TITLE: How does one show a set of axioms is independent? (of each other?)
QUESTION [18 upvotes]: I am being asked to show that the groups axioms of existence of an identity, of inverses and of associativity are independent. Does this mean that none of two of them imply the third? That is: do I have to exhibit a "bad" group which has two but fails to have the third? (for each case) Or do I have to find examples also where one succeeds and the other two fail?
ADD The axiom of inverses uses the axiom of the identity, so, how can the axiom of identity be independent of that of the inverses?  Ans One can state the axiom of inverses without resorting to the notion of a unity. 
What does it mean, more formally, that a set of axioms is independent?

REPLY [2 votes]: I will preface the following by saying that like you I don't currently understand how the axiom of inverses can exist independently of the axioms of identity for a group.  I can see how associativity can get shown as independent.
The concept of independence of axioms may get fruitfully compared with what we could call a non-independent axiom. A non-independent axiom of a system consists of an axiom which comes as a necessary consequence of the other axioms of the system.  In the theory of groups, statements like the cancellation law and the involutive law --x=x or equivalently x--=x in reverse polish notation, can get called non-independent axioms in the sense that given the 3 standard axioms used for group theory, they necessarily follow.  Within the theory of groups, those axioms cannot "stand alone" from the 3 axioms most commonly used to start group theory.  
As perhaps a better example, you can find authors like Nguyen and Walker in "A First Course in Fuzzy Logic" who will axiomitize lattices as commutative, idempotent, absorptive, and associative.  But, idempotence can get called as non-independent, since you can derive it from the other axioms.  In contrast, an axiom x qualifies as independent of a set of axioms S if it can "stand alone" from all the axioms of S, in the sense that we can't derive x from the axioms of S.
So, suppose we have a finite set of axioms {a, ..., n}.  If I want to show axiom n independent of the other axioms, then I need to show all axioms belonging to {a, ..., m} can hold true, while axiom n does not hold true.  Thus, we can't start with axiom set {a, ..., m} and derive axiom n, because then we could have a true statement that could lead to a false one.  If I want to show the set of axioms {a, b, c} independent, then I want to show that for all members of {a, b, c}, independence holds for each of the axioms, or that each axiom is independent, or can stand alone without the others.  So, I'll want to show that {a, b} can hold true while c does not hold true, {a, c} can hold true while b does not hold true, and {b, c} can hold true while "a" does not hold true.  Sometimes one does this by exhibiting concrete algebraic systems (similar to what you called ""bad" groups") which make the independence of each axiom (hopefully) clear.  These systems give us a model of all of the axioms except one of them.
As an example of how to demonstrate independence of a set of axioms, let's try and show the axioms for semilattices independent of each other.  A semilattice as an algebraic structure consists of a set S with a binary operation S which satisfies commutativity, associativity, and idempotence.  Or more symbolically for (S, S) it holds that for all x, y, z belonging to S we have:
1-xyzSS=xySzS associativity
2-xxS=x idempotence
3-xyS=yxS commutativity.
So, let's first consider algebraic structure ({0, 1}, E) with E defined by this table:
 E  0  1
 0  1  0
 1  0  1

Idempotence fails since 00E=1, while commutativity can get seen to hold by inspection, and associativity holds since this can get understood as the table for logical equivalence, which associates.  Thus, axiom 2 is independent of axiom set {1, 3}.  If I understand logical terminology correctly, we can say that ({0, 1}, E) models commutativity and associativity but not idempotency.
Next let's consider ({0, 1, 2}, A) with A defined by this table:
A  0  1  2
0  0  1  1
1  1  1  2
2  1  2  2

Commutativity and idempotence hold, while 012AA=02A=1 and 01A2A=12A=2 so associativity fails.  Thus, axiom 1 is independent of {2, 3}.
Finally consider structure ({0, 1}, B) with B defined by this table:
B  0  1
0  0  0
1  1  1

Idempotence holds, and for this structure for all x, xyB=x.  Thus, xyzBB=x, and xyBzB=xzB=x, and thus associativity holds.  Commutativity fails, and thus axiom 3 is independent of {1, 2}.
Consequently, it follows that all axioms for a semilattice are independent in the sense that for all of the axioms in the axiom set you can have all of the axioms hold true but one of them hold false, let's call the true axiom o and the set of axioms with without axiom o set T, and not have the ability to validly derive axiom o from set T.