2003 # 1: Prove that a group cannot be the union of two proper subgroups.
Proof- Let H and K be two proper subgroups of a group G such that H and K are not the same, otherwise their union would have to be G but they are the same group so the union would just be one group, the entire group and thus not a proper subgroup. Suppose their union is G. Let . Then for any we have or . In the first case we get since inverses are in subgroups. This contradicts k being in not H. In the second case we get that which is a similar contradiction. Thus G is not the union of H and K.
2011 #1A (Hungerford I.2 #18): Suppose G is a group and that are two distinct subgroups of G. State and prove a sufficient condition for to be a group.
Proof- In order for to be a group it must be closed. If we have one element from each distinct subgroup, we need the sum to be in the union. Let and . If then we get . Similarly if we get that . Thus the only way this can happen is if one is a subgroup of the other.
2011 1B: Give an example of a group G and 2 subgroups such that is strictly contained in the subgroup .
Example- Let and . Then we have while
Qual 2001 #1 (Hungerford I.8 #1): Show the symmetric group is not isomorphic to a direct product of two of its proper subgroups.
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.
Qual 2005 #3A: Prove that is not isomorphic to .
Proof- must have an element of order 8. Recall that if then . If and then because the order of is 2 and the order of is 8.
Hungerford #12 Sec I.6: 12) The center of the group is if n is odd and if n is even.
Proof- First we need to know all the normal subgroups in . If n is odd, the nontrivial subgroups are the group of rotations, and all the subgroups of the group of rotations. If n is even we have two more normal subgroups, each isomorphic to .
Let H be a normal subgroup. We have two cases, either H consists only of powers of the element a, or there is some . Suppose that for some k dividing n. Then:
because of the relation . And finally,
. Thus the subgroup is normal.
Now instead suppose there exists some . Then we have since H is normal. This gives that . This tells us that is a generator of H.
Next, . If is even let and if is odd let . Therefore either b is a generator or ab is a generator. So that or .
Now that we know all the normal subgroups we can find the center, because it must be a normal subgroup. Recall the center is the set of all elements which can commute with any other element.
First we check . We have , but n is larger than 2 so this is a contradiction. Thus this is not the center.
Next we check . We have that:
. But for n = 4, again a contradiction so this is not the center.
Lastly, assume that is the center. Then we have . Thus the center can only be which the same as but only when n is even. If n is odd then the center is just the identity.
2001 Qual #3 (Hungerford 2.4.9): Prove that if G is a group and is cyclic then G is abelian.
Proof- Suppose that is cyclic. It is a set of cosets so to be cyclic means to be generated by a coset, i.e, for some element g in G. Then for two elements we have that:
This gives that and . Then finally so that G is abelian.
1999 Qual #3 and 1995 #5 (Hungerford 2.5.13) – Prove that a group of order is abelian.
Proof- The center of the group is known to be a subgroup and therefore must divide so that . We can eliminate some choices though. Recall the class equation:
where are distinct conjugacy classes such that because the order of any conjugacy class must divide the order of G and so must be a power of p, up to . Thus, p must also divide so that is either p or . Now using Lagrange we can say:
is cyclic G abelian.
Alternate Proof- Since G is a p-group, the center . By Lagrange we then have either which gives the center as the entire group, thus making G abelian, or we have . If then we have that is cyclic being of prime order, which tells us that G is abelian by 2.4.9.
Hungerford 1.3 #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.
Hungerford 1.4 #11: Let G be a group of order 2n, then G contains an element of order 2. If n is odd and G is abelian, there is only one element of order 2.
Proof- Let the order of G be 2n. Consider the set . If there does not exist an element such that then the order of this set is , which is a contradiction so there must exist at least one element of order 2.
If G is abelian and n is odd, suppose we have two elements such that . G is abelian so that is a subgroup of G. But because n is odd, we know that 4 does not divide 2n, so this subgroup cannot exist by Lagrange. Thus there can be only one element of order 2.