In practice, you should be able to tell that a table is a group table by matching it with a group that you already know. Assuming that you don't look at tables with more than six elements, here are the possible groups:
- Cyclic groups: $C_2, C_4,\dots,C_6$
- The Klein four group $V_4$
- The symmetric group $S_3$.
So if you know what the multiplication tables of these groups look like, you'll be in good shape.
It's also good to know how to eliminate possibilities.
Here are three necessary criteria for a table to be a group multiplication table. When I say necessary, I mean that these can only tell you if the table is not a group table. In other words, if these criteria are satisfied, it is nonetheless possible that the table is not a group table.
Say the elements to be multiplied are $a_1,a_2,\dots,a_n$ (in that order). Then
- There is a row $R$ that reads $a_1, a_2,\dots, a_n$ (in that order), and a column $C$ that reads $a_1,a_2,\dots,a_n$ (in that order). Edit: $R$ and $C$ must intersect on the diagonal.
- Every row contains each of $a_1,a_2,\dots,a_n$ exactly once.
- Every column contains each of $a_1,a_2,\dots,a_n$ exactly once.
Condition 1 comes from the existence of an identity element, and conditions 2 and 3 come from the existence of inverses. (Having inverses basically means that we can always undo multiplication, but if $ax=bx$ for $a\neq b$ -- i.e., if a row has duplicates -- then we can't undo multiplication by $x$. I can give more details if you're interested.)
The condition I haven't mentioned yet is associativity. If I add the condition
- The multiplication is associative.
then conditions 1 - 4 become necessary and sufficient, which means that conditions 1 - 4 are true if and only if the table is a multiplication table.
No reasonable professor will ever ask you to verify that a multiplication table is associative. There are algorithms to do so, which this MSE post mentions, but they're beyond the scope of an introductory algebra class. If conditions 1 - 3 hold and the table is small, it's easier to see if the table matches the multiplication table of one of the groups I listed above.
Closure is most definitely part of the definition of a group.
It might be that in some texts, closure is part of the definition of the binary operation, e.g., implicit when writing $\star : G \times G \to G$. (Rather than reiterating it as a separate axiom.)
Best Answer
HINT: What is the multiplicative inverse of $0$?
If we replace $\mathbb Q$ with $\mathbb Q\setminus \{0\}$ to exclude $0$ from our set, under the operation of multiplication, we do then have a group: $(\mathbb Q\setminus \{0\}, \cdot)$