The notion of a group belongs to algebra rather than logic, so I would suggest that you take a proper algebra course. But the basic idea is that a group is a set equipped with a "multiplication" operation that satisfies a couple of axioms (associativity, existence of an identity, existence of inverses). So for example the integers are a group with respect to addition (I know I called it multiplication before, but any binary operation will do) because it is associative, there is an additive identity (zero) and there are additive inverses (negative numbers). But they do not form a group with respect to multiplication because though multiplication is associative and 1 is the multiplicative identity element, the integers don't contain multiplicative inverses for very many of its elements.
One important example of a group, and the one relevant to your question, is an object called a free group. Take a set S, and define the "free group whose generator set is S" to be the set of all "words" in the "alphabet" of S equipped with the concatenation operation. For example, typical elements of the free group on two generators are aaabbababbbab or aababaaa or aaaa. Every free group also has an empty word with no characters, usually denoted e. The concatenation operations just sticks two words back to back; for example, aaabbab * abaab = aaabbababaab. Thus the identity element is just the empty word, and to get multiplicative inverses you just have to assume that your generator set contains inverses for each of its elements (even though you don't usually write them when you specify the generators). So in the above example there are elements a' and b' such that a * a' is the empty word and b * b' is the empty word. Perhaps you can see that you can get inverses for arbitrary words this way; for example, if we multiply a'ba by a'b'a, then we get: a'baa'b'a = a'beb'a = a'bb'a = a'ea = a'a = e.
Given a free group, we can add "relations" to the group (this corresponds to forming the quotient by a certain normal subgroup). For example, if I take the free group on two generators {a,b} and declare that a^2 = e, b^4 = e, and aba = b^3, then you get what is called the dihedral group D_8 which corresponds to the set of symmetries of a square. Note that the notation a^2 is shorthand for a*a, b^4 is shorthand for b*b*b*b, etc. The generators allow you to simplify words; for example, the word a^3 b^5 a^3 is actually the same as b^3 in D_8 because a^3 = a^2 * a = e*a = a and likewise b^5 = b, which implies that a^3 b^5 a^3 = aba = b^3. You can check that there actually only 8 different words in the dihedral group, including the identity.
Note that it can actually be pretty hard to tell how many elements are in a group given by generators and relations. For example, let's say we took the group generated by a and b subject to the following relations: a b^2 = a, b*a = e, and a * b * a * b^2 = e. This looks perfectly innocuous, but if you combine the three relations then you will find that in this group a = e and b = e, so the group only has one element (the identity element)! The group whose only element is the identity element is called the "trivial" group.
We are finally ready to state the word problem. Simply put, the word problem asks if there is an algorithm which takes as input a group given by finitely many generators and relations and determines whether or not the group is trivial. The answer turns out to be no. I have seen a proof, but I have forgotten how it goes; I don't know enough logic to say anything intelligent about it. The word problem is important because it is one of the first examples of logic producing a theorem that other mathematicians were genuinely curious about. Groups given by generators and relations come up all over the place, especially in algebraic topology as the fundamental group of various spaces (for example, the fundamental group of the figure 8 is the free group with two generators). So the theorem also implies that given a finite CW-complex there is no way to determine if the underlying topological space is simply connected, even if you know a lot about how it is glued together. There are actually lots of other problems now in algebra and geometry that have been solved using methods from logic, but the word problem was I believe the first.