Download Algebra Homework 6 Solutions: Understanding p-Sylow Subgroups and Their Properties and more Assignments Abstract Algebra in PDF only on Docsity!
Algebra (647), Homework 6, solutions.
1. II.4: 8.
Actually since Z/ 6 is abelian, it has only the trivial inner automorphism. So 1 7 → − 1 is a non-inner automorphism.
- II.4: 9. Let g be a generator for G/C. Take g to be a lift of g to G. Suppose G/C is order n. Then gn^ ∈ C(G).
G = C ∪ gC ∪ g^2 C ∪ · · · ∪ gn−^1 C.
If we have h, k ∈ G, say h = gic and k = gkc′. Since c, c′^ ∈ C they commute with everything, and gi^ always commutes with gk, so
(gic)(gkc′) = gigkcc′^ = gkgic′c = gkc′gic.
3. II.4: 11.
If G is not abelian, conjugation by an element that isn't central is a non- identity automorphism. If G is abelian, x 7 → (−x) is an automorphism, which is the identity if and only if every element has order 2.
- II.5: 1. Pick g ∈ G. Since G/N is a p-group, gp i ∈ N for some i. Since N is a p-group, (gp
i )p
r = e for some r. So gp
ir = e.
- II.5: 5. Note that the image of a p-group is a p-group. So f (P ) is a p-group. By the Sylow theorems (I and II) it is thus a subgroup of some p-Sylow group, and since P is normal, by Sylow III it is the only p-Sylow group.
- II.5: 6. H < P for some p-Sylow group. If Q is an arbitrary p-Sylow, then Q = gP g−^1 for some g (Sylow III) so
H = gHg−^1 < gP g−^1 = Q.
- II.5: 7. S 3 : The 2 -Sylow group has order 2 , so is Z/(2). The 3 -Sylow group has order 3 , so is Z/ 3. There is only one 3 -Sylow, and there are 3 di erent 2 -Sylows generated by the various transpositions. S 4 : The 2 -Sylow group has order 8. It must contain the 4 -element subgroup N from last homework. (With the conjugacy classes of e and (12)(34).) If we add the element (12) and take the group generated by N and (12), we get a group of order 8. This group is isomorphic to D 4 , and is a 2 -Sylow. It contains two bicycles, (12) and (34). It is not normal, since
{(12), (34)} is not an entire conjugacy class (though the three non-identity elements of N do form a conjugacy class). Conjugating gives us the di erent possible pairs of disjoint bicycles. There are three possible pairs of disjoint bicycles, the other two are {(13), (24)} and {(14), (23)}. So the groups generated by N together with one of these subsets of the conjugacy class of (12) are the various 2 -Sylows, and there are 3 of them. 1
2
The 3 -Sylow group has order 3 , so is Z/(3). It is generated by any 3- cycle. There are are 8 possible 3-cycles, two occur in each group, so we get 4 di erent 3 -Sylow groups. S 5 : The 2 -Sylow group has order 8. The inclusion of S 4 into S 5 (as per- mutations xing 5 ) shows that the 2 - and 3 -Sylows of S 4 are 2 - and 3 -Sylows of S 5 as well. One obtains additional 2 -Sylow subgroups by mimicking the 2 -Sylows of S 4 using the other subsets of { 1 , 2 , 3 , 4 , 5 }, like { 1 , 3 , 4 , 5 }. There are 5 ways of doing this, which gives 15 di erent 2 -Sylows. Each of them is formed by taking 4 elements of {i, j, k, l} ⊆ { 1 , 2 , 3 , 4 , 5 } and then taking the group generated by {(ij)(kl), (ik)(jl), (il)(jk), (ij), (kl)} for some choices of i, j, k, l. For the 3 -Sylow, we note that there are 20 possible 3 -cycles, two in each group, so 10 di erent 3 -Sylows. Finally, there is a 5 -Sylow subgroup. It must have order 5 , and it is generated by any 5 -cycle. There are 24 di erent ones, 4 in each group, so 6 5 -Sylow groups.
- II.5: 8. Let Pp stand for the p-Sylow subgroup of G. In this situation we can do an unusual thing. We can construct maps G → Pp for each p. We do this by noting that |G| = pi 11... pi rr. So noting that each Ppi is normal, we map G → G/Pp 2 = G 2 → G 2 /Pp 3 = G 3 → · · · → Gr− 1 /Ppr^ ∼= Pp 1 and similarly for pj. We use several facts here. First Ppi is normal in G, so its image is normal in any quotient of G. Secondly, Ppi is the pi-Sylow subgroup of Gi− 1. This is easy to see inductively: if you assume (the image of) Ppi is the pi-Sylow subgroup of Gi− 2 , then since Gi− 1 is a quotient of Gi− 2 by a group of order prime to pi, no element of Ppi can become 0 in Gi. Finally, the same argument, together with the fact that the orders of the groups are the same, implies Gr− 1 /Ppr^ ∼= Pp 1. So since the direct product is a product, taking all these maps gives a map
φ : G →
∏^ r
i=
Ppi
such that φ composed with the inclusion of Ppi is the inclusion of the factor Ppi into the product. Now note that the two groups have the same order, and that φ is onto, since if we want to hit (x 1 ,... , xr ) where those are elements of the respective p-Sylow groups, it will get hit by the element x 1... xr ∈ G. Note as a corollary of this exercise, that while the p-Sylow groups them- selves need not be commutative, the elements from different p-Sylow sub- groups commute with each other.
- II.5: 9. Any subgroup of index q has order pn^ so it is p-Sylow subgroup. There are kp + 1 of these subgroups. Also, kp + 1 divides |G| = pnq, so kp + 1 divides q. So k = 0. So there is a unique subgroup of index q, which is a normal p-Sylow subgroup.
