Qual 1994 # 3(Hungerford 1.5.1): (a) If N is a subgroup of index 2 in a group G, then N is normal in G.
Proof- Let be the set of distinct left cosets of N in G. Let be the set of distinct right cosets of N in G. N itself is always both a right and left coset of a subgroup N of G. Also, cosets partition the group so we can write:
Thus, every left coset of N in G is also a right coset of N in G so that N is normal in G.
b) Give an example of a subgroup such that and .
Example- Let and . Since and by Lagrange. But we have that . So that H is not normal in G.
201A Midterm (2010) #5: What is the natural homomorphism of groups ? What is the kernal of this map? Use the 1st homomorpism theorem to conclude that there is a homomorphism map from . Prove that the kernel of this homomorphism is isomorphic to . Is it true that ?
Proof- Let by is a natural homomorphism with . Since is contained in the Kernel of the map we have by the first homomorphism theorem that there is homomorphism map such that where is the natural projection . Now let . Then we have . This says that so that . Now, these elements are all contained in so that . Lastly, because is cyclic while is not since 6 and 3 are not relatively prime.
Hungerford 1.8 #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.
Qual 2013 #1 (Hungerford 1.8 #12): A) Suppose that H is a normal subgroup of G and that . Prove that K is isomorphic to a normal subgroup of G.
Proof- Let be the isomorphism given. . Let be the canonical projection so that and recall that the kernel is always a normal subgroup. To finish the proof we need a side lemma. If is a surjective group hom, with N normal in A, then is normal in B. Proof- so that is normal in B.
Now, since is surjective and since we get that by the lemma. But .
B) Suppose G, H, K are as in part A). Prove that any homomorphism of groups gives rise to a canonical homomorphism from . Give an example to show that an isomorphism need not imply is an isomorphism.
Proof- Let be a group homomorphism. Let and be the unique element that is associated to g under the isomorphism, . Define by . Then clearly is a group homomorphism. Next, Let . Then we have . The only isomorphism possible is , which is the identity. So then . So that is not an isomorphism.
2011 Qual #2 (Hungerford 1.8.9): Suppose that G is a group and H, K are subgroups of G such that G is the internal direct product of H and K. Prove that . Deduce that there does not exist a subgroup H of , such that .
Proof- First, let and consider the projection map . Note that this map is surjective since K is a subgroup of G. If then which says that . The implications can be reversed so that . Then we have by the first isomorphism theorem that .
Suppose now that there is a subgroup H of such that Then we must have that but we have that so that . Now consider the element . This gives that the element is isomorphic to some element in of order 10, call it . There is no single cycle of order 10 so that , where . But because has order 5, it cannot be disjoint from which is a contradiction.
Qual 2011 #4, Qual 2001 #4, Qual 2003 #4 (Hungerford 2.1.4): Prove that free abelian group is a free group iff it’s cyclic.
Proof- Suppose that G is a free abelian group which is also a free group on the set X. Suppose that . Then there exist distinct . This gives that which is a contradiction to the abelian assumption. Therefore G is cyclic.
Suppose that G is a free abelian group which is cyclic. Then for some . Then each element of G has the form for some integer n. Define by is a group isomorphism so that G is free since is free.
Counterexample: is cyclic but not free because it has finite order, and all nontrivial free groups consist of elements of infinite order.
Hungerford 2.1.10b: Show that is not free abelian.
Proof- Suppose that was free so that it would have a nonempty basis, X. If then we have that is cyclic. Suppose it were generated by one element x, then which is a contradiction. So instead suppose that . Then let . We have that so that the arbitrary elements are not linearly independent. Thus and so the rationals do not have a basis and are not a free group.