Classification of Groups Classification of Groups of Order 24of Order 24

By: Tarek M. ElgindiBy: Tarek M. Elgindi

Mentor: Michael PenkavaMentor: Michael Penkava

TheoremsTheoremsThe Sylow Theorems: The Sylow Theorems:

Sylow Sylow Theorem 1Theorem 1: There exists a p-sylow subgroup of : There exists a p-sylow subgroup of GG, of order , of order ppnn,, where pwhere pn n divides the order of G butdivides the order of G but ppn+1n+1 does not. does not.

CorollaryCorollary: Given a finite group : Given a finite group GG and a prime number and a prime number pp dividing the order dividing the order of of GG, then there exists an element of order , then there exists an element of order pp in in GG . .

Sylow Theorem 2Sylow Theorem 2: All Sylow : All Sylow pp-subgroups of -subgroups of GG are conjugate to each are conjugate to each other (and therefore isomorphic), i.e. if other (and therefore isomorphic), i.e. if HH and and KK are p-sylow subgroup of are p-sylow subgroup of GG, then there exists an element , then there exists an element gg in in GG with with gg−1−1HgHg = = KK..

Sylow Theorem 3Sylow Theorem 3: Let : Let nnpp be the number of Sylow be the number of Sylow pp-subgroups of -subgroups of GG..

Then Then nnpp = 1 mod = 1 mod p.p.

More Definitions and TheoremsMore Definitions and Theorems

Thm: If HThm: If H11 and H and H22 are subgroups a group G, are subgroups a group G, and then we have the following:and then we have the following:

Theorem: Suppose H and K are subgroups, where H Theorem: Suppose H and K are subgroups, where H is normal, of a group G and H intersect K only is normal, of a group G and H intersect K only contains the identity. Then o(HK)=o(G) and G is contains the identity. Then o(HK)=o(G) and G is isomorphic to the semidirect product of H and K. isomorphic to the semidirect product of H and K.

21 HHP

G

HHP 21

More TheoremsMore Theorems

Theorem: Two different actions of H on K make two different Theorem: Two different actions of H on K make two different Semidirect Product structures. Semidirect Product structures.

Note: An action of a group H on a set X is a homomorphism Note: An action of a group H on a set X is a homomorphism from H from H Sym(X). In the case where X is a group, Sym(X). In the case where X is a group, Sym(X)=Aut(X)Sym(X)=Aut(X)

Theorem: Two conjugate automorphisms spawn the same Theorem: Two conjugate automorphisms spawn the same (isomorphic) semidirect product structures. (isomorphic) semidirect product structures.

Theorems(Cont.)Theorems(Cont.)

If H is the only subgroup of order n of a If H is the only subgroup of order n of a group G, then H is normal in G.group G, then H is normal in G.

If the index of a subgroup H of a group G If the index of a subgroup H of a group G is 2, then H is normal in G.is 2, then H is normal in G.

PreliminariesPreliminaries

Let G be a group of order 24. The only Let G be a group of order 24. The only distinct prime factors of 24 are 2 and 3.distinct prime factors of 24 are 2 and 3.So, we have a 2-sylow subgroup, H, of So, we have a 2-sylow subgroup, H, of order 8 and a 3-sylow subgroup, K, of order 8 and a 3-sylow subgroup, K, of order 3. order 3. But the problem is that neither H nor K is But the problem is that neither H nor K is necessarily normal in G, so we cannot necessarily normal in G, so we cannot invoke the Semidirect Product Theorem. invoke the Semidirect Product Theorem. So we have three cases.So we have three cases.

CasesCases

Case 1Case 1: H is normal in G.: H is normal in G.

Case 2Case 2: K is normal in G.: K is normal in G.

Case 3Case 3: Neither H nor K is normal in G.: Neither H nor K is normal in G.

More about H and KMore about H and KK is of order 3; therefore, it is isomorphic to ZK is of order 3; therefore, it is isomorphic to Z33..

H, unfortunately, can be any of the following groups:H, unfortunately, can be any of the following groups:

A. ZA. Z88

B.B. ZZ4 4 X ZX Z22

C.C. ZZ2 2 X ZX Z2 2 X ZX Z22

D.D. DD44

E.E. QQ(8)(8) (The Quaternion Group)(The Quaternion Group)

Both of the non-identity elements of K has order 3. On the other hand, non Both of the non-identity elements of K has order 3. On the other hand, non

of the elements of H has order 3. Therefore, H and K must be non-trivially of the elements of H has order 3. Therefore, H and K must be non-trivially

disjoint.disjoint.

Case 1, H is NormalCase 1, H is Normal We can find a non-trivial action of H on K if and only if we can find a We can find a non-trivial action of H on K if and only if we can find a

homomorphism from K into Aut(H). Since homomorphisms conserve the order of homomorphism from K into Aut(H). Since homomorphisms conserve the order of elements, Aut(H) must have some elements of order 3 for there to be a non-trivial elements, Aut(H) must have some elements of order 3 for there to be a non-trivial semidirect product.semidirect product.

• H is isomorphic to ZH is isomorphic to Z88 The automorphism of H has 4 elements.. But no group of order 4 has any element The automorphism of H has 4 elements.. But no group of order 4 has any element

of order 3. Therefore, there can be no homomorphism between K and Aut(H). of order 3. Therefore, there can be no homomorphism between K and Aut(H). So, we can make no non-trivial semidirect products of H and K, yielding the direct So, we can make no non-trivial semidirect products of H and K, yielding the direct

product Zproduct Z2424. .

B. H is isomorphic to ZB. H is isomorphic to Z44 X Z X Z22

The Aut(H) is the dihedral group of 8 elements, which has no elements of order 3.The Aut(H) is the dihedral group of 8 elements, which has no elements of order 3. As before, we only have the direct product ZAs before, we only have the direct product Z1212 X Z X Z22..

Case 1(cont.)Case 1(cont.)

C. H is isomorphic to ZC. H is isomorphic to Z2 2 X ZX Z2 2 X ZX Z22

Aut(H)= GLN(3, ZAut(H)= GLN(3, Z33), which has 54 elements of order 3. But all of these ), which has 54 elements of order 3. But all of these

elements are conjugate, so we only get one non-trivial semidirect product, as elements are conjugate, so we only get one non-trivial semidirect product, as well as the trivial direct product. well as the trivial direct product.

D. H is isomorphic to DD. H is isomorphic to D44

Interestingly, Aut(H)=DInterestingly, Aut(H)=D44. Again, D. Again, D44 has no elements of order 3, so we have the trivial has no elements of order 3, so we have the trivial direct product: Ddirect product: D44 X Z X Z33

E. H is isomorphic to QE. H is isomorphic to Q(8)(8) Aut(H)= SAut(H)= S44, which has 8 elements of order 3. But again, they are all conjugate so we only , which has 8 elements of order 3. But again, they are all conjugate so we only

get one non-trivial semidirect product: Qget one non-trivial semidirect product: Q (8)(8)XI ZXI Z33

And the trivial direct product: QAnd the trivial direct product: Q (8)(8)X ZX Z33..

Case 2, K is NormalCase 2, K is Normal

This part is very similar to the case where This part is very similar to the case where H is normal, and we actually get no new H is normal, and we actually get no new groups from this case.groups from this case.

Case 3, Neither H nor K is NormalCase 3, Neither H nor K is NormalBy the sylow theorems, the number of subgroups of order 8 of G must By the sylow theorems, the number of subgroups of order 8 of G must be 1 mod 2. Since H is not normal, the number of 2-sylow subgroups is be 1 mod 2. Since H is not normal, the number of 2-sylow subgroups is greater than or equal to 3. greater than or equal to 3.

We will now take two of the 2-sylow subgroups and call them HWe will now take two of the 2-sylow subgroups and call them H11 and and HH22. Let . Let Then Then

So, o(P)>2. Since P is a subgroup of HSo, o(P)>2. Since P is a subgroup of H1 1 and Hand H22 its order must divide 8. its order must divide 8. Therefore, o(P)=4.Therefore, o(P)=4. But since the index of P in H But since the index of P in H1 1 and Hand H22 is 2, P is is 2, P is normal in Hnormal in H1 1 and Hand H22. Thus its normalizer, N. Thus its normalizer, Npp, is at least of order , is at least of order 8+4=12.8+4=12.Thus, either P is normal in a subgroup of order 12, or P is normal in G. Thus, either P is normal in a subgroup of order 12, or P is normal in G. Now take K=<s> to be any 3-sylow subgroup of G. Let M=PK. Then M Now take K=<s> to be any 3-sylow subgroup of G. Let M=PK. Then M is of order 12. The index of M in G is 2, so M is normal in G.is of order 12. The index of M in G is 2, so M is normal in G.

G

HHP 21

21 HHP

Subcase 1Subcase 1

H is isomorphic to ZH is isomorphic to Z88

Since H is cyclic, it can be generated by some element p in G, so Since H is cyclic, it can be generated by some element p in G, so H=<p>. Then P=<pH=<p>. Then P=<p22> is isomorphic to Z> is isomorphic to Z44. So . So

Since K is the only subgroup of order 3 in M, K is normal in M.Since K is the only subgroup of order 3 in M, K is normal in M.

Further, since conjugation (a group automorphism) conserves the Further, since conjugation (a group automorphism) conserves the order of elements, o(s)=o(psporder of elements, o(s)=o(psp-1-1). So, psp). So, psp-1-1=s or psp=s or psp-1-1=s=s-1-1. If the . If the former were true, then we would get Case 2 again, so former were true, then we would get Case 2 again, so

psppsp-1-1=s=s-1-1. .

Now we check that H is not normal in G: Now we check that H is not normal in G: spssps-1-1=s=s-1-1p. So, H cannot be normal in G because conjugation of p by p. So, H cannot be normal in G because conjugation of p by s yields an element of G not in H. This gives us another possible s yields an element of G not in H. This gives us another possible group structure.group structure.

34 XZZM

Subcase 2Subcase 2

H is isomorphic to ZH is isomorphic to Z44 X Z X Z22

So, H=<p,t I pSo, H=<p,t I p44=t=t22=e, pt=tp>=e, pt=tp>

A. A. P is isomorphic to ZP is isomorphic to Z44. Then P=<p>.. Then P=<p>.

So, M is isomorphic to ZSo, M is isomorphic to Z44XZXZ33=<p,s I p=<p,s I p44=s=s33=e, ps=sp>.=e, ps=sp>.

Again, tst=s or tst=sAgain, tst=s or tst=s-1-1. And as before, the latter must be true. As in the previous subcase, H is . And as before, the latter must be true. As in the previous subcase, H is not normal in G, so we get another group of order 24.not normal in G, so we get another group of order 24.

B. B. P is isomorphic to ZP is isomorphic to Z22 X Z X Z22

We have two cases:We have two cases:

1. M is isomorphic to Z1. M is isomorphic to Z22 X Z X Z2 2 X ZX Z33

2. M is isomorphic to Z2. M is isomorphic to Z22 X Z X Z2 2 XI ZXI Z33

Subcase 2(cont.)Subcase 2(cont.)

1. M is isomorphic to Z1. M is isomorphic to Z22 X Z X Z2 2 X ZX Z33..

M is abelian; so we have the following:M is abelian; so we have the following:pp22s=sps=sp22, , ts=st, pt=tp.ts=st, pt=tp.As before, pspAs before, psp-1-1=s=s-1-1. So, sps. So, sps-1-1=s=s-1-1p. So, H and K are not normal in G, and we get another p. So, H and K are not normal in G, and we get another

group of order 24.group of order 24.

2. M is isomorphic to Z2. M is isomorphic to Z22 X Z X Z2 2 XI ZXI Z33. .

So, we have the following:So, we have the following:spsp22=ts, st=p=ts, st=p22ts, pt=tpts, pt=tpBut we notice that all of the elements of M have order 3, except those in P. So suppose But we notice that all of the elements of M have order 3, except those in P. So suppose

psppsp-1-1=p=p2k2kttlls. s.

Then Then pp22spsp22=pp=pp2k2kttllspsp-1-1=p=p2k2kttllpsppsp-1-1=p=p2k2kttllpp2k2kttlls=s. But this would imply s=s. But this would imply that pthat p22t=e. t=e.

This is a contradiction. This is a contradiction. Thus no group can be formed this way.Thus no group can be formed this way.

Subcases 3, 4, and 5Subcases 3, 4, and 5

H is isomorphic to ZH is isomorphic to Z2 2 X ZX Z2 2 X ZX Z22: we get 2 more : we get 2 more

groups.groups.

H is isomorphic to DH is isomorphic to D44: we get 2 more groups.: we get 2 more groups.

H is isomorphic to QH is isomorphic to Q(8)(8): we get 1 more group.: we get 1 more group.

Showing the above follows a similar procedure Showing the above follows a similar procedure as Subcases 1 and 2.as Subcases 1 and 2.

This concludes the classification of groups of This concludes the classification of groups of order 24 and we get 15 groups ( 3 abelian and order 24 and we get 15 groups ( 3 abelian and 12 non-abelian).12 non-abelian).