**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:

by cancelling.

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.
Proof-** Trivial

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 .

Then finally:

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.