Let and be elements of the group ring . Compute the following: , , , and .
Let and be elements of the group ring . Compute the following: , , , and .
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, .
Now we show that for all by induction on .
For the base case, let . Now . Let be arbitrary; there are 4 cases to consider.
For the inductive step, suppose that for some the conclusion holds, and consider . We have . Let and ; there are two cases to consider.
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 .
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 .
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 .
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 .
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 .
On these pages you will find a slowly growing (and poorly organized) list of proofs and examples in abstract algebra.
No doubt these pages are riddled with typos and errors in logic, and in many cases alternate strategies abound. When you find an error, or if anything is unclear, let me know and I will fix it.