Prove that an associative operation* is a structural property First off how does one prove that an operation $\ast$ is a structural property and how is this different from just proving an isomorphism?
My main question is: how can I prove that $\ast$ on a binary structure is a structural property. 
So far I have said:
Let $\langle S,\ast\rangle$ and $\langle T,\sharp\rangle$ be two arbitrary binary structures preserved by an isomorphism $f: S\to T$.
Assume $\ast$ is associative
Then we have that $f((a\ast b)\ast c)=f(a\ast b)\sharp c$ for all $a,b$ in $S$
Due to associativity we have that $f(a\ast (b\ast c))=f(a)\sharp f(b\ast c) \ldots$.
And I don't know if I can reach any end from here. 
How should I continue? Or restart?
 A: According to your description of "structural property", you seem to have started out correctly.  What you need to show is that, if $\langle S,\ast\rangle$ and $\langle T,\sharp\rangle$ are two binary structures and $f:S\to T$ is an isomorphism between them, then $\langle S,\ast\rangle$ is associative if, and only if, $\langle T,\sharp\rangle$ is associative.  By symmetry, it suffices to assume that $\langle S,\ast\rangle$ is associative, and show that $\langle T,\sharp\rangle$ is associative.
Suppose then that $x,y,z\in T$; we need to show that $(x\sharp y)\sharp z = x\sharp(y\sharp z)$.  Since $f$ is an isomorphism, it has an inverse map $f^{-1}$ that is also an isomorphism. Therefore, there exist $a,b,c\in S$ for which $x = f(a)$, $y = f(b)$ and $z = f(c)$. Then
$$(x\sharp y)\sharp z = (f(a)\sharp f(b))\sharp f(c) = f(a\ast b)\sharp f(c) = f((a\ast b)\ast c),$$
since $f$ is a homomorphism. But, since $\ast$ is associative, the argument $(a\ast b)\ast c$ of $f$ is equal to $a\ast(b\ast c)$, so you can unravel the right hand side to eventually get $x\sharp (y\sharp z)$.  (I'll leave this for you.)  That proves that $\langle T,\sharp\rangle$ is associative as well.
