Test 2 Review

Compute (1 2 5 4)^-1(1 2 3)(4 5)(1 2 5 4)

(1 2 5 4)^-1(1 2 3)(4 5)(1 2 5 4) = (4 5 2 1)(1 2 3)(4 5)(1 2 5 4) = (4 3 1)(5 2) = (3 7 4)

Let G be the internal direct product of subgroups H and K. Show that the map ϕ : G -> H x K defined by ϕ(g) = (h, k) for g = hk, where h ∈ H and k∈K is one - to - one and onto.

* I'm not too confident about this proof Assume G is the internal direct product of H and K. So H is a normal subgroup of G and K is a normal subgroup of G and H ∩ K = {e}. Define φ: H x K -> HK by φ(h, k) = hk. show φ is homomorphism: let (h1, k1), (h2, k2) ∈ H x K. Then we have: φ(h1, k1)(h2, k2)) = φ(h1h2, k1k2) = h1h2k1k2 = h1k1h2k2 = φ(h1, k1)φ(h2, k2). Therefore, φ is homomorphism. Now assume φ(h1, k1) = φ(h2, k2) so h1k1 = h2k2. Since H ∩ K = {e} then each element can be expresses uniquely in HK. Thus h1k1=h2k2 implies h1 = h2 and k1 = k2. Hence, φ is one-to-one and onto, and therefore φ is an isomorphism and H x K is isomorphic to HK. source: http://www.csus.edu/indiv/e/elcek/m210a/directprodn.pdf

Prove/disprove the following assertion: Let G, H and K be groups. If G x K is isomorphic to H x K, the G is isomorphic to H.

* I'm not too confident about this proof Take K = Π(to infinity, i = 1)Z and G = Z and H = Z x Z. then G x K is isomorphic to K is isomorphic to H x K but G is not isomorphic to H, and as such the assertion is false. source: http://www.pitt.edu/~roed/courses/430/PS4sol.pdf

Let G be a group and define a map λg : G -> G by λg(a) = ga. Prove that λg is a permutation of G.

A function is permutation of G, if f : G->G and f is a bijection. λg is a function from G to G, so it is necessary to prove that it is a bijection. Prove one-to-one: suppose g1, g2 ∈ G and λg(g1) = λg(g2). Then gg1 = gg2. Cancel out the g by left cancellation, and you're left with g1 = g2. Prove onto: suppose h ∈ G. Then g^-1h∈G because G is a group and λg(g^-1h) = g(g^-1h) = h. So λg is onto. Thus λg is a bijection and so λg is a permutation of G. source: http://campus.lakeforest.edu/trevino/Spring2016/Math330/Homework4Solutions.pdf

Prove that A x B is abelian iff A and B are abelian.

Assume that A and B are abelian. Then for (a, b) and (x, y) in AxB, (a,b)(x,y) = (ax, by) = (xa, yb) = (x,y)(a,b) and so AxB is abelian. source: http://www.chegg.com/homework-help/questions-and-answers/prove-x-b-abelian-iff-b-ablelian-q2709780

Let H and K be subgroups of a group G. Prove that gH ∩ gK is a coset of H ∩ K in G.

Because H and K each are subgroups of G, H ∩ K is also a subgroup. If xH ∩ yK = ∅, we are done. If not, let g ∈ xH ∩ K. Because g ∈ xH and yK, we have that gH=xH and gK = yK. Thus, xH ∩ yK = gH ∩ gK. Therefore, g(H ∩ K) = gH ∩ gK. source: https://math.la.asu.edu/~helene/mat444/04s/solhwk2.pdf

Prove that Q is not isomorphic to Z.

Claim: they are not isomorphic because Z is cyclic while Q is not cyclic. Z = <1> (as an additive group). The group (Q, +) is not cyclic. Thus, <0> = {0} ≠ Q. Also, if a ∈ Q, a≠0 then <a> = {na|n ∈ Z} ≠ Q since a/2 ∉ <a>. Thus for every element r ∈ Q we have <r> ≠ Q, and hence Q is not cyclic, as claimed. source: http://www.math.uiuc.edu/~kapovich/417-09/hw7s.pdf

Show that the matrices form a group. Find an isomorphism of G with a more familiar group of order 6. A = (1 0 0 0 1 0 0 0 1) B = (1 0 0 0 0 1 0 1 0) C = (0 1 0 1 0 0 0 0 1) D = (0 0 1 1 0 0 0 1 0) E = (0 0 1 0 1 0 1 0 0) F = (0 1 0 0 0 1 1 0 0)

Composition table (operation: multiplication): * | A B C D E F ------------------ A | A B C D E F B | B A F E D C C | C D A B F E D | D C E F B A E | E F D C A B F | F E B A C D A is the identity element Composition is well-defined Composition is unique and closed Composition is associative Inverse exists so this is a group It is isomorphic to P3, the permutation group of order 6 with mapping of F[A] = F1 = [I] = identity permutation F[B] = F2 = (1 2) F[C] = F3 = (2 3) F[D] = F4 = (3 1) F[E] = F5 = (1 2 3) F[F] = F6 = (1 3 2) The composition tables for the two are identical source: http://www.chegg.com/homework-help/questions-and-answers/show-matrices-form-group-find-isomorphism-g-familiar-group-order-6-q3975117

Show that Sn is isomorphic to a subgroup of An+2.

Consider Sn as a subgroup of Sn+2 in the natural way. Then, say that ψ : Sn -> An+2 by ψ(σ) = { σ if σ is even, or σ(n+1 n+2) if σ is odd}. This shows injectiveness, and also satisfies ψ(σ*tau) = ψ(σ)ψ(tau) for all σ, tau ∈ Sn. We use the fact that (n+1 n+2) commutes with everything in Sn and sqaures to the identity to compute ψ(σ)ψ(tau) = {σtau is σ and tau are both even or both odd, σtau (n+1 n+2) if one is even and the other is odd)}. If σ and tau are both even/odd, then σtau is even and so ψ(σtau) = σtau. If one of σ and tau is even and the other is odd, then σtau is odd and so ψ(σtau) = σtau(n+1 n+2). In either case, we have ψ(στ) = ψ(σ)ψ(τ) as required. source: http://www.math.cornell.edu/~riley/Teaching/Groups_and_Geometry2012/exams/prelim_with_solutions.pdf

Let G be a group and g∈G. Define a map ig:G→G by ig(x)=gxg^−1. Prove that ig defines an automorphism of G. Such an automorphism is called an inner automorphism. The set of all inner automorphisms is denoted by Inn⁡(G).

Consider ig: G -> G defined by ig(x) = gxg^-1. one to one: Let x, y ∈ G, then: ig(x) = ig(x) gxg^-1 = gyg^-1 g^-1(gxg^-1)g = g^-1(gyg^-1)g (g^-1g)x(g^-1g) = (g^-1g)y(g^-1g) exe = eye x = y onto: Let y ∈ G, then g^-1yg∈G such that ig(g^-1yg) = g(g^-1yg)g^-1 = (gg^-1)y(gg^-1) = y homomorphism: Let x, y ∈ G, then ig(xy) = g(xy)g^-1 = g(xey)g^-1 = g(xg^-1gy)g^-1 = (gxg^-1)(gyg^-1) = ig(x)ig(y). hence ig is an automorphism. source: http://www.chegg.com/homework-help/questions-and-answers/let-g-group-g-g-define-map-ig-g-g-ig-x-gxg-1--prove-ig-defines-automorphism-g-automorphism-q8969489

Prove that G x H is isomorphic to H x G.

Define f:G x H -> H x G by f(g, h) = (h, g). injective: if f(g1, h1) = f(g2, h2), then (h1, g1) = (h2, g2), so h1=h2, g1=g2. surjective: ∀(h, g) ∈ H x G, f(g, h) = (h, g) ∀(g1, h1), (g2, h2) ∈ G x H, f((g1, h1)(g2, h2)) = f(g1g2, h1h2) = (h1h2, g1g2) = (h1, g1)(h2, g2) = f(g1, h1)f(g2, h2) so G x H ≈ H x G source: http://www.math.wsu.edu/faculty/dzhang/320-spring13/320practicetestkey.pdf

Prove that ℂ∗ is isomorphic to the subgroup of GL2(ℝ) consisting of matrices of the form (a b -b a)

Define ϕ : C* -> GL2(R) by ϕ(a+bi) = (a b -b a). Given a+bi, c+di ⊂ C*, f(a+bi)f(c+di) = (a, b, -b, a) (c, d, -d, c) [by definition of f] = (ac-bd, ad+bc, -(ad+bc), ac-bd) [by matrix multiplication] = f((ac-bd) + (ad+bc)i) =f((a+bi)(c+di)). Therefore, f is a group homomorphism. The kernel of f is {1}. If a+bi is in ker(f) then f(a+bi) = (1, 0, 0, 1) which implies (a, b, -b, a) = (1, 0, 0, 1). So, a = 1 and b =0, i.e. a+bi = 1. Thus, f is one-to-one. Given Z⊂H, there exists real numbers a and b (≠0) such that Z =(a, b, -b, a). Then f(a+bi) = z which shows that f is onto. Because f is a one-to-one and onto homomorphism, it is an isomorphism. Hence, C* is isomorphic to H. source: http://www.chegg.com/homework-help/questions-and-answers/prove-c-isomorphic-subgroup-gl2-r-consisting-matrices-form-b-b-define-pi-c-right-arrow-gl2-q8805711

If [G:H] = 2 prove that gH=Hg.

Let xH be a coset of H ⊂ G. Since cosets partition G, either xH = H or it is the other coset G - H (the other coset is made of the elements not in the first coset, so it's the complement of H). If xH = H then x ∈ H so xH = H = Hx. Otherwise, x ∉ H, so Hx ≠ H. Thus xH = G - H = Hx. source: https://math.stackexchange.com/questions/715131/abstract-algebra-proof-that-if-mathith-is-a-subgroup-of-index-2-in-a-finit

Find two nonisomorphic groups G and H such that Aut(g) is isomorphic to Aut(H).

G = Z3 = {0, 1, 2} and H = Z4 = {0, 1, 2, 3}. Z3 has two automorphisms: 0<->0, 1<->1, 2<->2 and 0<->0, 1<->2, 2<->1 (exchanging one with two). Z4 has two automorphisms: 0<->0, 1<->1, 2<->2, 3<->3 and 0<->0, 1<->3, 2<->2, 3<->1 (exchanging one with three). Thus Aut(Z3) is isomorphic to Aut(Z4) is isomorphic to Z2. But in Z6 = {0, 1, 2, 3, 4, 5}, because only 1 and 5 are coprime with 6, there are also two automorphisms: the identity and automorphism that exchanges one with 5, so there are 3 groups with the desired property of Aut(Z3) isomorphic to Aut(Z4) isomorphic to Aut(Z6) isomorphic to Z2. source: https://in.answers.yahoo.com/question/index?qid=20101108080102AAFbGKE

Prove that D4 cannot be the internal direct product of two of its proper subgroups.

If D4 = H x K then because |D4| = 8, |H| = 4 and |K| = 2 (or vice versa). Then K ≈ Z2 and H ≈ Z4 or Z2⊕Z2. Particularly both H and K are abelian groups. So because D4 = H x K ≈ H⊕K, D4 must be abelian too. But D4 is not abelian, so it is not an internal direct product. source: http://www.hanbommoon.net/wp-content/uploads/2014/01/Homework_Solution_10.pdf

List the left and right coset of T in C*.

If T is the circle group, T = {z ∈ C* : zz* = 1}, then the left and right cosets are equal (because C* is the multiplicative group of a field and thus abelian). Therefore, the coset Tw is the circle of radius |w| in the complex plane. source: http://www.chegg.com/homework-help/questions-and-answers/list-left-right-cosets-following-a4-s4-b-sn-c-d4-s4-d-t-c-t-circle-group-given-z-element-c-q3866269

Prove that Z ≈ nZ for n ≠0.

If Z is a cyclic group and n is the order of Z, we define f: Z -> Zn = {0, 1, 2,...n-1} so that f(x^a) = a. Here, f is bijective. Then f(x^a*x^b) = f(x^(a+b)) = a+b = f(x^a) + f(x^b). source: http://www.chegg.com/homework-help/questions-and-answers/prove-z-isomorphic-nz-n-0-q3975093

Show that any cyclic group of order n is isomorphic to Zn.

Let G = <a>. So |a| = n. Define ϕ: G -> Zn by ϕ(a^k) = [k]. Show that ϕ is well-defined: Suppose that a^k = a^l. Then k ≡ l(mod n) since |a| = n. Thus [k] = [l], and hence ϕ(a^k) = ϕ(a^l), and therefore ϕ is well-defined. Show that ϕ is a homomorphism: Let a^k, a^l ∈ G. ϕ(a^k*a^l) = ϕ(a^k+l) = [k+l] = [k] + [l] = ϕ(a^k) + ϕ(a^l). Therefore, ϕ is a homomorphism. Show ϕ is one-to-one: Let a^k, a^l ∈ G such that ϕ(a^k) = ϕ(a^l). Therefore, [k] = [l] and hence k ≡ l(mod n). Thus, a^k = a^l, so ϕ is one-to-one. Show ϕ is onto: Let [k] ∈ Zn. Then, a^k ∈ G and ϕ(a^k) = [k]. Therefore ϕ is onto. Thus, ϕ is an isomorphism and so G is isomorphic to Zn. source: http://www.csus.edu/indiv/e/elcek/m110a/exam3s.pdf

Let G and H be isomorphic groups. If G has a subgroup of order n, prove that H must also have a subgroup of order n.

Let f: G -> H be an isomorphism, and let G1 be a subgroup of G of order n. Consider the subset of f(G1) in H of order n. Let f(g1), f(g2) be two elements of f(G1) where g1, g2 ∈ G1. f(g1)(f(g2))^-1 = f(g1)f(g2^-1) (since f is an isomorphism) = f(g1g2^-1). g1, g2 ∈ G1 and G1 is a subgroup of G => g1g2^-1 ∈ G1 =>f(g1g2^-1) ∈ f(G1). Therefore f(g1)(f(g2))^-1 ∈ f(G1). Hence f(g1) is a subgroup of H of order n. source: https://www.chegg.com/homework-help/questions-and-answers/let-g-h-isomorphic-groups-g-subgroup-order-n-prove-h-must-also-subgroup-order-n-q10802167

Let σ=σ1⋯σm∈Sn be the product of disjoint cycles. Prove that the order of σ is the least common multiple of the lengths of the cycles σ1,...,σm.

Let k be the order of σ and let σ = T1T2...Tl be the decomposition of σ into disjoint cycles of lengths k1, k2, ....kl. Now choose any integerh. Because T1, T2,...,Tl are disjoint, it follows that σ^h = T1^hT2^h...Tl^h. Also, we know that the RHS is equal to the identity iff each individual term is equal to the identity. As such, Ti^k = e and ki divides k. As such, the least common multiple m of k1, k2, ...kl divides k. But σ^m=T1^mT2^mT2^m...Tl^m = e. Thus m divides k and thus k = m. source: http://math.mit.edu/~mckernan/Teaching/12-13/Spring/18.703/l_5.pdf

If G is isomorphic to G¯ and H is isomorphic to H⁻, show that G x H is isomorphic to G⁻ X H⁻.

Let θ denote the isomorphism G → G¯ and let ψ denote the isomorphism H ∼= H¯ . Define a map φ : G × H → G¯ × H¯ specified by φ : (g, h) 7→ (θ(g), ψ(h)) ∀g ∈ G, h ∈ H. First show φ is 1-1. Suppose that φ(g1, h1) = φ(g2, h2), then by the definition of φ we have that (θ(g1), ψ(h1)) = (θ(g2), ψ(h2)), this means that θ(g1) = θ(g2) and ψ(h1) = ψ(h2). Since θ and ψ are bijective this means that (g1, h1) = (g2, h2) and hence φ is 1-1. Again since θ and ψ are bijective, for any ¯g ∈ G¯ and any h¯ ∈ H¯ we have that ¯g = θ(g) and h¯ = ψ(h), hence for all (¯g, h¯) ∈ G¯ × H¯ we have that (¯g, h¯) = (θ(g), ψ(h)) for some g ∈ G, h ∈ H. To see that φ is a homomorphism note that φ(g1, h1)φ(g2, h2) =(θ(g1), ψ(h1))(θ(g2), ψ(h2))=(θ(g1)θ(g2), ψ(h1)ψ(h2)) =(θ(g1g2), ψ(h1h2)) =φ((g1g2, h1h2)) =φ((g1, h1)(g2, h2)). source: https://math.berkeley.edu/~mhelmer/Teaching/Math113_2017/Assignment4_Sol.pdf

Show that A10 contains an element of order 15.

Let σ = (1 2 3 4 5)(6 7 8). Because σ = (1 5)(1 4)(1 3)(1 2)(6 8)(6 7), σ is an even permutation. Therefore, σ∈A10. Now, we show that σ has order of 15: because (1 2 3 4 5) and (6 7 8) are disjoint, they commute. Therefore σ^n = (1 2 3 4 5)^n(6 7 8)^n for all integers n. Because (1 2 3 4 5) is a cycle with 5 elements, (1 2 3 4 5)^n = id iff 5 | n. Likewise, (6 7 8)^n = id iff 3 | n. Therefore, σ^n = id iff 15 | n, and thus the order of σ is 15. source: http://campus.lakeforest.edu/trevino/Spring2016/Math330/Homework4Solutions.pdf

Let σ∈Sn be a cycle. Prove that σ can be written as the product of at most n−1 transpositions.

Let σ∈S, then by theorem 9.8 σ = µ1µ2...µr with µi's being disjoint cycles of length ni. Then we get n1+n2+...+nr = n. Because µi is a cycle, µi = (a1, a2,...,ani) then by the argument after definition 9.11 we have (a1, a2, ...ani) = (a1, ani) = (a1, ani-1)...(a1, a3)(a1, a2) which implies that each µi can be written as ni-1 transpositions. So σ can be expresses as n1-1+n2-1+...+nr-1 = n -r≤n-1 transpositions. source: http://andromeda.rutgers.edu/~xiaowwan/Teaching/Math451/mid/mid2p-sol-1.pdf

Let G be a group of order 20. If G has subgroups H and K of orders 4 and 5 respectively such that hk = kh for all h ∈ H and k∈ K, prove that G is the internal direct product of H and K.

Need to prove: G = HK and H ∩ K = {e}. Prove H ∩ K = {e}: Suppose x ∈ H ∩ K. then x∈H so x^4 = e and x∈K so x^5 = e. Since x^4 = e then x^5=x^4*x. Since x^5 = e and x^5 = x, then x = e. Therefore H ∩ K = {e}. Show that HK = G: H has 4 elements and k has 5 elements so there are 20 possible elements of the form hk with h∈H and k∈K. Because H and K are subsets of G, all of these hk are elements of G. Since G has 20 elements, if all the hk's are different, then HK = G. This only fails of we have h1k1 = h2k2 with either h1≠h2 or k1≠k2. So suppose h1k1=h2k2, then h2^-1*k1 = k2, so h2^-1*h1 = k2k^-1. Therefore h2^-1*h1∈H∩K and k2k1^-1. Since H∩K={e}, the h2^-1*h1 = e and k2k1^-1 = e. Therefore h1=h2 and k1 = k2. This means that all the hk's are different so G = HK. Therefore G ≈ H x K. source: http://campus.lakeforest.edu/trevino/Spring2016/Math330/Homework6Solutions.pdf

Find the order of (6, 15, 4) in Z30 X Z45 X Z24.

Order of 6 in Z30 = 5 because 6+6+6+6+6=30 Order of 15 in 45 = 3 because 15+15+15=45 Order of 4 in 24 = 6 because 4+4+4+4+4+4=24 so order of (6, 15, 4) in Z30XZ45XZ24 = lcm(5, 3, 6) = 30. source: http://www.chegg.com/homework-help/questions-and-answers/find-order-6-15-4-z30-x-z45-x-z24-q2746738

Use Fermat's Little Theorem to show that if p = 4n +3 is prime, there is no solution to the equation x^2 ≡ -1 (mod p).

P is prime and in the format of 4n+3 Fermat's little Theorem is a^p = a (mod p) or a^(p-1) = 1 mod(p) Given: x^2 ≡ -1 mod(p) x^2 + 1 ≡ 0 mod p x^2 + 2 ≡ 1 mod p so x^2 + 2 is in the form of a^(p-1) which means that a^(4n+2) = (a^(2n+1))^2 so a^(p-1) is a square number. However, x^2+2 cannot be a square number if x is an integer. source: http://www.chegg.com/homework-help/questions-and-answers/use-fermat-s-little-theorem-show-p-4n-3-prime-solution-equation-x-2-1-mod-p--case-means-co-q3810199

What are the possible cycle structures of elements of A5? What about A6?

S5 = 5, 4+1, 3+2, 2+2+1, 2+1+1+1, 3+1+1, 1+1+1+1+1 A5 = 5, 3+1+1, 2+2+1, 1+1+1+1+1 S6 = 6, 5+1, 4+2, 4+1+1, 3+3, 3+1+1+1, 1+1+1+1+2, 2+2+2, 2+2+1+1, 2+3+1, 1+1+1+1+1+1 A6 = 6, 4+1+1, 2+2+2, 2+3+1, 2+1+1+1+1 source: https://answers.yahoo.com/question/index?qid=20080305054432AATQjyF

List the left and right cosets of An in Sn.

Sn is the collection of all permutations on n elements, hence it has n! elements (n!). An is a subset of Sn that consists of all even permutations, so the cosets are x + An where "+" is a composition of cycles. source: http://www.chegg.com/homework-help/questions-and-answers/list-left-right-cosets-following-a4-s4-b-sn-c-d4-s4-d-t-c-t-circle-group-given-z-element-c-q3866269

Find the order of (3, 4) in Z4 X Z6.

The order of 3 in Z4 is 4 and the order of 4 in Z6 is 3. So the order (3, 4) = lcm(3, 4) = 12. source: http://campus.lakeforest.edu/trevino/Spring2016/Math330/Homework6Solutions.pdf

Let G be isomorphic to H. Show that if G is cyclic, then so is H.

Suppose G is cyclic where G = <a> for some a ∈ G. We claim that H = <ψa> for y ∈ H. Then y = ψ(x) for x∈G. Because G = <a>, we can write x=a^m for m∈Z, and so y = ψ(x) = ψ(a)^m = {ψ(a,a...a) if m > 0 or ψ(a^-1...a^-1) if m<0 = {ψ(a)^m if m>0 or ψ(a^-1)^-m if m<0 = {ψ(a^m) if m>0 or ψ(a^m) if m<0 *note that [ψ(a^-1)]^m = f(a)^-1 so y = ψ(a)^m for m∈Z. Therefore, H = <ψa>.

Let G be a cyclic group of order n. Show that there are exactly ϕ(n) generators for G.

Suppose that G = <a>. Then, we claim that <a^i>=G iff i is relatively prime to n. This finishes the problem because there are exactly ϕ(n) positive integers i < n with this property. Suppose first that i is relatively prime to n and also suppose that e = (a^i)^k = a^i*k. If follows that n divides ik. But if i is relatively prime to n, we also have that n divides k. Particularly, |a^i| ≥ n. However, |any element of G| is ≤ n so |a^i| is exactly n. Now suppose conversely that a^i generates G. Particularly, this means that |a^i| is exactly n. But suppose that k > 0 divides bot i and n and we will obtain a contradiction. Therefore, (a^i)^(n/k) = (a^n)^(i/k) = e^(i/k) = e. In particular, |a^i| is less than n/k, which is a contradiction. source: https://www.math.utah.edu/~schwede/math435/HW3Sols.pdf

Let Φ : G -> be a group isomorphism. Show that Φ(x) = eh iff e = eg, where eg and eh are the identities of G and H respectively.

Suppose that Φ(x) = eh. Then, Φ(x)(Φ(x))^-1 = eh(e(h))^-1 = eh since Φ is an isomorphism then Φ(x)((x))^-1=Φ(xx^-1)=Φ(eg) = eh. Therefore, eh = Φ(x) = Φ(eg) which implies that x = eg. Now, saw we declare that x = eg and let h = Φ(g) ∈ H(such a g exists for all h ∈ H since Φ is an isomorphism). Then hΦ(x) = Φ(g)Φ(x) = Φ(gx) = Φ(geG) = Φ(g) = h = Φ(eGg)=Φ(x)Φ(g)=Φ(x)h, and thus Φ(x)=eh by definition. source: https://math.berkeley.edu/~mhelmer/Teaching/Math113_2017/Assignment4_Sol.pdf

List the left and right cosets of the subgroup: <8> in Z24.

Supposing that G is abelian, g+G = {g+h : h∈G} and G+g={h+g : h∈G}. Therefore, g+G=G+g because G is abelian, and as such the left and right cosets are equal. Since <8> = {0, 8, 16}, 0+<8> = {0, 8, 16} 1+<8> = {1, 9, 17} 2+<8> = {2, 10, 18} 3+<8> = {3, 11, 19} 4+<8> = {4, 12, 20} 5+<8> = {5, 13, 21} 6+<8> = {6, 14, 22} 7+<8> = {7, 15, 23} source: https://answers.yahoo.com/question/index?qid=20120228095944AAc7TCl

List all the elements of Z4 X Z2.

The group Z4 X Z2 has 2*4 = 8 elements (0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3) source: https://www.chegg.com/homework-help/list-elements-2-x-4-find-order-elements-group-cyclic-chapter-s.11-problem-1e-solution-9780201763904-exc

Show that U(5) is isomorphic to U(10) but U(12) is not.

U(5) = {1, 2, 3, 4} U(10) = {1, 3, 7, 9} U(12) = {1, 5, 7, 11} First, note that x^2 =1 for all x in U(12). Suppose that f is an isomorphism from U(10) onto U(12). Then f(9) = f(3*3) = f(3)f(3) = 1 and f(1) = f(1 * 1) = f(1)f(1) = 1. Thus f(9) = f(1). However, 9≠1 and as such we have a contradictions to the assumption that f is one-to-one. Therefore, U(12) is not isomorphic to U(10). U(5) = {1, 2, 3, 4} mod 5 U (10) = {1, 3, 7, 9} mod 10. Both U(5) and U(10) are cyclic of order 4, we can define a homomorphism sending one generator to another. For example, because 2 is a generator for U5 and 3 is a generator for U10, we can define an isomorphism as follows: define f: U5 -> U10 by F(2) =3 and extend multiplicatively. F(4) = F(2^2) = (F(2))^2 = 3^2 = 9 mod 10 F(3) = F(2^3) = (F(2))^3 = 3^3 = 7 mod 10 F(1) = F(2^4) = (F(2))^4 = 3^4 = 1 mod 10. *F is automatically a bijection source: https://answers.yahoo.com/question/index?qid=20140410220422AAbqQXv

Prove or disprove: U(8) ≈ Z4.

We have (3+8Z)^2 = 1+8Z (7+8Z)^2 = 1+8Z (5+8Z)^2 = 1+8Z therefore, each element 3+8Z, 5+8Z, 7+8Z of U(8) is of order 2. |1+8Z (identity element)| = 1. Thus, each element of U(8) is of order 1 or 2. Because Z is cyclic with order 4, it has an element of order 4. But no element of U(8) is of order 4, therefore U(8) is not isomorphic to Z4. source: http://www.chegg.com/homework-help/questions-and-answers/prove-disprove-u-8-z4-q16780377 https://answers.yahoo.com/question/index?qid=20140410215434AA0fJTB

List the left and right cosets of the subgroup 3Z in Z.

Z is abelian, so the left and right cosets are equal. Therefore, 0+3Z = [0] = {..., -6, -3, 0, 3, 6,...} 1+3Z = [1] = {..., -5, -2, 1, 4, 7, ...} 2+3Z = [2] = {..., -4, -1, 2, 5, 8,...} source: http://homepage.lnu.se/staff/psvmsi/2ma105/111128.pdf

Compute [(1 2)(3 4)(1 2)(4 7)]−1

[(1 2)(3 4)(1 2)(4 7)]−1 = (4 7)(3 4)(1 2)(1 2) = (4 7)(3 4) = (3 7) = (3 7 4)

Find the order of (5, 10, 15) in Z25 X Z25 X Z25

order of 5 in Z25 = 5 order of 10 in Z25 = 5 order of 15 in Z25 = 5 so lcm(5, 5, 5) = 5. source: http://campus.lakeforest.edu/trevino/Spring2016/Math330/Homework6Solutions.pdf

Find the order of (8, 8, 8) in Z10 X Z24 X Z80.

order of 8 in Z10 is 5 order of 8 in Z24 is 3 order of 8 in Z80 is 10 lcm(5, 3, 10) = 30 source: http://www.people.vcu.edu/~rhammack/Math501/test/M501T2sol.pdf

Find all possible orders of elements in S7 and A7.

orders of elements in S7: 1, 2, 3, 4, 5, 6, 7, 10, 12 orders of elements in A7: 1, 2, 3, 4, 5, 6, 7 work found here: http://www.pitt.edu/~roed/courses/430/PS3sol.pdf

An automorphism of a group G is an isomorphism with itself. Prove that complex conjugation is an automorphism of the additive group of complex numbers; that is, show that the map Φ(a+bi) = a - bi is an isomorphism from C to C.

Φ is one to one: Let (a+bi_ and (c+id) ∈ C such that Φ(a+bi) = Φ(c+id) a - ib = c - id => a = c, b = d, so a+bi = c + id Φ is onto: let a - ib ∈ C => (a - ib) ∈ C. a+ib∈C such that Φ(a-ib); for every element in C, there exists complex conjugates of the element such that the image of the complex conjugate is equal to the element. so the map Φ is an isomorphism. source: http://www.chegg.com/homework-help/questions-and-answers/automorphism-group-g-isomorphism--prove-complex-conjugation-automorphism-additive-group-co-q16155917