Selected exercises from Abstract Algebra Dummit and Foote

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Selected exercises fromAbstract AlgebrabyDummit

and Foote(3rd edition).

Bryan Felix

Abril 12, 2017

Section 4.1

Exercise 1.LetGact on the setA. Prove that ifa;b2Aandb=gafor someg2G, thenGb=gGag1. Deduce that ifGacts transitively onAthen the kernel of the action isT g2GgGag1: Proof.For the rst part we use the usual containment criterion. i) We show thatg1GbgGa(equivalentlyGbgGag1).

Letx2g1Gbg, then

x=g1~bgfor some~b2Gb xa=g1~bga =g1~b(ga) =g1~b(b) =g1(~bb) =g1b =a

ThereforeGbgGag1.

ii) Now we showgGag1Gb.

Letx2gGag1, then

x=g~ag1for some ~a2Ga xb=g~ag1b =g~a(g1b) =g~a(a) =g(~aa) =ga =b

Hence,gGag1Gbas desired.

1 For the second part, recall that the kernel of the action is given by the intersection of the stabilizers of the elements inA. Since the action is transitive, we have that the orbit ofais equal to the entire setA. Then, by means of the previous part we have g2GgG ag1=\ g2GG ga=\ a2AG a as desired.Exercise 2.LetGbe a permutation group on the setA(i.e.,GSA), let2Gand let a2A:Prove thatGa1=G(a):Deduce that ifGacts transitively onAthen 2GG a1= 1 Proof.The rst part is trivial using exercise 1. Observe thataand(a) are elements ofA, with the identity (a) =a: For the second part, observe that if the action is transitive, then there is only one orbit. Using

Lagrange's we see that

jGjjGaj=jOaj=jGj thereforejGaj= 1 for anya2G. HenceGa= 1 for anya2G. It follows that 2GG a1=\ 2GG (a)=\

2G1 = 1Exercise 3.Assume thatGis abelian, transitive subgroup ofSA:Show that(a)6=afor all

2G1and alla2A:Deduce thatjGj=jAj:

Proof.IfGis abelian we have

2GG a1=\ 2GG a=Ga= 1 as desired. For the second part, we use Lagrange's and we have that [G:Ga] =jOaj=jGj as desired.Section 4.2 Exercise 7.LetQ8be the quaternion group of order 8 a) Prove thatQ8is isomorphic to a subgroup ofS8 Proof.This is trivial using Cayley's Theorem (Corollary 4).2 b) Prove thatQ8is not isomorphic to a subgroup ofSnfor anyn7. Proof.It suces to show thatQ8is not isomorphic to any subgroup ofS7(sinceS1< S2< < S6< S7). LetQ8act on a setAof order 7. Then we inspect the order of the orbit and stabilizer of an arbitrary elementa2A. From Lagrange's, the order of the stabilizer divides the order of the groupQ8. Therefore the order is either 1;2;4 or 8. If the order of the stabilizer equals

1 then, from

[G:Ga] =jOaj we have that the orbit has order 8. This is a clear contradiction since the setAhas seven elements. Therefore the order of the stabilizer is either 2;4 or 8. In any case we can see form the lattice ofQ8that any such subgroup contains the grouph1;1i. It follows that the kernel of the action ker(action) =\ a2AG a contains the subgrouph1;1i. Therefore the action is not faithful, then not injective, and

hence, no isomorphism exists.Exercise 9.Prove that ifpis a prime andGis a group of orderpfor some2Z+, then

every subgroup of indexpis normal inG. Deduce that every group of orderp2has a normal subgroup of orderp. Proof.We assume that2 otherwise the group is cyclic and the only subgroup of indexp is the identity (which is normal). We know that groups of orderpnare nilpotent. The following lemma is taken from Isaacs

Algebra(2008):

Lemma.The following are equivalent:

i)Gis a nilpotent group. ii) Every maximal proper subgroup ofGis normal. The result then follows for all2. For groups of orderp2we have thatGis cyclic, then it suces to show that a subgroup of orderpexists. Letxbe the generator ofG, thenhx2iis a subgroup ofGof orderp.Section 4.3 Exercise 6.AssumeGis a non-abelilan group of order 15. Prove thatZ(g)=1. Proof.We inspect the order ofZ(G). SinceZ(G) is a subgroup ofG, it's order divides 15.

Then we have four cases:

i) The order ofZ(G) is equal to 1.

Then the conclusion is trivial.

ii) The order ofZ(G) is equal to 3. Then [G:Z(G)] = 5 and thereforeG=Z(G) is cyclic. By exercise 36 on section 3.1 we have thatGis abelian arriving at a contradiction. 3 iii) The order ofZ(G) is equal to 5. Like the previous caseG=Z(G) is cyclic and thenGis abelian. iv) The order ofZ(G) is equal to 15.

ThenGis abelian by denition. A clear contradiction.Exercise 30.IfGis a group of odd order, prove for any non identity elementx2Gthatx

andx1are not conjugate inG. Proof.We proceed by contradiction and we assume thatx1is a conjugate ofx. Then, we look at the action ofGwith itself by conjugation and inspect the orbitOxofx. If the only elements of the orbit arexandx1then the order ofOxequals two and we have a contradiction (by Lagrange's theorem) since 2 does not dividejGj. Then, there must bey2Ox such thaty6=x. We make a remark and note thaty6= 1 either. Otherwise (by the denition of conjugate elements) x=g1g1 x= 1: is a contradiction.

Now takeyand observe that

x=gyg1 ,x1=gy1g1 Thereforey1is in the same orbit asx1and hence bothyandy1are inOx. Again, we observe that the order ofOxis even, and we arrive at a contradiction again. By the principle of indenite exhaustion the existence ofOxcontradicts the assumption as desired.Section 4.4 Exercise 7.IfHis the unique subgroup of a given order in a groupGproveHis characteristic onG. Proof.Letbe an element of Aut(G). Recall thatis an isomorphism:G!G. By properties of isomorphisms, group orders are preserved; i.e. jHj=j(H)j SinceHis the unique subgroup ofGof orderjHj, this forcesto mapHto itself. Remark.It is not necessary that(h) =hforh2H.Exercise 8.LetGbe a group with subgroupsHandKwithHK: a) Prove that ifHis characteristic inKandKis normal inGthenHis normal inG. 4 Proof.LetGact onKby conjugation and letgbe the associated permutation of a xed element inGacting onK. Observe that sinceKis normalg(K) =Kfor allg2G. Fur- thermoregis an isomorphism ofK, thereforeg2Aut(K). Then, sinceHis characteristic

inKwe have that(H) =Hfor allg2G. EquivalentlygHg1=Has desired.b) Prove that ifHis characteristic inKandKis characteristic inGthenHis characteristic

inG. Proof.Letbe an element in Aut(G). SinceKis characteristic inG,(K) =K. Further- morerestricted toKis an isomorphism ofK, thereforeK2Aut(K). It is easy to see that

K(H) =H

and therefore(H) =Has desired.Exercise 9.Ifr;sare the usual generators for the dihedral groupD2n, use the preceding two

exercises to deduce that every subgroup ofhriis normal inD2n. Proof.We will show thathriis normal inD2nand then show that any subgroup ofhriis characteristic. For the rst part note that the only generator outsidehriiss. Then it suces to show that hri=hsrs1i. Observe that hsrs1i=hr1i =hri as desired. Thereforehriis normal inD2n. For the second part, recall that a group is cyclic if and only if no two (distinct) subgroups have the same order. Therefore, by problem 7, the subgroups ofhriare characteristic. The result then follows.Exercise 16.Prove that(Z=24Z)is an elementary abelian group of order 8. Proof.It suces to show that for every numbern <24 relative prime to 24,n21 mod 24.

Observe that

2=11 mod 24

2=251 mod 24

2=491 mod 24

2=1211 mod 24

2=1691 mod 24

2=2891 mod 24

2=3611 mod 24

2=5291 mod 24

Hence, (Z=24Z)is an elementary abelian group of order 8.5

Section 4.5

Exercise 13.Prove that a group of order56has a normal Sylow p-subgroup for some primep dividing its order.

Proof.Using Sylow's theorems we see that

2= 1 or 7

and n

7= 1 or 8:

We proceed by contradiction and we assume that neither of the Sylow subgroups is normal. Therefore we necessarily haven2= 7 andn7= 8. Since the Sylow 7-subgroups only intersect at the identity we have 8(71) = 48 non-identity elements of order 7 inG. Observe that none of these elements can belong to the Sylow 2-subgroups by Lagrange's. Then we are left with 8 elements belonging to the 8distinctSylow 2-subgroups. This is impossible and we have our contradiction.Exercise 16.LetjGj=pqr, where p,q, and r are primes withp < q < r. Prove thatGhas a normal Sylow subgroup for either p,q,r. Proof.We inspect the values ofnrandnq. By the Sylow theorems we must have thatnr satises: n r1 modr and n rpq: The latter restricts the options tonrbeing either 1;p;qorpq. Note that ifnr= 1 then the Sylow subgroup is normal and we are done. Otherwise note that the assumptionp > q > r, forcesnrto be equal topq(otherwise the congruence is not satised). Likewise we inspect the possible values ofnqand conclude thatnqis eitherrorpr. Now, we make the standard count element. ThepqSylow r-subgroups contribute withpq(r1) elements while the Sylow q-subgroups contribute with at leastr(q1) elements. In total we have pq(r1) +r(q1)> pq(r1) +p(q1) =pqrpq+pqp =pqrp Recall the t the order ofGispqr. Therefore we only havepelements to distribute to the renaming Sylow subgroups. This forces the uniqueness of the Sylow p-subgroup, making is a

normal subgroup ofGas desired.Exercise 30.How many elements of order7must there be in a simple group of order168?

Solution.Note that 168 = 2337. If we inspect the number of Sylow 7-subgroupsn7we see that n

71 mod 7

and n 724
6 The latter restrictsn7to either 1;2;3;4;6;8;12 or 24. Together with the congruence and the fact that the group is simple we have thatn7= 8. Since these Slow 7-subgroups are of prime order, they all intersect only at the identity, therefore we have

8(71) = 48

elements of order 7 in the group.Exercise 33.LetPbe a normal Sylow p-subgroup ofGand letHbe any subgroup ofG.

Prove thatP\His the unique Sylow p-subgroup ofH.

Proof.SincePis normal inG,His a subgroup ofN(P) and we may use the second isomorphism theorem. Then,H\Pis normal inHand byCorollary 20H\Pis the unique Sylow subgroup ofH.Exercise 34.LetP2Sylp(G)and assumeNEG. Use the conjugacy part of Sylow's Theorem to prove thatP\Nis a Sylow p-subgroup ofN. Deduce thatPN=Nis a Sylow p-subgroup of G=N. Proof.Take any Slyow p-subgroupHofNand observe thatHis a p-subgroup inG, therefore there existg2Gsuch that

HgPg1:

FurthermoreHis also a subgroup ofgNg1(by the normality of N). Then

HgPg1\gNg1

Hg P\N g 1 and gHg 1P\N: Note that bothgHg1andPTNare p-subgroups inG. SincejgHg1j=jHjandHis a Sylow p-subgroup ofNit follows thatPTNhasp-powerorder at least as large asH. Therefore

PTNis a Sylow p-subgroup ofN.

For the second part we use thesecond isomorphism theorem. SinceNis normalPN(N) and thereforePNis a subgroup ofG. We only need to show thatpdoes not divide the order of the index [G=N:PN=N]:

By the second isomorphism theoremPN=N=P=P\N, then

[G=N:PN=N] = [G=N:P=P\N] =jGjjP\NjjPjjNj: By assumption (P2Sylp(G))pdoes not dividejGjjPjand by the rst part of the problem (P\N2Sylp(N))pdoes not divideP\NN . Thus,pdoes not divide [G=N:PN=N] as desired.7quotesdbs_dbs13.pdfusesText_19

[PDF] Selected exercises from Abstract Algebra Dummit and Foote