Establish a finite presentation for using two generators.

We know that is generated by an element of order 2 and an element of order 3. Let and . Evidently, these generators satisfy the relations . To show that these relations provide a presentation for , it suffices to show that the group has order at most 24. To that end, we will compute the reduced words in this group.

- There is one word of length 0: 1.
- There are two words of length 1: and . Both are reduced by definition.
- There are at most four words of length 2: , , , and . Since , it is not reduced. Thus 3 words of length 2 might be reduced.
- There are at most six words of length 3: , , , , , and . Since and , these words are not reduced. The remaining 4 words might be reduced.
- There are at most eight words of length 4: , , , , , , , and . Since , , , , , and , these words are not reduced. The remaining two words might be reduced.
- There are at most four words of length 5: , , , and . None of these are reduced.

Thus any group presented by has cardinality at most 12, and we have the presentation .

Show that a group of order 12 with no subgroup of order 6 is isomorphic to .

Note that , so that Sylow’s Theorem forces and . Suppose . If , then by the recognition theorem for direct products, , where and are Sylow 2- and 3-subgroups of , respectively. Let have order 2 and have order 3 by Cauchy; then has order 6, a contradiction. Now suppose . Since mod 4, there exist such that is nontrivial and is divisible by and another prime; thus we have normal in . Recall that every normal group of order 2 in a finite group is central; thus we have an element of order 2 in the center of . Let have order 3 by Cauchy; then has order 6, a contradiction.

Thus . Let be a Sylow 3-subgroup and let . Now has index 4 and is self-normalizing by a previous result; in particular, is not normal in . Now acts on by left multiplication, yielding a permutation representation whose kernel is contained in . Note that , so that either or . If , then for all and , we have , so that . Thus, for all , ; since is finite, this implies that is normal, a contradiction. Thus and in fact . Since contains elements of order 3, it contains all elements of order 3 in . The subgroup generated by the three cycles in is , so that ; since these subgroups are finite and have the same cardinality, in fact .

Compute , , , , and .

We begin with some lemmas.

Lemma 1: If is a subgroup of order 12, then . Proof: By Cauchy, contains an element of order 2 and an element of order 3. If some element of order 2 is a product of two 2-cycles, then , and by counting elements we have . Suppose no element in is a product of two 2-cycles; then also no element of is a 4-cycle, since the square of a 4-cycle is a product of two 2-cycles. Suppose now that contains two 3-cycles which are not inverses of one another; from the subgroup lattice of , then, ; since also contains a 2-cycle, we have , a contradiction. Suppose then that contains exactly two 3-cycles, and . then must contain nine 2-cycles, a contradiction since contains only six 2-cycles. Thus we see that .

Lemma 2: If has order 6, then is maximal. Proof: If is not maximal, then is contained in some subgroup of order 12 by Lagrange. By Lemma 1, , but contains no subgroups of order 6, a contradiction. Thus is maximal.

Now to the main results.

- and are distinct maximal subgroups, since each has prime index. Thus , so that .
- We know from the subgroup lattice of that and are distinct maximal subgroups of prime order. Thus .
- Note that and are distinct subgroups of order 6 in , which are maximal by Lemma 2. Evidently . Now is also maximal, and is an odd permutation. Thus , and we have .
- is normal and proper. Since is simple, .
- We found in a lemma to the previous exercise that .

Find the upper and lower central series for and where .

- First we consider .
Recall that the higher centers of are characteristic, hence normal. Since is simple for , and is not abelian, we have and . Thus the upper central series of , , is .

Now and . Since , contains all 3-cycles, and thus . Hence , and the lower central series of is .

- Now we consider .
We have seen previously that for . Thus Z_1(S_n) = Z_0(S_n)$, and the upper central series of is .

Now , and we saw in §5.4 #5 that . Now , so that the lower central series of is .

Find the commutator subgroups of and .

We start with . Note that is normal in and is abelian, so that by Proposition 7, . Note that ; that is, . Now is a characteristic, hence normal, subgroup, and thus is a union of conjugacy classes. Since the conjugacy classes in are precisely the cycle shapes, all 3-cycles are in .

Since , all products of two 2-cycles are also in . With the identity, this exhausts the elements of ; hence .

Now we consider . We know that is characteristic, hence normal, and that (by Lagrange) is abelian. By Proposition 7, . Note moreover that since , every order 2 subgroup is not normal in , hence not characteristic, and thus cannot be the commutator subgroup. Thus .

Let (as in this previous exercise). Let be the subgroup of consisting of permutations which move only a finite number of elements of (described in this previous exercise) and let be the set of all elements that act as even permutations on the (finite) set . Prove that is an infinite simple group. [Hint: show that every pair of elements of lie in a finite simple subgroup of .]

We begin with a lemma.

Lemma: If , then there is a finite simple group with . Proof: If , then and are finite. Let be five elements in not moved by either or . Now is finite, and by Theorem 24 in the text, is simple. Clearly .

Now suppose is a normal subgroup. Suppose ; let be a nonidentity. Suppose that there exists . Now there exists a finite simple group with . Moreover, is a nontrivial normal subgroup (containing in particular ) so that, since is simple, . Thus . But then , a contradiction. Thus no such element exists, and we have .

Thus if is a normal subgroup, either or . Hence is simple.

Finally, it is clear that is infinite since, for example, is a subgroup for every integer .