**1) Let a,b be elements of a group G. Show that , , and
Proof-** We show the first equality by showing and we show this by subset inclusion. Let . This says that for some . The other direction is similar so that both sets contain each other and thus are the same set, so by definition the order of their generators is the same.

For the second equality, suppose . By definition this gives that Thus:

. This conclusion tells us that the order of the element ba can be no larger than n, when we are given the fact that the order of ab is n. Thus Repeating this argument for we get so that . The case when either side is infinite automatically gives the other side to be infinite too.

For the last equality, suppose . Then consider:

Again this tells us the that . Suppose . Then which gives

which is a contradiction since the order of a is n. Thus n = m. If either side is infinite then again the other side must be also.

**2) Let G be an abelian group with elements a,b having respective orders m,n. Show that G contains an element with order lcm(m,n)
Proof-** First we assume that gcd(m,n)=1. Then we have lcm(m,n) = mn. We have that:

so that . Suppose Then we have:

. Then since and the order of b is m, we have that since gcd(m,n) =1. Similarly Then since gcd(m,n)=1 we have , but so that . Thus the element ab has order lcm(m,n) when gcd(m,n) = 1.

Now suppose that . Then by definition . Note that:

and if for some then where qr < m which contradicts the order of a being m. Thus:

and . Therefore Thus the element has order lcm(m,n)

**3) Let G be an abelian group of order pq with gcd(p,q) = 1. Assume there exists such that , and show that G is cyclic.
Proof- **From number 2 we know that G must contain an element g, with order pq. Because pq is also the order of the group it must be true that g generates G. Thus G is cyclic.

**4) If is a group homomorphism, , and has finite order in H, then is infinite or divides .
Proof- **Let . Suppose that the order of a is finite, . Then we have:

. Then n must divide the order of f(a) so that n divides m.

**The case when k is 1 is trivial so assume . Let be cyclic. We can write . This by definition means that . Because we get that . Thus is a subgroup of G with order k. Now show it is unique. Suppose H is a subgroup of G also with order k. Since k is not 1, H is nontrivial. Being a subgroup of a cyclic group, where m is the least positive integer such that From the division algorithm we have for some with This gives:**

6) If G is a cyclic group of order n and then G has exactly one subgroup of order k.

Proof-

6) If G is a cyclic group of order n and then G has exactly one subgroup of order k.

Proof-

. However m is the least positive integer such that , and we have now that also, with which means that r must be 0. Thus so that which gives that gcd(m,n) = m. Then since we have that for the same reason as before. Then:

, which gives that so that , so the subgroup is unique.

**8) A group that has only a finite number of subgroups must be finite.
Proof-** Let a group G have a finite number of subgroups but suppose G is infinite. Let not the identity. Consider . If it is infinite then it must be cyclic and isomorphic to the integers under addition, which has infinitely many subgroups. Thus cannot be infinite and in fact every element of G must have finite order. Next, let and again form . Since G is infinite, but is finite we have that must be infinite as well. Choose another . Form . Note that . Again we can pick and form . Continuing in this manner we obtain an infinite sequence of subgroups of G which are not all equal. A contradiction to the finite number of subgroups. Thus G is not finite.

**9) If G is an abelian group, then the set T of all elements with finite order is a subgroup of G.
Proof-** Since the order of the identity element is 1, so that T is nonempty. Let have orders m, n respectively. Let lcm(m,n) = q. Then for some . Note that . This gives that the order of must divide q, which means the order of is finite so that so by the subgroup test, T is a subgroup of G.

**10) An infinite group is cyclic iff it is isomorphic to each of its proper subgroups.
Proof-** Assume that G is cyclic so that for some a in G. Let H be a proper subgroup of G, then we can say that , where m is the least positive integer such that . Define by . If then so that is well defined.

Let . Then . Since G is cyclic write then since G is infinite, a cannot have finite order so we get:

since m was positive. Thus so that is injective.

Now for any we can write for some integer k. Note that and this element also maps to h, so is onto.

Lastly, let . Then we can write and so that:

so we have an isomorphism.

Assume that for every proper subgroup H of G. Since G is infinite, let a be an element which is not the identity. Then , so let be an isomorphism. Want to show that by subset inclusion to show G is cyclic. Let . Since f is onto, there is some such that Thus . Now let then for some integer k. This shows that h is in G. Thus the two sets are equal and so G is cyclic.