A useful way of thinking of free groups is that they are the fundamentalgroups of (connected) graphs with base points. Given such a graph we canobtain an independent set of generators for its free fundamental group by pickinga maximal tree. The remaining edges then correspond to generators of thefundamental group as follows: given such an edge, start at the basepoint, travelalong the tree to one end of the edge, go along the edge, then go back along thetree to the basepoint.Lemma 66 Any subgroup of a free group is free. More precisely, an index msubgroup of a free group on n generators is free on m(n − 1) + 1 generators.Proof Represent the free group as the fundamental group of a graph with nloops. Then a subgroup of index m is the fundamental group of the correspond-ing m-fold connected cover. Since this is also a tree, its fundamental group isalso free.To count the numb er of generators, observe that the number of generatorsof the fundamental group of a graph is 1 − χ where χ is the Euler characteristic(numb er of vertices minus number of edges). Since the Euler characteristicgets multiplied by m when we take an m-fold cover, this gives the number ofgenerators of a subgroup. Exercise 67 Consider the action of the free group on 2 generators a, b on 3points 1, 2, 3 such that a and b act as the transposition (12) and (13). Find aset of four generators for the free subgroup fixing 1. (Draw the graph with threevertices 1, 2, 3 and four edges giving the actions of a and b on the vertices, thenpick a maximal tree (with 2 edges) then find the four generators by starting at1, running along the tree, across and edge, and back along the tree.)Exercise 68 How many subgroups of index 3 does the free group on 2 genera-tors have? (The subgroups correspond to transitive actions on 3 points, one ofwhich is marked.) How many triple covers does a figure 8 have?Now we show that free Lie algebras and free Lie groups are closely related.This may be a little surprising, because these correspond to connected anddiscrete groups, which in some sense are opposite to each other. Given a freegroup F , we can form its descending central series F0⊇ F1⊇ · · · , with Fi+1=[Fi, F ], the group generated by commutators.If a group has a descending central series G0⊃ G1· · · we can construct agraded Lie ring from it as follows. The Lie ring will be G0/G1⊕ G1/G2⊕ · · · .The additive structure of the ring is just given by the (abelian) group structureon each quotient. The Lie bracket is given by the commutator [a, b] = a−1b−1ab.The key point is to check that the Jacobi identity holds. This follows from PhilipHall’s identity:Exercise 69[[x, y−1], z]y· [[y, z−1], x]z· [[z, x−1], y]x= 1Exercise 70 Check that G0/G1⊕ G1/G2⊕ · · · is a Lie ring.35Theorem 71 The Lie ring of the descending central series of the free Lie groupon n generators is the free Lie ring on these generators.Proof First, there is an obvious homomorphism from the free Lie ring to theLie ring of the free group, by the universal property of the free ring. To provethis is an isomorphism we want to construct a map in the other direction. Wedo this as follows.We map each generator A of the free group to exp(a) in the rational com-pleted universal enveloping algebra of the free Lie ring, where a is the generatorof the free Lie ring corresponding to the generator A of the free group. This ex-tends to a homomophism f of groups by the universal property of a free group.If A is in Fnthen f(A) is of the form 1 + an+1+ an+2+ · · · where aihas de-gree i in the universal enveloping algebra. We define the image of A to be theelement an+1. We can check that this is primitive (as the log of a group-likeelement is primitive) and integral, so an element of the free Lie algebra. Wecan also check that this preserves addition and the Lie bracket and so gives aLie algebra homomorphism from the Lie ring of the free group. This gives thedesired inverse map, so proves that the Lie ring of the free group is the free Liealgebra. Exercise 72 Show that free groups are residually nilpotent. Show that free Liealgebras are residually nilpotent.So the relation between the free group and the free Lie algebra on somegenerators is given as follows. The Lie ring of the free group is the free Liering on the generators. The group generated by the elements exp(a), as a runsthrough generators for a free Lie ring, is the free group.6 Nilpotent Lie groupsThe main result about nilpotent Lie algebras is Engel’s theorem, due to FriedrichEngel (not to be confused with the philosopher Friedrich Engels).Theorem 73 (Engel) Suppose that g is a Lie algebra of nilpotent endomor-phisms of a non-zero finite dimensional vector space V . Then V has a nonzerovector fixed by g.Proof We use induction on the dimension of g. The main step is to show thatg has an ideal h of codimension 1 (unless g is 0). So fix any proper nonzerosubalgebra h of g. Then h acts on g by nilpotent endomorphisms, and so actson the vector space g/h by nilp otent endomorphisms. By induction there is anonzero element of g/h killed by h, so if h has codimension greater than 1 wecan add this to h and repeat until h has codimension 1. In this case h is anideal of g.Now lo ok at the subspace W of V fixed by all elements of h, which is non-zero by induction. This is acted on by the 1-dimensional Lie algebra g/h as his an ideal, and as g/h acts by a nilpotent endomorphism of W there must bea non-trivial fixed vector. This theorem s hows that if g is a Lie algebra of nilpotent endomorphisms ofV , then there is a flag 0 = V0⊂ V1⊂ · · · ⊂ Vn= V such that g acts trivially36on each Vi/Vi−1. (Take V1to be the vectors fixed by g and apply inductionto V/V1). In other words V has a basis so that g is strictly upper triangu-lar. Conversely any strictly upper triangular Lie algebra consists of nilpotentendomorphisms.We would like to say that a Lie algebra is nilpotent if all elements are rep-resented by nilpotent matrices, but there is a slight problem that this dep endson the choice of representation: an 1-dimensional abelian Lie algebra can berepresented by either a nilpotent or a non-nilpotent matrix. So instead we usethe following definition:Definition 74 A Lie algebra g is called nilpotent if it has a central series 0 =g0⊂ g1⊂ · · · ⊂ gn= g. This means that each giis an ideal, and g fixes allelements of gi/gi−1(or equivalently that gi/gi−1is in the center of g/gi−1).There