By James Wilson
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36 37 37 37 38 38 39 39 40 40 40 41 41 42 42 44 44 44 45 Homomorphisms. If f : G → H is a homomorphism of groups, then f (eG ) = eH and f (a−1 ) = f (a)−1 for all a ∈ G. Show by example that the first conclusion may be false if G, H are monoids that are note groups. Proof: Assuming f : G → H is a homomorphism of groups, then f (a) = f (aeG ) = f (a)f (eG ) and likewise on the left, f (a) = f (eG a) = f (eG )f (a).
Certainly given any two elements σ and τ , (στ )0 = ε = εε = σ 0 τ 0 . Likewise (στ )1 = στ = σ 1 τ 1 thus for two consecutive integers all elements have the property (ab)n = an bn . However clearly S3 is non-abelian since (12)(123) = (23) and (123)(12) = (13). Of course any nonabelian group can serve in this counter example. 12 Cyclic Conjugates. Hint(1/5): Use the standard trick for conjugation: insert b−1 b between powers of a to create conjugates of lower powers where information is given. i If G is a group, a, b ∈ G and bab−1 = ar for some r ∈ N, then bi ab−i = ar for all i ∈ N.
47 47 48 48 48 49 49 51 51 52 Order of Elements. Let a, b be elements of a group G. Show that |a| = |a−1 |; |ab| = |ba|, and |a| = |cac−1 | for all c ∈ G. Proof: Consider the cyclic group generated by an element a. 3, |a| = |a−1 |. Suppose the order, n, of ab is finite, so that (ab)n = e. We re-associate the product as follows: (ab) · · · (ab) = a(ba) · · · (ba)b = a(ba)n−1 b. So a(ba)n−1 b = e which implies (ba)n−1 = a−1 b−1 = (ba)−1 , and thus finally (ba)n = e.
A Hungerford’s Algebra Solutions Manual by James Wilson