How to define a group structure on a given arbitrary set? I am doing a course in Abstract Algebra, and my teacher gave me a question: "Find all possible group structures on a set X whose cardinality is ≤ 4."
I know basic group theory, but I am unable to understand what exactly does the question expect us to do (meaning of the question in simple language, and a possible technique or way to answer such questions, exact answer not needed.)
 A: As you are specifically wanting 


*

*to understand what the question asks, and

*a hints.


I will address these and (in view of (2)) not give a worked solution.
1) As I read it, the question essentially wants you to fix a set with $4$ elements and find all group structures on this set. A different (easier) question is "find all groups with four elements" (but then the question should have read "Find all possible group structures up to isomorphism on a set $X$ whose cardinality is $\leq 4$).
Possibly this second interpretation is what was intended when the question was written, but I do not think this is what the question is actually asking. [This is discussed in the comments to the question.]
2) It is well-known that there are two groups of order $4$, up to isomorphism. If you are unaware of this fact then you should start by verifying it. Lets fix the set $X=\{a, b, c, d\}$. Every bijection from the set $X$ to the group $\mathbb{Z}_4$ defines a group structure on $X$. So, for example, $a\mapsto 0$, $b\mapsto 1$, $c\mapsto 2$, $d\mapsto 3$ gives a group structure, while $a\mapsto 1$, $b\mapsto 2$, $c\mapsto 3$, $d\mapsto 0$ gives a different group structure. There are $4!$ bijections between $X$ and $\mathbb{Z}_4$, and similarly $4!$ between $X$ and $\mathbb{Z}_2\times\mathbb{Z}_2$ (the Klein $4$-group). Hence, there are at most $4!+4!$ group structures which we can put on $X$. However, there is some double-counting going on. The actual number of group structures is $4!/2+4!/6=16$. Why?
