By James Wilson

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**Extra resources for A Hungerford’s Algebra Solutions Manual **

**Sample text**

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).

Thus Rp is closed under addition. Lastly −(a/pi ) = (−a)/pi which is in Rp by definition. Therefore Rp is a (sub)group. 1. For the infinity of the group consider the elements 1/pi . How many are there? 1. 9 to show (i),(ii), and (iii) are equivalent;(i), (ii), and (iii) imply (iv); and (iv) implies and (v). Conclude the equivalence by showing (v) implies (i) by showing first the relations abn = bn a and abn+1 = bn+1 a are true. As a counter example of part (v) with only two consecutive integers, consider the two consecutive powers 0 and 1 in a non-abelian group.

Thus H ∨ K ⊆ HK. 8. Therefore HK = H ∨ K. Now suppose we have a finite collection of subgroups H1 , . . , Hn of G. Since multiplication is associative, H1 · · · Hn = (H1 · · · Hn−1 )Hn , as the elements a1 · · · an = (a1 · · · an−1 )an . Suppose H1 · · · Hn = H1 ∨ · · · ∨ Hn for some n ∈ Z+ . Then H1 · · · Hn+1 = (H1 · · · Hn )Hn+1 = (H1 ∨ · · · ∨ Hn )Hn+1 = n n (H1 ∨· · ·∨Hn )∨Hn+1 . Finally i=1 Hi ∨Hn+1 is defined as ( i=1 Hi )∪Hn+1 Which is simply H1 ∨ · · · ∨ Hn+1 . Therefore by induction, H1 ∨ · · · ∨ Hn = H1 · · · Hn , for all n ∈ Z+ .