Disclaimer: these are my personal solutions; they have not been reviewed, so be careful relying on them as there might be errors. Some competitions may have official solutions from the contest publishers which you may wish to refer to instead if you’re looking for verified solutions.
Problem 2
Let $S$ be a set which is closed under the binary operation $\circ$, with the following properties:
there is an element $e \in S$ such that $a \circ e = e \circ a = a$, for each $a \in S$,
$(a \circ b) \circ (c \circ d) = (a \circ c) \circ (b \circ d)$, for all $a, b, c, d \in S$.
Prove or disprove:
- $\circ$ is associative on $S$
- $\circ$ is commutative on $S$
Hints
Hint #1
What can you do with the two given properties? How can you apply them?
Hint #2
What if we apply property #1 more than once, i.e., to more than one variable in the given property #2?
Solution
(a) Is $\circ$ associative on $S$?
Open to see solution; try solving it first or use the hints!
Let’s consider the given equation $(a \circ b) \circ (c \circ d) = (a \circ c) \circ (b \circ d)$ and consider the case where $b = e$:
$$ \begin{align*} (a \circ b) \circ (c \circ d) & = (a \circ c) \circ (b \circ d) & \\ (a \circ e) \circ (c \circ d) & = (a \circ c) \circ (e \circ d) & \qquad \textrm{substitute $b = e$} \\ a \circ (c \circ d) & = (a \circ c) \circ d & \qquad \textrm{given: $x \circ e = e \circ x = x$} \\ \end{align*} $$This proves that $\boxed{\circ\ \text{is associative on} \ S}$. $\quad \blacksquare$
Alternatively, we can set $c = e$ with similar results:
$$ \begin{align*} (a \circ b) \circ (c \circ d) & = (a \circ c) \circ (b \circ d) & \\ (a \circ b) \circ (e \circ d) & = (a \circ e) \circ (b \circ d) & \qquad \textrm{substitute $c = e$} \\ (a \circ b) \circ d & = a \circ (b \circ d) & \qquad \textrm{given: $x \circ e = e \circ x = x$} \\ \end{align*} $$This proves that $\boxed{\circ \ \text{is associative on} \ S}$. $\quad \blacksquare$
However, we can’t prove the same result if we set $a = e$ or $d = e$, why is that? Exercise for the reader.
(b) Is $\circ$ commutative on $S$?
Open to see solution; try solving it first or use the hints!
For this case, let’s take the same given equation $(a \circ b) \circ (c \circ d) = (a \circ c) \circ (b \circ d)$ and consider the case where $a = e$ and $d = e$:
$$ \begin{align*} (a \circ b) \circ (c \circ d) & = (a \circ c) \circ (b \circ d) & \\ (e \circ b) \circ (c \circ e) & = (e \circ c) \circ (b \circ e) & \qquad \textrm{substitute $a = e$, $d = e$} \\ b \circ c & = c \circ b & \qquad \textrm{given: $x \circ e = e \circ x = x$} \\ \end{align*} $$This proves that $\boxed{\circ \ \text{is commutative on} \ S}$. $\quad \blacksquare$