Steinberg group
The elementary matrices obey certain commutation rules, and you can ask: what group do you get if you take symbols obeying only those rules, forgetting that they are matrices at all? The answer is the Steinberg group. It is the most efficient group that maps onto the elementary group while respecting the known relations — and the surprise is that it is slightly bigger than the elementary group. That extra slack, the kernel of the map back to the matrices, is precisely K_2.
For a ring R and n at least 3, the Steinberg group St(n, R) is generated by symbols x_{ij}(r), one for each pair of distinct indices i, j in {1, ..., n} and each r in R, subject to: x_{ij}(r) x_{ij}(s) = x_{ij}(r + s); the commutator [x_{ij}(r), x_{jk}(s)] = x_{ik}(rs) when i, j, k are distinct; and [x_{ij}(r), x_{kl}(s)] = 1 when j is not k and i is not l. The stable Steinberg group St(R) is the direct limit over n. The map x_{ij}(r) -> e_{ij}(r) defines a surjection St(R) -> E(R).
The central fact is that St(R) is the universal central extension of the perfect group E(R), and the kernel of St(R) -> E(R) is its center, equal by definition to K_2(R). Equivalently 1 -> K_2(R) -> St(R) -> E(R) -> 1 is the universal central extension. So the Steinberg group is the precise algebraic mechanism that produces K_2: it builds the relations among elementary matrices abstractly, and the unexpected ones constitute the second K-group.
The Steinberg relation [x_{ij}(r), x_{jk}(s)] = x_{ik}(rs) faithfully mirrors the matrix identity [e_{ij}(r), e_{jk}(s)] = e_{ik}(rs) in E(R). The elements that map to the identity in E(R) but are not forced to be trivial in St(R) generate K_2(R).
St(R) abstracts the relations of elementary matrices.
A perfect group (one equal to its own commutator subgroup) has a unique universal central extension, and its kernel is the Schur multiplier H_2 of the group. Since E(R) is perfect, K_2(R) is exactly H_2(E(R), Z), tying K_2 to group homology.