Given an infinite group which is finitely generated, is there a proper maximal normal subgroup?

If you mean nontrivial maximal normal subgroup (not 1 or the whole group), then the answer is no. Higman constructed a finitely generated infinite group G with no subgroups of finite index. You then get a finitely generated group with no nontrivial normal subgroups by taking the quotient by a maximal normal subgroup. Higman's group G is < a,b,c,d  a^1 b a = b^2, b^1cb = c^2, c^1dc=d^2, d^1ad=a^2 > See Higman, Graham. A finitely generated infinite simple group. J. London Math. Soc. 26, (1951). 6164. Edit: If you mean does it have a proper maximal normal subgroup, then the answer is yes: Finitely generated groups have a (possibly trivial) maximal normal subgroup. Higman's reference for this is B.H. Neumann, "Some remarks on infinite groups ", Journal London Math. Soc, 12 (1937), 120127. 


Check out the Tarski monster. It is 2generated and simple. 


So many answers! I'm comletely lost. The paper of "B.H. Neumann, "Some remarks on infinite groups ", Journal London Math. Soc, 12 (1937), 120127" stated results for the existence of maximal subgroups, not maximal normal subgroup. Is this existence question of nontrivial normal subgroup still unsolved? 


Assuming you mean "does a maximal normal subgroup always exist?" (and that you don't care about computing it), here is a way to restate the problem. Notice that if G has no maximal normal subgroups, that means that every proper normal subgroup H of G is contained in a larger proper normal subgroup K of G. In particular, this means that the group G/H must not be finite; if it were, we could only find a finite chain of normal subgroups between H and G. So the question "does a maximal normal subgroup always exist" is the same as "must a finitely generated group have any finite nontrivial quotients?" I'm not sure what the answer to that is, but it seems like a useful restatement. 

