Consider the following elements of the group ring : and . Compute , , , , and .

Evidently,

unnecessary lemmas. very sloppy. handwriting needs improvement.

Consider the following elements of the group ring : and . Compute , , , , and .

Evidently,

Consider the following elements of the integral group ring : and . Compute the following elements: , , , , and .

Evidently,

This exercise exhibits an automorphism of that is not inner (hence it shows that . Let , , , , and . Show that satisfy the following relations:

- for all ,
- for all with , and
- for all .

Deduce that and that the map , , , , extends to an automorphism of (which is clearly not inner since it does not preserve cycle shape.

Showing that satisfy the given relations is a straightforward computation. Moreover, they generate since and .

Prove that .

We already know that by counting the possible images of, say, and . By a previous theorem, it suffices to show that no automorphism of has order 6.

Let be an automorphism of . Now has 6 elements of order 4, one element of order 2, and one identity; must fix the identity and the element of order 2, so that we can consider as an element of , where . Suppose now that some automorphism has order 6. Now , where , and , where . Clearly and . If , then and has order 3; thus , and we have . Now . Note that . If , then . Similarly, if , then . Both cases yield a contradiction, so that no automorphism of order 6 exists.

In addition, we record this handy lemma.

Lemma: Let be a group and let be characteristic. Then acts on by . Proof: First we need to show well-definedness: if , we have . Then since is characteristic in , so that . Moreover, and .

Establish a finite presentation for with 2 generators.

Recall that is generated by and . Evidently, these elements satisfy the relations . We will now show that any group presented by has at most 24 elements.

- There is one reduced word of length 0: 1.
- There are two reduced words of length 1: and .
- There are at most four reduced words of length 2: , , , and . We see that at most three of these are reduced.
- There are at most six reduced words of length 3: , , , , , . We see that at most five of these are reduced.
- There are at most ten reduced words of length 4: , , , , , , , , , and . At most five of these are reduced.
- There are at most ten reduced words of length 5: , , , , , , , , , and . We see that at most five of these are reduced.
- There are at most ten reduced words of length 6: , , , , , , , , , and . We see that at most three of these are reduced.
- There are at most six reduced words of length 7: , , , , , and . We see that none of these is reduced.

Thus any group with this presentation has at most 24 elements, and in fact we have .

Show that a group of order 24 with no element 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. By Cauchy, there exist elements and of order 2 and 3, so that has order 6, a contradiction. Suppose now that . Since mod 4, there exist such that is nontrivial and its normalizer has order , by a previous theorem. Thus is normal. Note that is either 2 or 4.

Suppose . Then at most nonidentity elements of are contained in Sylow 2-subgroups, and 3 elements are contained in Sylow 3-subgroups. This leaves 4 elements not of prime power order, one of which must have order 6- a contradiction.

Suppose now that . By the N/C Theorem, , so that is central in . By Cauchy, there exist elements and of order 2 and 3, so that has order 6, a contradiction.

Thus we may assume . Let be a Sylow 3-subgroup and let . The action of on yields a permutation representation whose kernel is contained in . Recall that normalizers of Sylow subgroups are self normalizing, so that is not normal in . Moreover, we have . We know from the classification of groups of order 6 that is isomorphic to either or ; however, in the first case we have an element of order 6, a contradiction. Thus . We know also that the normal subgroups of have order 1, 3, or 6. If , then is normal in , a contradiction. If , then by the N/C theorem we have . In particular, contains an element of order 2, so that contains an element of order 6, a contradiction.

Thus , and in fact . Since is finite, .

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 .

Prove that there is no group whose commutator subgroup is .

Suppose is a group such that . We computed in §5.4 #4 that and . Note that and are cyclic; by this previous exercise, we have , a contradiction.

Thus no such group exists.

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 upper and lower central series of and .

Recall that .

- First we consider . Since and , . Thus the upper central series of is .
Now . In §5.4 #4, we found that . Using the computations above, we can see that contains all products of two 2-cycles and is contained in ; thus . Thus the lower central series of is .

- First we consider . We have seen previously that ; thus the upper central series of is .
Now . We computed that . Now is contained in ; moreover, since , contains all 3-cycles, so that . Thus the lower central series of is .