo Magidin Oct 9 '11 at 3:12
@ArturoMagidin: I did not know that; thanks! – jwodder Oct 9 '11 at 3:15