Exercises on the fundamental theorem of group homomorphism

  1. Prove that if G is infinite cyclic, then Aut(G) is isomorphic to ℤ2, if G is cyclic with order n, then Aut(G) is isomorphic to (U(Zn), ⋅). the group of the invertible element of ℤn.

  2. Prove that 𝓘(G) ⊲̲ Aut(G).

  3. Prove that the center of a group is a normal subgroup.

  4. Prove that a group, (G, °) other than the trivial group {e}, without proper subgroups (that is there are no subgroups of G other than {e} and G itsels) is necessarily a cyclic group with order a prime number.

  5. Determine the Aut(ℤ9) specifying whether the group is cyclic or not.

  6. Let G the abelian additive group made by all continuos functions defined on [0,1] with real values. Consider the subset

    N = {f ∈ G | f(1/7) = 0}

    Prove that N ⊲̲ G. Determine to which known group the quotient G/N is isomorphic to.

  7. In the linear group GL2(C) consider the subgroup H generated by two matrices (0, i; i,1 0) and (0, 1; −1, 0). Determine the order of this subgroup, write its elements and their orders and say whether the group is cyclic or abelian. Find the element g having two as order; the subgroup generated by g is normal in H or in GL2(C)?.

  8. Find all group homomorphisms from the symmetric group S3 into (Z15, +).

  9. Prove that the function that maps an element of a group G to its inverse is an automorphism iff G is abelian.

  10. Let H the subset of S4:

    H = {id, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)}

    Show that H is a normal subgroup of S4. Determine the quotient group S4/H.

  11. Determine the inner automorphism group of A4 (alternating subgroup of S4).

  12. Determine the automorphism group of Z15. Is it a cyclic group? Which automorphism is inner?.

  13. Let D4 the group of the symmetries of the square and N its only normal subgroup with order 2. Determine the quotient D4/N.

  14. Let G be any group. If a, b ∈ G, then the commutator of a and b is the element aba−1b−1. The commutator subgroup (also called a derived group) of a group G is the subgroup generated by the commutators of its elements, and is commonly denoted G' or [G,G]. Prove that G' ⊲̲ G. Prove that G' = {e} ⇐⇒ G is abelian.

  15. Is it true that cyclic groups are always abelian?

Solutions

  1. Let G = ⟨g⟩ be infinite cyclic. An automorphism on a cyclic group is completely determined by its action on the generator g, hence we have φ(g) = gi for some i ∈ ℤ. Since φ is surjective, we must have g = φ(gk) for some k ∈ Z. Thus g = φ(gk) = φ(g)k = gik ⇒ ik = ±1. Each of these two values give origin to an autormorphism thus Aut(G) ≃ ℤ2

    Let now G = ⟨g⟩ a cyclic group with order n. An automorphism φ of G maps the generator g to another generator gk of G, for some k ∈ ℤ such that (k,n) = 1, 0 < k < n. The application φ ∈ Aut(G), φ(g) = gk, we the restrictions mentioned before, is an automorphism of G. The map

    Ψ: Aut(G) ⟶ (U(Zn))
    Ψ(φ) = [k]

    is bijective and preserves the operations, hence an isomorphism and Aut(G) ≃ (U(Zn)).

  2. Let φ any automorphism and let fx = x ⋅ g ⋅ x−1 be an inner automorphism. Then

    (φ ⋅ fx ⋅ φ−1)(g) = φ(fx(φ−1(g)) = φ(x ⋅(φ−1 (g)) ⋅ x−1).

    Since φ is a homomorphism recalling that φ(x−1) = φ(x)−1, this will simplify to

    φ(x ⋅(φ−1 (g)) ⋅ x−1) = φ(x) ⋅ φ((φ−1 (g)) ⋅ φ(x−1) = φ(x) ⋅ g ⋅ φ(x)−1 = fφ(x)

    So φ ⋅ fφ(x) ⋅ φ−1 = fφ(x), is an inner automorphism of G and by Proposition 7.11.7 𝓘(G) is a normal subgroup of Aut(G).   ■

  3. Let z1, z2∈ Z(G), then z1x = xz1 and z2x = xz2 for all x ∈ G.

    We have z2x = xz2, for all x ∈ G

    z2–1(z2x)z2–1 = z2–1(xz2)z2–1
    xz2–1 = z2–1x   ∀x ∈ G
    z2–1 ∈ Z(G)

    Now

    (z1z2–1)x = z1(z2–1x) = z1(xz2–1) = (z1x)z2–1 = (xz1)z2–1 = x(z1z2–1) ⇒ z1z2–1 ∈ Z(G)

    Thus, z1, z2 ∈ Z(G) ⇒ z1z2–1 ∈ Z(G). Therefore, Z(G) is a subgroup of G.

    Now, we shall show that Z(G) is a normal subgroup of G. Let x ∈ G and z ∈ Z(G), then

    xzx–1 = (xz)x–1 = (zx)x–1 = z ∈ Z(G)

    Thus, x ∈ G, z ∈Z(G) ⇒ xzx–1 ∈ Z(G). Therefore, Z(G) is a normal subgroup of G.   ■

  4. Suppose G has no proper subgroups. Suppose that g ∈ G, g ≠ e, it must necessarily be G = ⟨g⟩, because is given that G has no proper subroups. It cannot be an infinite cycle group since it would be G ≃ ℤ, since the group (ℤ, +) contains the subgroups ℤn, thus it is a finite cycle groups. It is prime order, otherwise it would count for every divisor of its order a subgroup. (Notice that we excluded G = {e} because is infinte cyclic of non-prime order.   ■

  5. Being ℤ9 a finite cyclic group of order 9 it is isomorphic to U9 = {1,2,4,5,7,8}; the automorphisms are α1,α2,α4,α5,α7,α8, a cyclic group with order six. The automorphism defined as αk(a) = ak e.g. α4 sends 1 → 4, 2→8, 3→3, 4→7, 5→2, 6→6, 7→1, and 8→5 (2→8 since 24 = 2+2+2+2 = 8).   ■

  6. We define a homomorphism φ: G → ℝ as φ(f) = f(1/7), this map is an epimorpshim since injectivity does not hold: two functions can assume the same value in 1/7 and be still different. All functions belonging to the same class assume the same value in 1/7. Ker φ = N. By the fundamental isomorphism theorem G/N is isomorphic to ℝ.  ■

  7. The elements of the group are

    1. (0, i; i, 0)

    2. (0, 1; -1, 0)

    3. (0, i; i, 0)−1 = (0, -i; -i, 0)

    4. (0, 1; -1 0)−1 = (0, -1; 1 0)

    5. (0, i; i, 0)2 = (0, -1; -1, 0)2 = (-1, 0; 0 -1)

    6. (-1, 0; 0 -1)2 = (1, 0; 0, 1)

    7. (0, i; i, 0) (0, 1; -1, 0) = (-i, 0; 0, i)

    8. (0, 1; -1, 0) (0, i; i, 0) = (i, 0, 0 -i)

The groups has thus order 8. The generators have both order 4. It's not abelian: ab ≠ ba, where a,b are your two generators).

(-1, 0; 0 -1) is the element having order 2. The inverse of (-1, 0; 0 -1) is the elements itself, H = {e,c} i'm calling it c to avoid writing it all out. We have that c = −1*I, and A*I=I*A for all A ∈ H. N is normal in H if ∀h ∈ H, hN = Nh. Thus {e,c} is normal both in H and GL2(C). ■

  • Let the homomorphism φ: S3 → (Z15,+). The order of ϕ(a) divides the order of a. The three elements of S3 of period 2 are the transpositions (1,2), (1,3), (2,3). Since there are no elements of Z/15Z of order 2, then they are mapped to the null element of Z15, φ(a) = [0], since [0] is the only element of this additive group of order 1. Furtheremore any permutation can be expressed as the product of transpositions, suppose c is an element of order 3 in S3 i.e. one of {(1,2,3), (1,3,2)}, then φ must fulfill the requirement c = ab, with a,b transposition of S3, thus φ(c) = φ(ab) = φ(a) ⋅ φ(b) = [0] ⋅ [0]. This implies that the only homomorpshim possible is the trivial one.   ■

  • Given an automorphism ϕ: G → G defined by ϕ(x) = x−1 we have ϕ(xy) = ϕ(x) ϕ(y) = (xy)−1 = x−1y−1 which is true only if the group is abelian. Conversely suppose that the application ϕ, defined by ϕ(x) = x−1 is an automorphism. Then ϕ(xy) = (xy)−1 = x−1y−1, ∀x,y ∈ G, which implies that G is abelian.   ■

  • We could check directly whether, σδσ−1 ∈ H, ∀σ ∈ S4 and δ ∈ H. We use the fact that for every σ,δ ∈ Sn, the permutations δ and δ = σδσ−1 have the same cycle type (when they are conjugates); δ is of cycle type 2,2, and hence σδσ−1 is also of cycle type 2,2. But there are only three elements of cycle type 2,2 in S4 and H contains all of them. Thus σδσ−1 ∈ H. The cosets are H,

    (123)H = {(123), (134), (324), (142)},   (132)H = {(132), (234),(124),(143)},  (12)H = {(12),(34),(1324),(1423)},
    (13)H = {(13),(1234),(24),(1432)   (23)H = {(23),(1342),(1243),(14)}

    Since every element of S4 has already appeared in one coset above, we know that there are no more cosets. Alternatively, we know the number of cosets of H is |S4 | /|H| = 6, and we have already found 6 distinct cosets.).

    Define φ:S3→S4/H by φ(σ) =σH, where we are thinking of an element σ ∈ S3 as an element of S4 via its cycle notation. For instance, φ((12)) = (12)H. From the calculation of distinct cosets, it clear that this is an isomorphism.  ■

  • We've that 𝓘(A4) ≃ G/Z(A4). By the definition of center, If we fix a non-identity element of A4, we find at least one element of A4 that does not commute with it; This means that Z(A4) = {e} is trivial thus 𝓘(A4) ≃ A4.   ■

  • From exercise 1) we know that a finite group is isomorphic to U(Zn), then Aut(Z15) = U(Z15) = {[1]15, [2]15, [4]15, [7]15, [8]15, [11]15, [13]15, [14]15}. It's not a cyclic group since there's no element with order 8. Being Z15 abelian, the identical automorphism is the only inner automorphism.   ■

  • N = {e, r2}. D4 = {1, r, r2, r3, s, rs, r2s, r3s}. First notice that Z(D4) = N (see Cayley table of D4). Second by Lagrange's theorem we have |G|/|H = [G : H] = 8/2 = 4. Since there are only two gorups of order 4 namely: Z/4Z and V4 (klein group), the first is cyclic and the second is not, the quotient D4/Z(D4) is isomorphic to one of them. To find out, let's list the right cosets of Z(D4) in D4:

    Z(D4)r = {r, r3} = Z(D4)r3 = {r, r3}
    Z(D4)s = {s, r2s} = Z(D4)r2s = {s, r2s}
    Z(D4)rs = {rs, r3s}
    Z(D4)r3s = {rs, r3s}

    It is now apparent that D4/Z(D4) is not cyclic (neither of the cosets without s can generate the ones with s, and both the ones with s are of order 2).   ■

  • We must prove that ∀x ∈ G' and ∀g ∈ G it results gxg−1 ∈ G' this means that the conjugate of a commutator is yet a commutator. Let x = aba−1b−1. Then

    gxg−1 = g(aba−1b−1)g−1 = gag−1gbg−1ga−1g−1gb−1g−1 = (gag−1)(gbg−1)(ga−1g−1) (gb−1g−1) = (gag−1)(gbg−1)(gag−1)−1)(gbg−1) ∈ G'

    b) G' = {e} ⇐⇒ xyx−1y−1 = e ∀x,y ∈ G ⇐⇒ xy = yx ∀ x,y ∈ G that is iff G is abelian.  ■

  • Yes, a cyclic group is abelian. Here is why. A cyclic group is generated by one generator, let's call this g. Now if a = gm and b = gn are two elements of the group, then ab = gm gn = gn gm = ba (since gmgn = gm+n = gn+m = gngm).

    The converse is not true; Klein four group is an example of group that is abelian but not cyclic.

    K4 = {e,a,b,c}
    ⟨e⟩ = {e} ≠ K4
    ⟨a⟩ = {e,a} ≠ K4​
    ⟨b⟩ = {e,b} ≠ K4
    ⟨c⟩ = {e,c} ≠ K4​

      ■

  • . Then as ϕ(a+nZ) = a+mZ. Since m∣n, then nZ ⊆ mZ we see that ϕ is surjective. U(Z_n) = {[x] ∈ ℤn: gcd(x,n) = 1} U(Z_m) = {[y] ∈ ℤn: gcd(y,m) = 1} For any a+nZ ∈ Z/nZ we pick a+mZ ∈ U(Z/mZ) such that (a,n)=1 and so the canonical projection ψ:Z→Z/mZ induces a homomorphism ψ¯:Z/nZ→Z/mZ that has the same image as ψ, in particular it is surjective. m∣n. Then nZ ⊆ mZ

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