Let and be elements of the group ring . Compute the following: , , , and .

Evidently,

unnecessary lemmas. very sloppy. handwriting needs improvement.

Let and be elements of the group ring . Compute the following: , , , and .

Evidently,

Prove that if then is nonabelian.

First we show that the free group of rank 2 is nonabelian. Suppose to the contrary that is abelian where ; say . Define by , . By the universal property of free groups, there exists a unique group homomorphism . This homomorphism is clearly surjective, so that by the First Isomorphism Theorem, . However, every quotient of an abelian group is abelian, but if , is not abelian. Thus is nonabelian.

Clearly every free group of rank at least 2 contains a subgroup which is free of rank 2. (For instance, use the universal property.) Thus no free group of rank at least 2 can be abelian.

Prove that for all and , .

We first consider the case . Now is abelian, so that and . Thus for all .

Now let . Recall that is nilpotent of nilpotency class , so that and . We now consider the case by induction on ; in particular, we compute , by two separate induction arguments.

First we show that by induction on .

For the base case, note that since , by §2.2 #7, .

For the inductive step, suppose where . By definition, . Using the isomorphism defined in this previous exercise, and the calculation of the center in this previous exercise, we see that . Thus .

Now we show that for all by induction on .

For the base case, let . Now . Let be arbitrary; there are 4 cases to consider.

- If and , then .
- If and , then .
- If and , then .
- If and , then .

Thus .

For the inductive step, suppose that for some the conclusion holds, and consider . We have . Let and ; there are two cases to consider.

- If and , then .
- If and , then .

Thus .

In particular, for all and , we have .

Prove that is nilpotent if and only if is a power of 2.

Suppose is nilpotent. Let be an odd prime dividing . Then is an element of order in ; in particular, . Now and are relatively prime, so that, by §6.1 #9, ; a contradiction. Thus no odd primes divide , and we have .

We proceed by induction on , where .

For the base case, , note that is abelian, hence nilpotent.

For the inductive step, suppose is nilpotent. Consider ; we have , and by §3.1 #34, is nilpotent. By §6.1 #6, is nilpotent.

Moreover, we show that for , is nilpotent of nilpotence class , by induction on .

For the base case, , is abelian, and thus of nilpotence class 1.

Suppose now that is nilpotent of nilpotence class . Since , by a previous lemma, is nilpotent of nilpotence class .

Let and check that the construction of the two nonabelian groups of order (in Example 7) is valid in this case. Prove that *both* of the resulting groups are isomorphic to .

For the first construction, let and . Proposition 4.17(a) in the text does not apply for as it does for odd ; however, the comments after Proposition 4.17 demonstrate that for this special case, we still have . Since (by Cauchy) has an element of order 2, we can define by . If , say and . Then the semidirect product has order and is nonabelian since while .

For the second construction, let and . Now since is abelian, as in Example 1 we can define by . Then if , we have a nonabelian group of order 8.

Now it remains to show that is isomorphic to . To that end, it suffices to show that there exist elements such that , , and .

Let and . A straightforward calculation verifies that , , and .

Prove that is not isomorphic to .

Suppose to the contrary that . Then there exist subgroups such that (1) and are normal in , (2) , (3) , (4) , and (5) .

Recall that has a unique cyclic subgroup of order ; namely . It must be the case that .

Now the subgroups of order 2 in are of the form for and . Recall that if , then . In particular, because is noncentral in , is not normal in . Thus cannot be of this form. But then , so that , and cannot be of this form either. Thus we have a contradiction, and no such subgroups and exist.

Thus is not isomorphic to .

Prove that , , , , , and are nonisomorphic non-abelian groups of order 16. ( is described here and and are described here and here, respectively.)

In §5.1 #6, we saw that all subgroups of are normal. Note that is not normal since , so that ; that is not normal since , so that ; and that is not normal since , so that .

The maximal subgroups of are all abelian, while the maximal subgroups of and are not. Thus and . Similarly, is a nonabelian maximal subgroup, so that , and is a nonabelian maximal subgroup, so that .

Note that , , , and are distinct maximal subgroups of , so that this group has at least 4 maximal subgroups. In particular, and .

We previously computed the subgroup lattices of and ; has 9 subgroups of order 2, while has 5. Thus .

The (relatively) hairy group to work with is .

Recall from §5.1 #13 that has the presentation . We can see that every element of this group can be written in the form , where , , and . This gives at most 16 "representations" of elements in , and we know by other means that this group has exactly 16 elements. Thus these representations are distinct.

In the following table, we compute the orders of the elements in .

Reasoning | ||

1 | ||

4 | ||

2 | ||

4 | ||

2 | ||

4 | ||

2 | ||

4 | ||

4 | ||

4 | ||

4 | ||

4 | ||

2 | ||

4 | ||

2 | ||

4 |

Note that contains no elements of order greater than 4. Thus, since has order 8, ; since has order 8, ; and since has order 8, .

Note also that contains 5 elements of order 2. Now , , , , , and are 6 distinct elements of order 2 in , so that .

Finally, note that has order 2 only if . Thus the only elements of order 2 in are , , and . In particular, .

Thus these groups are nonisomorphic nonabelian groups of order 16.

Give presentations for the groups and constructed in a previous example.

Note that is generated by , , and . Hence is generated by the images of these elements under the natural projection; moreover, the relations satisfied by these also hold in the quotient. We have an additional relation: note that . Thus .

Similarly, is generated by the natural images of , , and , and we have an additional relation because . Thus

Let and be groups. Suppose and are subgroups and that there exists an isomorphism . Define by . Note that is normal in since , using a previous theorem. We denote by the quotient . (In particular, depends on .) Think of as the direct product “collapsed” by identifying each element with its -image in .

- Prove that the images of and in are isomorphic to and , respectively, and that these images intersect in a central subgroup isomorphic to . Find .
- Let . Let and as usual. Let be the central product of and which identifies and . (I.e. , , and .) Let be the central product of and which identifies and . (I.e. , , and .) Prove that .

- Let denote the canonical projection, and identify with and with in .
Let , and suppose . Then , so that . Thus , and hence is injective. Similarly, is injective.

Note that the restriction is also injective. Suppose ; then for some and we have . Thus ; by definition, then, and . Note that for all we have , so that is in the center of . Moreover, . Conversely, if , we have , so that . Then , and by the First Isomorphism Theorem, . In addition, is a central subgroup of .

Finally, by Lagrange write and . Note that . Now . We may also write this equation in the form .

- Define as follows: , , and . Because generates and these images satisfy the relations , , , and , extends to a homomorphism .
If we let denote the natural projection, is a mapping . We claim that is contained in the kernel of this mapping; it suffices to show this for the (unique) nonidentity element of , namely . In fact, = . Thus . By the remarks on page 100 in the text, there exists a unique group homomorphism such that , and in particular, .

We now show that . Suppose . Then ; note that for some integers , and that . Then .

If , then mod 2 and mod 4. Now if mod 4, then mod 4 and . If mod 4, then mod 2 and .

If , then mod 2 and mod 4. If mod 4, then mod 4 and . If mod 4, then mod 4 and .

Thus , hence is injective.

Now by part (a), we see that both and have order 16. Thus is an isomorphism.

Let where is odd. Prove that the number of Sylow 2-subgroups of is .

We begin with several lemmas.

Lemma 1: If , then . Proof: The homomorphism defined on the generators of by and is injective.

Lemma 2: Let be a finite group and a prime. If is a subgroup such that does not divide , then every Sylow -subgroup of is a Sylow -subgroup of . Proof: We have and , with not dividing and and ; since does not divide the index of in , . Hence any Sylow subgroup of has order , and thus is a Sylow -subgroup in .

Now to the main result.

Suppose is odd. Then and every Sylow 2-subgroup has order 2. In this case, the Sylow 2-subgroups correspond to the elements of order 2 in . There are exactly of these: elements of the form for . Thus .

Suppose now that is even; then . We proceed by induction on the width of . (Recall that the *width* of an integer is the number of prime factors which divide it, including multiplicity.) Let be a Sylow 2-subgroup of .

For the base case, if the width of is 0, we have . Then . In this case, is the unique Sylow 2-subgroup, so that .

For the inductive step, suppose that for any integer of width at most , the number of Sylow 2-subgroups of is . Let be an integer of width . Then for some prime and some width integer ; without loss of generality, we may assume that is the smallest prime divisor of , so that . By the induction hypothesis, has distinct Sylow 2-subgroups. By Lemma 1, . Note also that 2 does not divide , so that by Lemma 2, has at least distinct Sylow 2-subgroups.

Now by Sylow’s Theorem, divides . Thus either or . Note that there exist elements of order 2 in which are not in (the image of) ; for instance, , since otherwise we have in the image of in , a contradiction. By Sylow’s Theorem, every 2-subgroup (For instance ) is contained in some Sylow 2-subgroup. Thus there must exist at least one other Sylow 2-subgroup in ; hence .

By induction, the result is proved for all .