Problem source

The set $S$ consists of $N(>2)$ elements $a_{1},a_{2},\ldots,a_{N}.$ $S$ is acted upon by a binary operation $\circ,$ defined by 
	\[
	a_{j}\circ a_{k}=a_{m},
	\]
	where $m$ is equal to the greater of $j$ and $k$. 
	Determine, giving reasons, which of the four group axioms hold for $S$ under $\circ,$ and which do not. 
	Determine also, giving reasons, which of the group axioms hold for $S$ under $*$, where $*$ is defined by 
	\[
	a_{j}*a_{k}=a_{n},
	\]
	where $n=\left|j-k\right|+1$.

Solution source

\begin{enumerate}
\item (Closure) This operation is clearly closed by construction
\item (Associative) $a_j \circ (a_k \circ a_l) = a_j \circ a_{\max(k,l)} = a_{\max(j,k,l)} = a_{\max(j,k)} \circ a_l = (a_j \circ a_k) \circ a_l$, so it is associative
\item (Identity) $a_1 \circ a_k = a_{\max(1,k)} = a_k = a_{\max(k,1)} = a_k \circ a_1$ so $a_1$ is an identity.
\item (Inverses) There is no inverse, since $a_N \circ a_k = a_N$ for all $k$, and hence $a_N$ can have no inverse.
\end{enumerate}

\begin{enumerate}
\item (Closure) $n = |j-k|+1 \geq 1$ so we need to show that $n \leq N$ to ensure closure. This is true since the largest $j-k$ can be is if $j = N$ and $k = 1$, and this also satisfies $|j-k| + 1 \leq N$. Hence the operation is closed.
\item (Associative) $a_j * (a_k * a_l) = a_j * (a_{|k-l|+1}) = a_{|j-|k-l|-1|+1}$. $(a_j * a_k) * a_l = a_{|j-k|+1}*a_l = a_{|l-|j-k|-1|+1}$. $a_2 * (a_2 * a_3) = a_2 * a_2 = a_1$. $(a_2 *a_2)*a_3 = a_1 * a_3 = a_3 \neq a_2$ therefore this isn't associative for any $N > 2$
\item (Identity) $a_1$ is an identity, since $a_1 * a_k = a_{|k-1|+1} = a_{k-1+1} = a_k$.
\item (Inverse) Every element is self-inverse since $a_k * a_k = a_{|k-k|+1} = a_1$
\end{enumerate}