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!
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!
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.)
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: $\begin{eqnarray} ba & = & a^{-1}a \\ ba & = & 1 \end{eqnarray}$