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MODERN ALGEBRA 1 - FALL 2012 ASSIGNMENT 7 - Hungerford 1.9 - free groups, generators and relations - Solutions 1) Show that every non-identity element in a free group has infinite order Given a, b, | we have irreducible words of arbitrary length for each generator. 2) Show that a, b|a 5 b 2 1, ba a 2 b 2 Note that b b 1 , so bab a 2 . Also, a a 6 , so bab 3 a. Moreover, bab 3 babbabbab ba 3 b. From a ba 3 b we get ba a 3 b. The relation ba a 2 b implies a 3 b a 2 b, whence a 1. Now observe that b|b 2 1 2 . The key to all problems of this type is to realize that the finite group defined by a presentation is the largest group (for the given set of generators) that satisfies the relations given, and any other group with the same generators that is no smaller than the one with the presentation must actually be isomorphic to it. This is the content of van Dyck’s Theorem, which says if you have two groups with the same generators and one group has all of the relations that the other one has plus more, the one with more relations is the epimorphic image of the one with fewer relations. Images of homomorphisms can only be the same size or smaller than their domains. The trick in this problem is to get rid of a by showing it is really the identity, then the presentation of 2 has the same generator as a, b|a 5 b 2 1, ba a 2 b , allowing van Dyck’s Theorem to be applied....remember the generator sets have to be the same. We see that | 2 | a, b|a 5 b 2 1, ba a 2 b . It follows that 2 a, b|a 5 b 2 1, ba a 2 b . 3) Show that a, b|a 2 b 2 ab 2 1 Q 8 Example 5 in Gallian (p.442). 4) Show that a, b|a 2 b 3 a 1 b 1 ab 1 6 From a 1 b 1 ab 1, we have ab ba, so the group is commutative. The cyclic group 6 is generated by a 3 and b 2 with the same relations as the given group, so by the argument above, a, b|a 2 b 3 a 1 b 1 ab 1 6 . Another approach would be to argue that the only distinct elements of the given group are a, a 2 , b, ab, a 2 b, and 1. The only commutative group of order six is 6 . 5) Define the infinite dihedral group D a, b|a 2 b 2 1. Find the order of each element of D . Either a or b could have order 1 or 2 to satisfy the relations. The group is not given a relation that forces it to be abelian, so the element ab can have infinite order...the relations don’t allow any contraction of abab, likewise for baba. Clearly all the elements of infinite order are chains of successive ab or ba, but they can’t be mixed without an a 2 or b 2 appearing and allowing a contraction. 6) Let G a, b|a 4 b 4 aba 1 b 1 1. Show |G| 16. We have commutativity, so possible distinct elements are of the form a i b j , where 0 i, j 3, which gives at most 16.