Math 402 Assignment 4. Wednesday, October 23, 2013.

1. Let Cn be a cyclic group generated by an element a of order n, and let Cm be a cyclic group generated by an element b of order m. Suppose that m divides n.

(a). Describe an injective homomorphism φ : Cm → Cn.

Solution. Let Cm be a cyclic group of order m generated by an element b. Let G be a group and let c be an element of G. t t A mapping φ : Cm → G, defined by φ(b ) = c , for t between 0 and m-1, is a homomorphism if and only if the order of the element c is a divisor of m, and the homomorphism is injective when the order of c is exactly m.

So, if m divides n, and G = Cn is the cyclic group of order n, generated by an element a, n/m then to get an injective homomorphism φ : Cm → Cn, we can take c to be a , an element of order m.

(b). Describe a surjective homomorphism δ : Cn → Cm, and find a generator of the kernel of δ, a cyclic subgroup of Cn.

Solution. t t To get a homomorphism δ : Cn → Cm when m divides n, δ(a ) = c for some element c of Cm, we need the order of c to be a divisor of the order of a, as we said in part (a). Since b has order m which divides n, we can take c to be b, i.e., δ(at) = bt. That homomorphism is surjective since the image of δ consists of the powers of b, which gives all of Cm since b generates Cm.

To find the kernel of δ: at ∈ Ker(δ) if and only if δ(at) = bt = e if and only if m|t, since m b has order m. Therefore, the kernel of δ is the cyclic subgroup of Cn generated by a , an element of order n/m.

1 2. Suppose that G = {g1, g2, ..., gn} is a finite group of n elements. Let Sn be the permutation group of the set {g1, g2, ..., gn}. For each gj ∈ G, define a mapping φ(gj): G → G by φ(gj)(gi) = gjgi, i.e., φ(gj) combines gj with each element gi of G.

This problem is Cayley’s Theorem.

(a). Show that φ(gj) is a bijection, i.e., an element of Sn.

Solution.

For the group G of n elements, let φ : G → Bij(G, G) = Sn be the homomorphism from G to the group of bijective mappings from G to G, defined at each element g of G by the rule φ(g): G− > G is φ(g)(h) = gh, for h in G. φ(g) is bijective since its inverse map is φ(g−1).

(b). Show that φ : G → Sn is an injective homomorphism of groups.

Solution. To show that φ is a homomorphism, we need to show that φ(gigj) = φ(gi)φ(gj) as permutations of G. For each element g ∈ G, we have

φ(gigj)(g) = (gigj)(g) = gi(gjg) = φ(gi)(gjg) = φ(gi)(φ(gj)(g)) = (φ(gi)φ(gj))(g).

To show that φ is injective, we need to show that the kernel of φ consists of just the identity element e of G. Let g ∈ G be an element of the kernel of φ, i.e., φ(g) = I. Applying that equation to e, we get φ(g)(e) = I(e), i.e., ge = e, i.e., g = e.

3. Take D4, the symmetry group of the square, which consists of 4 rotations and 4 reflections. Take the set of the four lines through the origin at angles 0, π/4, π/2 and 3π/4, numbered 1,2,3,4. Let S4 be the permutation group of the set of four lines. Each symmetry s of the square in particular gives a permutation φ(s) of the four lines, as you should confirm.

Define a mapping φ : D4 → S4 whose value at the symmetry s ∈ D4 is the permutation φ(s) of the four lines. Such a process always defines a homomorphism.

(a). Find the kernel of φ.

(b). Find the image of φ.

Solution. D4, the eight element symmetry group of the square, permutes the set T of four lines consisting of the two axes and the two diagonal lines. Those permutations gives us a homomorphism φ : D4− > S4, where S4 is the group of bijections from T to T. Let us

2 number the x-axis as line 1, the diagonal from the first to the third quadrant as line 2, the y-axis as line 3, and the diagonal from the second to the fourth quadrant as line 4.

Let us see what the elements of D4 do to those numbered lines: the 90-degree rotation exchanges lines 1 and 3 (the axes) and exchanges lines 2 and 4 (the diagonals). Hence, it gives the element (13)(24) of S4. The 270-degree rotation gives the same permutation. The 180-degree rotation gives the identity permutation.

Now for the reflections in D4: the reflection in the x-axis take the x-axis to itself and the y-axis to itself, and it exchanges the two diagonals, i.e., it give the element (24) of S4. Similarly, the reflection in the y-axis gives the element (24) of S4. The reflection in the diagonal numbered 2 gives the element (13) of S4, as does the reflection in the diagonal numbered 4. From these calculations, we see that the image of φ is the subgroup {I, (13), (24), (13)(24)} o of S4, and the kernel of φ is the subgroup {I, 180 } of D4.

4.

(a). Find a surjective homomorphism φ : S4 → S3 by permuting the 3-element conjugacy class of (12)(34) in S4. How many elements of S4 are in the kernel of φ? Which elements of S4 form the kernel of φ?

Solution. If G is a group and a ∈ G, then the elements of G permute the elements of the −1 conjugacy class Ea via conjugation since if g ∈ G and b = cac ∈ Ea, for c ∈ G, then −1 −1 −1 −1 gbg = gcac g = (gc)a(gc) ∈ Ea. Thus, we get a homomorphism φ : G → P erm(Ea) −1 where φ(g) is the permutation of Ea defined by φ(g)(b) = gbg .

Consider the three element conjugacy class E(12)(34) = {(12)(34), (13)(24), (14)(23)} of S4. Via conjugation, we get a homomorphism φ : S4 → P erm(E(12)(34)) = S3, where P erm(E(12)(34)) is the group of permutation of the conjugacy class of (12)(34).

Using the fact that H = I, (12)(34), (13)(24), (14)(23) is an abelian subgroup (you can check that it is abelian), it follows that H is contained in the kernel of φ. Below , we will see that it is the whole kernel of φ.

Let S3 be the subgroup of the elements of S4 that fix the number 4. Let us check that no element of S3 lies in the kernel of f except for the element I: the element (12) of S3 conjugates ((13)(24) to (14)(23), so it is not in the kernel. Similarly, (13) and (23) are not in the kernel. (123) conjugates (12)(34) to (14)(23), and so, it is not in the kernel. Since (123) is not in the kernel (a subgroup), its inverse (321) is not in the kernel either.

Thus, φ gives an injective homomorphism from the subgroup S3 of S4 to S3. Hence, that injective homomorphism from S3 to S3 must be surjective, and so, φ : S4 → S3 is surjective

3 since φ is surjective on the subgroup S3 . We have the general rule for a homomorphism 0 f : G → G ◦(G) = ◦(kerf) ◦ (Imf). Here, ◦(S4) = ◦(kerφ) ◦ (Imφ) = ◦(kerφ) ◦ (S3), by the surjectivity of φ. Hence, there are 4 elements in the kernel, those we gave earlier in this discussion.

In general, if H and K are subgroups of a group G such that H ∩ K = {e}, then ◦(HK) = ◦(H) ◦ (K), since all the left-H cosets kH, k ∈ K, are distinct. In S4, K and S3 above intersect in I; hence, KS3 has 6 · 4 = 24 elements, i.e., S4 = KS3. That shows how S4 can be built from subgroups.

(b). Let S3 ⊂ S4 be the 6 elements of S4 that fix the number 4, and K ⊂ S4 be the subgroup K = {I, (12)(34), (13)(24), (14)(23)}. Using the 6 subgroups of S3, find the 6 subgroups of S4 that contain the subgroup K.

Solution. We showed in class that if K is a normal subgroup of G and H is a subgroup of G, then HK is a subgroup of G. Here, K = {I, (12)(34), (13)(24), (14)(23)} is a normal subgroup of S4 since {(12)(34), (13)(24), (14)(23)} is a conjugacy class (and so its elements −1 are permuted by the elements a of S4, which gives us the condition aKa = K for normal- ity). Then for S3, the subgroup of elements of S4 that fix the number 4, if we take the six subgroups H of S3, then HK gives us 6 subgroups of S4.

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