1) If and N, are both p-groups, then G is a p-group.
Proof- Since N and are p-groups let for some i and j. Then by Lagrange we have . Thus G is a p-group.
2) If G is a finite p-group, and H is nontrivial, then
Proof- Note that the center of G is . Let G be a finite p-group so that . Then since H is a subgroup we have where . We can let G act on H by conjugation and let Then we have by theorem that . We can also write . Now if , and . Therefore . We also have that p divides the order of so that . Thus must have more than one element and is therefore nontrivial.
3) Let . For each G has a normal subgroup of order .
Proof- First note that any subgroup of the center of a group is automatically normal in the group. Now we prove that G has a normal subgroup of order . Since G is a p-group it has a nontrivial center so that the center is in fact a p-group as well. Then being of order a prime power, we can say by Cauchy’s Theorem that there exists such that . Then is a subgroup of the center of G, so is automatically normal in G, and has order p.
Now we show for each G has a normal subgroup of order by induction on k. If we have the trivial subgroup which is automatically normal in G. Now assume that G has normal subgroups for each where and call them . Consider the quotient group . We have from Lagrange that so that is a p-group. In the previous step we’ve shown that a p-group must contain a normal subgroup of order p, call it . Let be the canonical projection map. Then let . The preimage of a normal subgroup under a homomorphism is also normal, so that H is normal in G. Note that because is the identity in so that so that . Then using Lagrange we have but . Therefore
4) If G is an infinite p-group, then either G has a subgroup of order for each or there exists such that every finite subgroup of G has order .
Proof- Suppose that no such m exists. Then for each positive integer n, there exists another positive integer such that G has a subgroup H of order . But then by Sylow we know that H has a subgroup of order .
5) If P is a normal Sylow p-subgroup of a finite group G and is an endomorphism, then .
Proof- Note that . Let . Then there exists a such that . Since P is a p-group, for some i.Then we have . This tells us that divides so that for some . Since y was arbitrary, is a p-subgroup of G. Therefore by Sylow’s Second Theorem, there exists such that .
6) If H is a normal subgroup of order of a finite group G, then H is contained in every Sylow p-subgroup of G.
Proof- H is a p-subgroup of a finite group G. Let P be any Sylow p-subgroup of G. From Sylow’s Second Theorem we know there exists such that . We also have that is nonempty. Let . Then we can write for some . Since for all , we can write for some . Then rewrite . Therefore . Lastly, let . Then we can write for some . Then we have so that in fact is a subgroup of P. Now, since H is normal in G we can say that , so that H is contained in any arbitrary Sylow p-subgroup P.
7) Find the Sylow 2-subgroups and Sylow 3-subgroups of
Proof- . Thus contains at least one Sylow 2 subgroup of order 2 and one Sylow 3 subgroup of order 3 by the first Sylow Theorem. In fact, from Sylow’s Third Theorem we know that there is either 1 Sylow 2 subgroup or 3 Sylow 2 subgroups. We also know that there can only be one Sylow 3 subgroup.
In the same manner we can say that has either 1 or 3 Sylow 2 subgroups of order 8 and either 1 or 4 Sylow 3-subgroups of order 3. This is because .
Lastly, has either 1,3,4, or 15 Sylow 2 subgroups of order 8 and either 1,4, or 10 Sylow 3-subgroups of order 3. This is because .
8) If every Sylow p-subgroup of a finite group G is normal for every prime p, then G is the direct product of its Sylow subgroups.
Proof- First we write . where we have distinct primes. Then by Sylows first theorem, for each prime , G contains a Sylow subgroup of order . From the second Sylow theorem we have that any two Sylow subgroups are conjugate, but we are given that they are all normal so in fact there is a unique Sylow subgroup for each . Denote each Sylow subgroup by . Then we have that is a family of normal subgroups of G so that for any .
Next we need to show that for each that , for all . This is done by induction on n. Lastly we have to show that , but we already have that so that we need only show that . Then we will have that .
9) If with primes, then G contains a unique normal subgroup of index q.
Proof- From Sylow’s First Theorem, G has a Sylow p-subgroup of order . Let m be the number of Sylow p subgroups of G. By Sylows Third Theorem we have that and . Then since we have to have that either or for some . If then for some . Then which is a contradiction. If instead we have that then , which is another contradiction. Therefore so that there is the only 1 Sylow p subgroup of G, call it H, which must be normal in G. Then by Lagrange, . Thus H has index q.
10) Every group of order 12, 28, 56, and 200 must contain a normal Sylow subgroup and hence is not simple.
Proof- i) . By 3rd Sylow Theorem we know that there is either 1 Sylow 3 subgroup, or 4 Sylow 3 subgroups. If there is only one, then know it must be normal. Thus, suppose there are 4, each with order 3. Each of these subgroups has 3 elements, 1 being the identity. Thus the other 2 elements are of order 3. Together then we have elements of order 3 in G. Then G can only have one Sylow 2-subgroup which must be of order 4 because this gives us 12 total elements. Again, being the unique Sylow 2-subgroup gives that it is also normal.
ii) . The Third Sylow Theorem tells us that the number of Sylow 7 subgroups is either 1 or 4, and that it must also be 1 mod 7, therefore it cannot be 4. So there is only one Sylow 7-subgroup and thus it is normal.
iii) . By the Third Sylow Theorem we know that the number of Sylow 7-subgroups is either 1 or 8. If it is 1 then we are done, so suppose it is 8. Call them . Then similar to part (i) we have that there are elements of order 7 in G. Then G can only have one Sylow 2-subgroup as it must be of order . Therefore it is normal, being unique.
iv) . By the Third Sylow Theorem we have that the number of Sylow 5-subgroups is a factor of or equal to 1, and must also be congruent to 1 modulo 5. This can only happen if there is exactly one Sylow 5-subgroup. Thus it is normal.
11) How many elements of order 7 are there in a simple group of order 168?
Proof- Let . By the Third Sylow Theorem, the number of Sylow 7-subgroups must divide and also must be congruent to 1 modulo 7. Thus the only possibilities are that the number of Sylow 7-subgroups is 1 or 8, but since G is simple we cannot have only one Sylow 7-subgroup otherwise it would be normal. Therefore there are 8 of them, each with 7 elements, with each non identity element having order 7 since a group with prime order must be cyclic. Thus there are elements of order 7.
13) Every group G of order is abelian.
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.