**1) Give examples of semigroups and monoids which are not groups**

**Solution-** is a semigroup, but neither a monoid nor a group. This is because the number 0 is not in the natural numbers, thus there is no additive identity.

is a monoid because 1 is the identity, however it has no inverses.

**2) Let G be a group, S a nonempty set, and M(S,G) the set of all functions . Define addition in M (S,G) as follows: is given by . Prove that M (S,G) is a group, which is abelian if G is.
Proof-** We must check that the operation is associative. We have that

Now, G is a group so it has an identity element by definition, call it 0. Define a function by . This is by definition a function so it belongs to M (S,G). Then we have so that e is the identity of M (S,G).

Lastly, define for any function the function defined by . Then we have .

If G is abelian then we have Therefore M(S,G) is abelian.

**7) If p is prime, then the nonzero elements of form a group of order p-1 under multiplication (mod p).
Proof-** inherits associativity from multiplication of integers, as well as the identity element 1. Since p is prime, for any we have that gcd(a,p) = 1. Thus, there exists such that . Then by doing mod p on both sides we have (mod p), so that . Thus we have an inverse for each element and hence a group of order p-1 elements.

**8) a) The relation given by is a congruence relation on the additive group
Proof-** Let . First, since we have that is reflexive. If then for some so then so that the relation is symmetric. If and then we have and for some integers, m and n. Then putting them together we have so the relation is transitive. Lastly, suppose that and . Then and for some integers r and s. To show congruence we must show that . Consider Thus it is a congruence relation.

**b) The set of equivalence classes is an infinite abelian group.
Proof-** From part a) we know that the relation is a congruence relation. Theorem 1.5 in Hungerford gives us that the set of equivalence classes of an abelian group under a congruence relation is an abelian group.

**9) Let p be a fixed prime. Let be the set of all those rational numbers whose denominator is relatively prime to p. Let b the set of rationals whose denominator is a power of p. Prove both and are groups under ordinary addition.
Proof-** First we consider and we show closure. Let . Then We have that so that and which gives that . Thus so the operation is closed.

Associativity is inherited from the ordinary addition of rationals.

The identity 0, can be written as where p` is some other prime so that .

The inverse of an element is

is very similar except easier.

**11) The following conditions on a group G are equivalent:
(i) G is abelian
(ii)
(iii)
(iv)
(v) for 3 consecutive integers n and
Proof-** Let G be abelian. The case when n is 0 is trivial. Let Then Assume it is true for some natural number k and then use induction. If then we have so we can use the previous case to show it is true.

follows immediately.

Let . Assuming for some n the hypothesis is true, we have , , . We use these facts to say first that

We also have

Then using the fact that , we get

So that G is abelian.

Trivial

Assume that . Then we have:

Assume that . Then

so that G is abelian.

**13) If , then G is abelian.
Proof-** The fact that gives us that . Let . Then we know

This is true for any arbitrary elements a and b, so G is abelian.

**14) If G is a finite group of even order then G contains an element such that .
Proof-** Suppose that no element satisfies the given condition. So that there is no element a such that . Then each element can be paired with it’s inverse to give 2n elements in G for some n. All of these pairs together with the identity element e, gives 2n + 1 elements in G, which is a contradiction to G being of even order. Therefore such an element exists.