MATH 4361 (Modern Algebra) – Assignment 4 Solutions Due on December 4, 2015
Please write in a neat and organized fashion. In addition to being mathematically correct, your answers should also be easy for me to read and contain enough detail for me to be able to follow your thought process. Correct use of words and notation are essential. 1. The dihedral group D4 is the group of symmetries of a square.
The elements of D4 are
R0 do nothing ∘ R1 rotate clockwise 90 ∘ R2 rotate clockwise 180 ∘ R3 rotate clockwise 270
FA reflect across line A
FB reflect across line B
FC reflect across line C
FD reflect across line D.
a. Write the elements of D4 as permutations. b. Construct the Cayley table for D4. c. Find all of the subgroups of D4. d. Determine which of the subgroups of D4 are normal subgroups and which are not.
Solution: We write the elements of D4 as permutations as follows:
1 R0 1
R1 1234
R2 1324
R3 1432
FA 24
FB 13
FC 1234
FD 1423.
Here is the Cayley Table for D4.
∘ R0 R1 R2 R3 FA FB FC FD
R0 R0 R1 R2 R3 FA FB FC FD
R1 R1 R2 R3 R0 FC FD FB FA
R2 R2 R3 R0 R1 FB FA FD FC
R3 R3 R0 R1 R2 FD FC FA FB .
FA FA FD FB FC R0 R2 R3 R1
FB FB FC FA FD R2 R0 R1 R3
FC FC FA FD FB R1 R3 R0 R2
FD FD FB FC FA R3 R1 R2 R0
There are 10 distinct subgroups of D4. They are
R0 which is normal
R0,R2 which is normal
R0,FA which is not normal
R0,FB which is not normal
R0,FC which is not normal
R0,FD which is not normal
R0,R1,R2,R3 which is normal
R0,R2,FA,FB which is normal
R0,R2,FC,FD which is normal
D4 which is normal.
2. The dihedral group D5 is the group of symmetries of a regular pentagon.
2 The elements of D5 are
R0 do nothing ∘ R1 rotate clockwise 72 ∘ R2 rotate clockwise 144 ∘ R3 rotate clockwise 216 ∘ R4 rotate clockwise 288
FA reflect across line A
FB reflect across line B
FC reflect across line C
FD reflect across line D
FE reflect across line E.
a. Write the elements of D5 as permutations. b. Construct the Cayley table for D5. c. Find all of the subgroups of D5. d. Determine which of the subgroups of D5 are normal subgroups and which are not.
Solution: We write the elements of D5 as permutations as follows:
3 R0 1
R1 12345
R2 13524
R3 14253
R4 15432
FA 2534
FB 1235
FC 1345
FD 1423
FE 1524.
Here is the Cayley Table for D5.
∘ R0 R1 R2 R3 R4 FA FB FC FD FE
R0 R0 R1 R2 R3 R4 FA FB FC FD FE
R1 R1 R2 R3 R4 R0 FB FC FD FE FA
R2 R2 R3 R4 R0 R1 FC FD FE FA FB
R3 R3 R4 R0 R1 R2 FD FE FA FB FC
R4 R4 R0 R1 R2 R3 FE FA FB FC FD .
FA FA FE FD FC FB R0 R4 R3 R2 R1
FB FB FA FE FD FC R1 R0 R4 R3 R2
FC FC FB FA FE FD R2 R1 R0 R4 R3
FD FD FC FB FA FE R3 R2 R1 R0 R4
FE FE FD FC FB FA R4 R3 R2 R1 R0
The subgroups of D5 are
R0 which is normal
R0,FA which is not normal
R0,FB which is not normal
R0,FC which is not normal
R0,FD which is not normal
R0,FE which is not normal
R0,R1,R2,R3,R4 which is normal
D5 which is normal. 3. If G is a group and b ∈ G, then a square root of b is an element a ∈ G such
4 that a2 b. For example 3 is a square root of 6 in 〈Z, because 2 3 3 3 6 and 123 is a square root of 132 in 〈S3,∘ because 1232 123123 132. a. Suppose that G and H are groups and that f : G → H is a homomorphism of G onto H. Prove that if every element of G has a square root then every element of H also has a square root. b. Let 〈Q, be the group of rational numbers with the operation of addition and let Qpos, be the group of positive rational numbers with the operation of multiplication. Use the fact that you proved in part a to explain why there does not exist any homomorphism of Q onto Qpos. Proof of Part a: Suppose that G and H are groups and that f : G → H is a homomorphism of G onto H. Also suppose that every element of G has a square root. We want to show that every element of H has a square root. To this end, let h ∈ H. Then, since f is onto H, there exists g ∈ G such that h fg.Sinceg has a square root, there exists a ∈ G such that g a2. Then, since f is a homomorphism, we see that y fg fa2 faa fafa. Since fa ∈ H and fa2 y, we see that y has a square root. This proves the result. Explanation of Part b: Observe that every element of Q has a square root. (Remember that the operation of Q is addition.) To be specific, if b ∈ Q, then b p/q where p and q are integers. This means that a p/2q ∈ Q and we observe that p p p a2 a a b 2q 2q q which means that a is the square root of b in Q. Now observe that not every element of Qpos has a square root. (Remember that the operation of Qpos is multiplication.) In particular 2 ∈ Qpos but there is no number, x,inQpos that satisfies x2 2. Since every member of Q has a square root but not every member of Qpos has a square root, then we see by Part a that there does not exist any homomorphism of Q onto Qpos. 4. Let G D4 and let H R0,R2 . a. In problem 1, you should have found that H is a normal subgroup of G (so you don’t have to do this again). b. List the elements of the quotient group G/H. c. Construct the Cayley table for G/H. Solution: The cosets of H are
5 H R0,R2 R2H
R1H R1,R3 R3H
FAH FA,FB FBH
FCH FC,FD FDH. Thus G H,R1H,FAH,FCH H with Cayley Table
HR1HFAHFCH
HHR1HFAHFCH
R1HR1HH FCHFAH .
FAHFAHFCHH R1H
FCHFCHFAHR1HH Observe that G/H is the Klein 4–group. 5. a. Suppose that G and K are finite groups and suppose that f : G → K is a homomorphism. Prove that for each element a ∈ G, the order of the element fa in K is a divisor of the order of a. b. Suppose that G is a finite group and that H is a normal subgroup of G. Prove that for each element a ∈ G, the order of the element aH in G/H is a divisor of the order of the element a. (Hint: Use what you proved in part a.) Proof of Part a: Suppose that G and K are finite groups and that f : G → K is a homomorphism. Let a ∈ G and suppose that orda n. Also suppose that ordfa m. We want to prove that m is a divisor of n. First observe that since f is a homomorphism then n n fa fa feG eK. n Since fa eK, then by Theorem 5 in Chapter 10, n must be a multiple of ordfa. Thus n is a multiple of m (which is the same as saying that m is a divisor of n). Proof of Part b: Suppose that G is a finite group and that H is a normal subgroup of G and let a ∈ G. We want to show that ordaH is a divisor of orda.Todothiswedefinef : G → G/H by fx xH. We have shown in class that this mapping is well–defined and that it is a homomorphism. Since fa aH, then by Part a above, we can conclude that ordaH is a divisor of orda.
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