Test 2 Review

Let G be a group and let H be a subgroup of G. Give the definition of a left coset of H in G.

A left coset of H in G is any set of the form aH = {ah : h ∈ H} for some a ∈ G.

Let H, K be subgroups of G, with H ⊂ K. Use Lagrange's theorem to prove that |G : H| = |G : K| · |K : H|.

By Lagrange we have |G : K| · |K : H| = |G|/|K| · |K|/|H| = |G|/|H| = |G : H|.

Prove that xH has the same cardinality as H.

Define ψ : H → xH by ψ(h) = xh. Let h, k ∈ H be such that ψ(h) = ψ(k). Then xh = xk, and hence h = k by cancellation. Therefore ψ is one-to-one. Let y ∈ xH. Then there is h ∈ H such that y = xh. Hence y = ψ(h), so ψ onto. Since we have shown that ψ : H → xH is a bijection, we get |xH| = |H|.

Give the definition of the inner automorphism φa : G → G associated to a.

For x ∈ G we have φa(x) = axa⁻¹

Let A = ℤ10 = {0, 1, 2, . . . , 9} and define σ : A → A by σ(a) = a +10 4. Express σ as a product of 2-cycles.

It follows from (b) that σ = (0 4) ◦ (4 8) ◦ (8 2) ◦ (2 6) ◦ (1 5) ◦ (5 9) ◦ (9 3) ◦ (3 7).

Give an example of a left coset L of h(1 2)i in S3 and a right coset R of <(1 2)> in S3 such that L ∩ R ≠ { } and L ≠ R.

Let G = S3 and H = <(1 2)> = {id,(1 2)}. Let L = (1 3)H = {(1 3),(1 2 3)} and R = H(1 3) = {(1 3),(1 3 2)}. Then L ∩ R = {(1 3)} ≠∅ but L ≠ R.

Let H be a subgroup of the symmetric group Sn. Prove that either all the elements of H are even permutations or exactly half of the elements of H are even permutations.

Let H0 = {σ ∈ H : σ is even} and H1 = {σ ∈ H : σ is odd}. If H1 = { } then every element of H is even. Suppose H1 6= { } and let τ ∈ H1. Define f : H0 → H1 by f(σ) = τ ◦ σ. Then f is well-defined, since H is closed under multiplication and the composition of an odd permutation with an even permutation is an odd permutation. If f(σ1) = f(σ2) then τ ◦ σ1 = τ ◦ σ2, and hence σ1 = σ2 by cancellation. So f is one-toone. Let ρ ∈ H1. Since τ is odd, so is τ⁻¹, so τ⁻¹◦ ρ is even. Hence τ⁻¹◦ ρ ∈ H0 and f(τ⁻¹◦ ρ) = τ ◦ τ⁻¹◦ ρ = ρ. So f is onto. It follows that f is a bijection, so |H0| = |H1|. Since {H0, H1} is a partition of H we get |H| = |H0| + |H1|

Prove Lagrange's Theorem: |G| = |G : H| · |H|.

Let S be the set of distinct left cosets of H in G. Given x ∈ G we have x ∈ xH and xH ∈ S. Hence S S = G. By (a) the members of S are pairwise disjoint, so S is a partition of G. Write S = {x1H, x2H, . . . , xkH}. Then k = |G : H| and |xiH| = |H| for 1 ≤ i ≤ k. It follows that |G| = X k i=1 |xiH| = X k i=1 |H| = k|H| = |G : H| · |H|.

Prove that the set Inn(G) = {φa : a ∈ G} is a subgroup of Aut(G).

Let a ∈ G and let x, y ∈ G. If φa(x) = φa(y) then axa⁻¹ = aya⁻¹ Hence by cancellation on the left and right we get x = y. Therefore φa is one-to-one. Note that a⁻¹xa ∈ G, and that φa(a⁻¹xa) = a(a⁻¹xa)a⁻¹ = x. Therefore φa is onto. Finally, we have φa(xy) = a(xy)a⁻¹= (axa⁻¹)(aya)) = φa(x)φa(y). Therefore φa ∈ Aut(G). Since this holds for every a ∈ G, we deduce that Inn(G) ⊂ Aut(G). Since e ∈ G we have idG = φe ∈ Inn(G), so Inn(G) 6= { }. Let φa, φb ∈ Inn(G), with a, b ∈ G. Then φa ◦ φb = φab ∈ Inn(G). Also note that φa ◦ φa⁻¹ = φaa⁻¹ = φe = idG and φa⁻¹ ◦ φa = φa⁻¹a = φe = idG. Hence φ⁻¹a = φa⁻¹ ∈ Inn(G). By the two-step subgroup test we conclude that Inn(G) ≤ Aut(G).

Prove that if xH ∩ yH is nonempty then xH = yH.

Let g ∈ xH ∩ yH. Then there are h, k ∈ H such that xh = g = yk. We get y−1x = kh−1 ∈ H. Set ℓ = kh−1. Then ℓ ∈ H and x = yℓ. Now let a ∈ xH. Then there is m ∈ H such that a = xm. Hence a = yℓm with ℓm ∈ H. Thus a ∈ yH, so xH ⊂ yH. By symmetry we also get yH ⊂ xH. Therefore xH = yH.

Prove that φab = φa ◦ φb.

Let x ∈ G. Then (φa ◦ φb)(x) = φa(φb(x)) = a(bxb⁻¹)a⁻¹ = (ab)x(ab) = φab(x). Since this holds for all x ∈ G we have φa ◦ φb = φab.

Prove that if x ∈ G then |x| divides |G|.

Let x ∈ G. Then <x> ≤ G, so we get |<x>| divides |G| by Lagrange. Since |x| = |<x>| it follows that |x| divides |G|.

Let φ : G → G be an isomorphism and let x ∈ G. Prove that φ(x⁻¹) = φ(x)⁻¹

Let x ∈ G. Then φ(x)φ(x⁻¹) = φ(xx⁻¹) = φ(1G) = 1Ḡ = φ(x)φ(x)⁻¹ Hence by left cancellation in G we get φ(x⁻¹) = φ(x)⁻¹

Let R denote the group of real numbers with the operation of addition. Prove that the function φ : R → R defined by φ(x) = 3x is an automorphism of R.

Let x1, x2 ∈ R. If φ(x1) = φ(x2) then 3x1 = 3x2, so x1 =1/3 3x1 =1/3· 3x2 = x2. Hence φ is one-to-one. Also φ(x1 + x2) = 3(x1 + x2) = 3x1 + 3x2 = φ(x1) + φ(x2), so φ is operation-preserving. Finally, let y ∈ R. Then 1/3y ∈ R, and φ(1/3y) = 3 ·1/3y = y, so φ is onto. Hence φ is an isomorphism.

Let H ≤ G. State Lagrange's theorem for G and H.

Let |G : H| denote the number of distinct left cosets of H in G. Then |G| = |G : H| · |H|, and hence |H| divides |G|.

Let S4 be the symmetric group and let D12 be the dihedral group. Prove or disprove that S4 is isomorphic to D12. (Note that both groups have order 24.)

Let ρ ∈ D12 denote the 30◦ counterclockwise rotation. Then |ρ| = 12. On the other hand, the elements of S4 are either k-cycles with 2 ≤ k ≤ 4, which have order k; elements of the form (a b)(c d), which have order 2; or the identity, which has order 1. So S4 has no elements of order 12. Hence D12 ≃ S4.

Let A4 be the alternating group and let D6 be the dihedral group. Prove or disprove that A4 is isomorphic to D6.

Let ρ ∈ D6 denote the 60◦ counterclockwise rotation. Then |ρ| = 6. On the other hand, the elements of A4 are either 3-cycles, which have order 3, elements of the form (a b)(c d), which have order 2, or the identity, which has order 1. So A4 has no elements of order 6. Hence D6≃A4.

Let G = <x> be a cyclic group of order 20. Determine all automorphisms of G. In particular, for each σ ∈ Aut(G) give the value of σ(x).

Since G is cyclic of order 20, Aut(G) = {σn : n ∈ U(20)}, where σn(g) = g n for all g ∈ G. Thus Aut(G) = {σ1, σ3, σ7, σ9, σ11, σ13, σ17, σ19}, with σ1(x) = x, σ3(x) = x³, σ7(x) = x⁷, σ9(x) = x⁹, σ11(x) = x¹¹, σ13(x) = x¹³, σ17(x) = x¹⁷ and σ19(x) = x¹⁹

Determine the number of elements of order 4 in A5.

Since a 4-cycle is an odd permutation, there are 0 elements of order 4 in A5.

Let A = ℤ10 = {0, 1, 2, . . . , 9} and define σ : A → A by σ(a) = a +10 4. Prove that σ is a permutation of A.

Since each element of Z10 occurs once and only once on the second row, σ is onto and one-to-one. Therefore σ is a permutation of Z10.

Determine which of the groups Z2 ⊕ Z3, S3, A4, Z6 are isomorphic to which, with some explanation.

Since gcd(2, 3) = 1 we see that Z2 ⊕ Z3 is cyclic of order lcm(2, 3) = 6. Hence, Z2 ⊕ Z3 ≃ Z6. The nonabelian groups S3 and A4 are not isomorphic to the abelian groups Z2 ⊕ Z3 and Z6. Also S3 6≃ A4 since |S3| = 6 ≠ 12 = |A4|. Hence the only isomorphism is between Z2 ⊕ Z3 and Z6.

Let G be a group of order n, with 1 < n < ∞. Prove that if the only subgroups of G are {eG} and G then n is prime.

Since |G| > 1 there is x ∈ G with x 6= 1G. Thus <x> ≠ {1G}, so by assumption <x> = G. Hence G is a cyclic group of order n, and |x| = n. If n is not prime then since n > 1 we see that n is composite, so n = ab with 1 < a, b < n. Hence |<xa>| = |xa| = b, so the subgroup <xa> cannot be equal to either {1G} or G. This is a contradiction, so n is prime.

Let A = ℤ10 = {0, 1, 2, . . . , 9} and define σ : A → A by σ(a) = a +10 4. Determine whether the permutation σ is even or odd.

Since σ is a product of 8 2-cycles, σ is even

Let φ be an automorphism of G. Prove that the set H = {x ∈ G : φ(x) = x} is a subgroup of G.

Since φ : H → H is an isomorphism we have φ(e) = e. Therefore e ∈ H, so H 6= ∅. Let x, y ∈ H. Then φ(x) = x and φ(y) = y. Hence φ(xy⁻¹) = φ(x)φ(y⁻¹) = φ(x)φ(y)⁻¹ = xy⁻¹ so xy−1 ∈ H. Therefore by the one-step subgroup test we get H ≤ G.

Let φ : G → G be an isomorphism, let e be the identity of G, and let ē be the identity of Ḡ. Prove from first principles that φ(e) = ē.

Since φ(e)φ(e) = φ(ee) = φ(e) = φ(e)ē, by cancellation we get φ(e) = ē

Let φ : G → G be an isomorphism. Prove that for all x ∈ G we have φ(x⁻¹) = φ(x)⁻¹

Since φ(x)φ(x⁻¹) = φ(xx⁻¹) = φ(e) = e and φ(x⁻¹)φ(x) = φ(x⁻¹x) = φ(e) = e we see that φ(x⁻¹) = φ(x)⁻¹

Let A be a set, let x ∈ A, and let G be a subgroup of the symmetric group SA. Define the stabilizer StabG(x) of x in G.

StabG(x) = {σ ∈ G : σ(x) = x}.

Determine the number of elements of order 4 in S5.

The only elements of order 4 in S5 are 4-cycles. Hence there are 5 · 4 · 3 · 2/4 = 30 elements of order 4 in S5.

List the distinct left cosets of H = <(1 2)> in S3.

There are a, b ∈ G such that L1 = aH and L2 = bH. Suppose L1 ∩ L2 6= ∅ and let c ∈ L1 ∩ L2. Then c = ah and c = bk for some h, k ∈ H. Thus ah = bk, so a = bkh−1. Let x ∈ L1. Then x = am for some m ∈ H. Hence x = bkh−1m with kh−1m ∈ H. Thus x ∈ bH = L2. It follows that L1 ⊂ L2. By symmetry we get L2 ⊂ L1, and hence L1 = L2. We conclude that either L1 ∩ L2 = ∅ or L1 = L2.

Let A = ℤ10 = {0, 1, 2, . . . , 9} and define σ : A → A by σ(a) = a +10 4. Find a cycle decomposition for σ.

Using the table in (a) we get σ = (0 4 8 2 6)(1 5 9 3 7).

Let φ : G → Ḡ be an isomorphism and let x ∈ G. Prove that if |x| = n then |φ(x)| = n.

We have φ(xn) = φ(e), and hence φ(x)n = ē. Therefore |φ(x)| ≤ n. Suppose 1 ≤ k < n and φ(x)ᵏ = ē. Then φ(xᵏ) = φ(e). Since φ is one-to-one it follows that xᵏ= e. Since |x| = n this is a contradiction. Therefore |φ(x)| = n.

Let φ : G → Ḡ be an isomorphism and let H be a subgroup of G. Prove that φ(H) = {φ(x) : x ∈ H} is a subgroup of Ḡ.

We use the one-step subgroup test. Since H 6= { } we see that φ(H)≠{ }. Let a, b ∈ φ(H). Then there are x, y ∈ H such that φ(x) = a and φ(y) = b. Since H ≤ G we get xy⁻¹ ∈ H. It follows that ab⁻¹ = φ(x)φ(y)⁻¹ = φ(x)φ(y⁻¹) = φ(xy⁻¹) lies in φ(H). Hence φ(H) is a subgroup of G.

Give an example of each of these, or explain why no example exists. i. An element σ of S5 which has order 6. ii. An element τ of S6 which has order 7. iii. An element ρ of A9 which has order 10

i. Let σ = (1 2 3)(4 5) ∈ S5. Then |σ| = lcm(3, 2) = 6. ii. Since 7 is prime any τ ∈ S6 which has order 7 must have a 7-cycle in its cycle decomposition. This is impossible since 7 > 6. iii. Let ρ = (1 2 3 4 5)(6 7)(8 9). Then |ρ| = lcm(5, 2, 2) = 10. Since ρ is a product of 4 + 1 + 1 = 6 2-cycles, ρ ∈ A9.