TITLE: Extension of a binary operation
QUESTION [5 upvotes]: Could you help me with the following problem?
Let $X$ be a set with a (not necessarily associative) binary operation. Let $o$ be an element such that $o \not \in X$ and let $X^o=X \cup \{o\}$.
Show that there exists only one extension of the operation on $X$ to an operation on $X^o$ such that $o$ is it's neutral element.
I would appreciate any help.
Thank you.

REPLY [4 votes]: An operation on $X$ is a function $X\times X\to X$, so when you've specified the action on all pairs, you have it.
Let's denote by $\cdot$ the original operation on $X$ and by $\bullet$ the operation on $X^o=X\cup\{o\}$.
Then, for $x$ and $y$ in $X$, we must have
$$x\bullet y=x\cdot y,$$
because we want an operation that extends the original one, while we want that $o$ is the neutral element in $X^o$, which forces
\begin{gather}
o\bullet x=x\\
x\bullet o=x\\
o\bullet o=o
\end{gather}
for all $x\in X$. There's no missing pair, because a pair of elements in $X^o\times X^o$ either has $o$ or not, so it must be of one of the forms
$$(x,y),\quad (o,x),\quad (x,o),\quad (o,o)$$
with $x,y\in X$.
Therefore the extension is unique as required.