1) If is a group homomorphism, then and
Proof- First we have that so that . Since we are in a group H we can multiply the inverse to both sides, so whatever element happens to be we get:
Next we show that the inverse of is .
Since the inverse of an element f(a) is by definition and the inverse in any group of an element is unique, we must have that:
2) A group G is abelian if and only if the map given by is an automorphism.
Proof- First assume that G is abelian and suppose . We show the map is an automorphism. First, let then we have
so the map is well defined.
Next, so the map is a homomorphism.
Assume Then so the map is one to one.
For any we have that since G is a group. Then so that the map is surjective. Thus is an automorphism.
Assume that is an automoprhism. Show G is abelian. Let By definition of homomorphism:
Then by definition of the map we have:
by property of inverse.
Multiplying on both sides we get
so that G is abelian.
5) Let S be a nonempty subset of a group G and define and equivalence relation on G by . Show that is an equivalence relation iff .
Proof- Assume that is an equivalence relation. Show that S is a subgroup of G. S inherits associativity from G. Since S is nonempty it must contain some element, call it a. Then since we have an equivalence relation So S contains the identity. Since we can write we have that and since we have symmetry this gives . This says that for any element a in S we have the inverse element is also in S. Thus S is a subgroup of G.
This direction is done in the book in section 4 of chapter I.
6) A nonempty finite subset of a group G is a subgroup iff it is closed under the product in G.
We assume that S is closed under the product in G. That is, assume . Let then we need only show . If then we are done, so suppose . Note that the set is finite. Being finite tells us that there exists some numbers with the property such that . Then . Then since we have that
. Then .
so we have found the inverse of b to be which was shown to be in S. Since the product is closed in S, we have so by the subgroup test, S is a subgroup of G.
13) If is a cyclic group and H is any group, then the homomorphism is completely determined by the element
Proof- Suppose we know the value of . Then for any , x can be written as for some integer n, since G is cyclic. Then because f is a homomorphism we have:
, so what f does to any element x is completely determined by what it does to a.
15) b) (p prime)
Proof- Because is cyclic, we know from 13) that any homomorphism on is determined by where the generating element 1 is sent. So for any map write . Then we have:
. Then the map given by is an isomorphism.