Can closure of quaternions under multiplication be shown with a cayley table? Unsure about my understanding of groups and quaternions. I'm trying to figure out if just using a cayley table (specifically this one) can show closure of quaternions under multiplication, is something more needed or am I completely wrong?
 A: The axiom of "closure" is the odd one. This is because you need not take it as an axiom. It depends on your definition. Personally, I like to define my groups as follows.

A group is a set $S$ with a binary operation (multiplication) $\ast: S\times S\rightarrow S$ such that

*

*(Associativity) $(g\ast h)\ast k=g\ast(h\ast k)$


*(Identity) There exists $1\in S$ such that $g\cdot1=1\ast g=g$ for all $g\in S$.


*(Inverses) For all $g\in S$ there exists $h\in S$ such that $g\ast h=1$.

Notice that there is no closure axiom here. This is because the "binary operation" statement takes care of closure - I can never leave the set $S$ using the operation $\ast$. This definition is analogous to a Cayley table - a Cayley table describes a binary operation, and it is the other three axioms that you need to check for.
In summary: if you have a Cayley table for your group then closure is immediate. Otherwise it isn't a Cayley table. This is the case here. On the other hand, if your lecturer is trying to trick you then they might stick in a symbol not from your group, and so closure does not happen. But then this isn't a Cayley table...
