**1) (Qual 2001 #1) is not the direct product of any family of its proper subgroups. The same is true of and
Proof- **Note that the order of is 6, so any subgroup must be of order 2 or 3, which are prime numbers. Any group of prime order must be cyclic, and cyclic groups are abelian. The direct product of abelian groups is abelian, but is not abelian so it cannot be ismorphic to an abelian group, and hence is not isomorphic to a direct product of proper subgroups.

Now, let p be prime and . We have is cyclic of order . Thus any proper subgroup of must be of the form with . Suppose that:

. Consider the subgroups of this direct product:

and

Both have order p. This is because in and in . Thus we have distinct subgroups of which is impossible if is cyclic because subgroups of cyclic groups are unique for each order. So we have a contradiction, thus

**4). Give an example to show that the weak direct product is not a coproduct in the category of all groups.
Proof-** Consider the weak direct product with the canonical injections and . Next define the group homomorphisms, by and by . Now then we have a family of groups and maps into the weak direct product along with maps into a different group. In order for the weak direct product to be a coproduct we would now need a map from the weak direct product into the different group such that the diagram commutes. So suppose there exists such that . Then we have that . But these elements generate all of so that . But which is abelian, while is not abelian. Therefore such a map does not exist.

**5) Let G, H be finite cyclic groups. Then is cyclic iff
Proof-** First we need to prove another fact. The order of an element is Note that . Suppose there is some s. t. . Then and , so the order of g divides m and the order of h divides m. This gives m as a common multiple of the orders of g and of h, but l is the least common multiple so that which is a contradiction. Thus . Now we can do the proof.

Assume that is cyclic so that for some . Let . Then we have:

. This tells us that divides but which means that d = 1.

Assume now that . We know that and for some g, h. By the fact above we have

Therefore (g,h) generates the entire group so that is cyclic.

**6) Every finitely generated abelian group in which every non identity element has prime order p is isomorphic to (n summands) for some natural number n.
Proof-** Suppose G is finitely generated abelian and let be a minimal generating set of G. Since G is abelian all subgroups are normal so that . Also . We need only show that . Let and suppose . Then it must be the case that , but every element has order p by assumption so that which gives that . This tells us that anything in is also in which gives that is redundant so G can be generated without the element . This contradicts the minimality of the generating set. Thus . Now from theorem 8.6 we can say But each is cyclic of order p and hence isomorphic to . Cyclic groups are abelian so we can rewrite this as (n summands)

**9) If a group G is the (internal) direct product of its subgroups H, K then and .
Proof-** Let and consider the canonical projection maps:

and by and . Then note that iff so that . Clearly also is surjective. Then by the first isomorphism theorem we have:

The other isomorphism is shown in the exact same manner.