Given that the binary operation on N given by $a \ast b = $ HCF $(a,b)$:

$\textbf {Step 1: Checking if the operation is commutative}$:

We know $ HCF $(a,b) = $HCF $(b,a)$.

$\Rightarrow a*b=b*a$ for all $a,b \in N \rightarrow \ast $ is commutative.

$\textbf {Step 2: Checking if the operation is associative}$:

For an operation $\ast$ to be associative $a\ast ( b \ast c) = (a \ast b) \ast c\; \forall \; a, b, c \in N$

$(a*b)*c = $ HCF $(a,b) \ast c =$ HCF $(a,b,c)$

$ a*(b*c)= a*($ HCF $(b,c)) =$ HCF $(a,b,c)$

$\Rightarrow (a*b)*c=a*(b*c) \forall a,b,c \in N \rightarrow \ast $ is associative.

$\textbf {Step 2: Checking if the operation has an identity}$:

We know that the element $e \in N $ is an identify element for operation * if $a*e=e*a$ for all $a \in N$

We can clearly see that this is NOT true for all $a \in N$ through a small example:

Let $a=5$, $ 5 \ast 1 = 1$ and so is $6 \ast 1 = 1$.

So, this relation is not true for al $a \in N$. Therefore $\ast$ does not have an identity element in $N$.