sign in sign up
<<
Home , Canonical map, Coset, Group homomorphism, Group isomorphism, Quotient group

(c)Karen E. Smith 2018 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. Math 412. Groups and the First Theorem

Fix a (G, ◦). Let N denote a normal of G. Let G/N denote the set of all N-.

DEFINITION/THEOREM: The operation gN ?hN = ghN on G/N is a well-defined , that defines a group structure on G/N. This is called the “G N.”

Remember! The elements of G/N are sets—more precisely, they are N-cosets (subsets of G). Our operation ? combines two of these sets into a third; be sure you know exactly how. The operation ? is not well-defined if N is not normal!

π THEOREM/DEFINITION: The map G −→ G/N sending g 7→ gN is a surjective group homomor- phism, called the canonical quotient map.

φ FIRST ISOMORPHISM THEOREMFOR GROUPS: Let G → H be a surjective group with K. Then K is a of G and G/K ∼= H. More precisely, the map

φ G/K −→ H gK 7→ φ(g) is a well-defined .

A.WARMUP: (1) Make sure everyone in your group can define normal subgroup. (2) Explain a good general strategy for showing a given subgroup K is normal in G that does not involve writing out all left and right cosets. [Hint: Remember Theorem 8.11.] (3) In the definition of quotient group above, are we working with left or right cosets? How do you know? Is there a difference? (4) Prove the Theorem above stating that the canonical quotient map is a surjective group homomor- phism. Find its kernel. (5) The two theorems above, put together, say that normal are the same as kernels of group . Explain.

B. Let G = (Z12, +). Let H = {[0], [3], [6], [9]}. (1) Give a one-line proof that H is a normal subgroup of G. (2) Before computing anything, use Lagrange’s theorem to predict the structure of the quotient group G/H. (3) List out all twelve elements of G, partitioned in an organized way into H-cosets. [Why have I dropped the adjective left modifying ?] (4) Make a table for the operation of group G/H. Good notation is critical: be sure to choose a reasonable name for each element in G/H. (5) Describe the canonical quotient map π : G → G/H explicitly in this example. That is, say where each of the 12 elements of G = Z12 is sent. What is the kernel of π? (1) First, we should check it is a subgroup: This is easy, since its elements are of the form [3k]12 where k ∈ Z. This is obviously non-empty, and closed under sums and (additive) inverses. To see that it is normal, we use the fact that G = Z12 is abelian: since gh = hg for all g, h ∈ G, this in particular means that gh = hg for all g ∈ G and all h ∈ H. So the sets gH and Hg are the same. (2) There are twelve elements of G, which (abusing notation) we denote by the digits 0, 1,..., 11. We can list them out grouping them by cosets as follows: 0 3 6 9 1 4 7 10 2 5 8 11 where each row here consists of the elements in a coset. The first row consists of the elements in the coset H, we we denote He, the second row consists of the elements in the coset H+, 1 which we denote H1 and the third row consists of the elements in the coset H + 2 which we denote H2. [One group used the notation [[0]], [[1]], [[2]], respectively, which I thought showed a decent understanding of the situation. What are they getting out with the “double bracket”?]. [Another group used the notation H, L, M, which I also liked, because it removes the artificial bias in my notation—note that I could have just as well used H4 = H7 = H10 instead of H1 to denote the coset {1, 4, 7, 10}. It is important to realize that all the elements of this set are on equal standing as far as the group structure goes.] (3) The table is ? H0 H1 H2

H0 H0 H1 H2 H1 H1 H2 H0 H2 H2 H0 H1 (4) The G → G/H sends each element of G to the coset it lives in. Pictorially: 0 3 6 9 {0 3 6 9} 1 4 7 10 −→ {1 4 7 10} 2 5 8 11 {2 5 8 11} where the left side above shows the 12 different elements of the source G and the right side above show the THREE elements (each is a set but each is ONE ELEMENT) of the target G/H. The map takes the elements of G to the set they belong to. We can think of the canonical quotient map as the process of drawing the set-brackets.

× C. Let G = (Z15, ×). Let N = h[−1]i. Let K be the subgroup generated by [4]. (1) Compute |G|, |N|, |K|. Compute [G : N] and [G : K]. Are N and K normal? (2) List out all eight elements of G, partitioned in an organized way into N-cosets. (3) Make a table for the operation in the group G/N. (4) List out the elements of G again, but now partitioned into K-cosets. (5) Make a table for the quotient group G/K. (6) Is G/K ∼= G/N? (7) Describe the canonical quotient maps G → G/N and G → G/K. List each out explicitly where each elements goes. Describe the kernel of each.

(1) |G| = 8, |N| = |K| = 2. By Lagrange’s theorem, [G : N] = [G : K] = 2. Both N and K are normal because G is abelian, and all subgroups of an are normal. (2) The group G contains the 8 elements (abusing notation) {±1, ±2, ±4, ±7}. This way of writing them organizes them pretty well into cosets, since each coset contains one element and its (additive) inverse. (3) The table is ? ±1 ±2 ±4 ±7 ±1 ±1 ±2 ±4 ±7 ±2 ±2 ±4 ±7 ±1 ±4 ±4 ±7 ±1 ±2 ±7 ±7 ±1 ±2 ±4 (4) {{1, 4}, {2, 8}, {7, 13}, {11, 14}}. The four cosets will be called [[1]], [[2]], [[7]], [[11]]. (5) The table is ? [[1]] [[2]] [[7]] [[11]] [[1]] [[1]] [[2]] [[7]] [[11]] [[2]] [[2]] [[1]] [[11]] [[7]] [[7]] [[7]] [[11]] [[1]] [[2]] [[11]] [[11]] [[7]] [[2]] [[1]] (6) The quotient groups both have 4 but are not isomorphic. We see that G/K has every element its own inverse (witness the diagonal of identity elements). So it is a Klein 4-group. The other is cyclic of order 4. (7) In both case, the quotient map “globs together” the elements of G that are in the same coset. So for for G → G/N for example, each element of G is taken to the symbol ±g denoting the two-element set {g, −g}. Working mod N is awesome for people who tend to make sign errors, because modulo N identifies each element of g with its negative. Likewise, the map G → G/K just assigns each element to the coset containing it: it makes all the elements of each fixed coset “the same”.

D. Use the first isomorphism theorem to prove the following:

(1) For any field F, the group SLn(F) is normal in GLn(F) and the quotient GLn(F)/SLn(F) is isomorphic to F×. (2) For any n, the group An is normal in Sn and the quotient Sn/An is cyclic of order two.

× (1) Easy peasy: The map GL2(F) → F is a surjective . Its kernel is SL2(F). ∼ × So by the first isomorphism theorem, the quotient GL2(F)/SL2(F) = F . ∼ (2) The sign map Sn → {±1} is a homomorphism with kernel An. So the quotient Sn/An = {±1}, which is a of order two.

E. Let H be the subgroup of Z × Z generated by (5, 5). (1) Find an element of finite order and an element of infinite order in the quotient Z × Z/H. (2) Prove that ψ : Z × Z → Z × Z5 (m, n) 7→ (m − n, [n]5) is a group homomorphism. (3) Prove that ψ is surjective and compute its kernel. (4) Show that (Z × Z)/H is isomorphic to Z × Z5.

(1) The coset (1, 1) + H has order five, so finite. The coset (1, 0) + H has infinite order, since (n, 0) ∈/ H for any n. (2) We check ψ((a, b)+(m, n)) = ψ(a+m, b+n) = (a+m−(b+n), [b+n]5) = ((a−b)+(m−n), [b]5 +[n]5) = ((a − b), [b]5) + ((m − n), [n]5) = ψ(a, b) + ψ(m, n). (3) Take any (m, [a]5) in the target. Then ψ(m + a, a) = (m, [a]5), so the map is surjective. The kernel is H. Say (5n, 5n) ∈ H. Then ψ(5n, 5n) = (5n − 5n, [5n]5) = (0, [0]5), so H ⊂ ker ψ. Conversely, say ψ(m, n) = 0. Then (m−n, [n]5) = (0, [0]5), so m = n, and also n (hence both) is a multiple of 5. So (m, n) = (5d, 5d) ∈ H. QED. (4) Immediate from First Isom! ∼ (5) The sign map Sn → {±1} is a homomorphism with kernel An. So the quotient Sn/An = {±1}, which is a cyclic group of order two.

× F. Consider the map (C , ×) → (R>0, ×) sending z to |z|. Show that this is a surjective group homo- with kernel the S1. Describe the quotient C×/S1. We’ve checked before that |zw| = |z||w|, which shows the map is a group homomorphism. It is surjective, since for × 1 any a ∈ R>0, |a + 0i| = a. The kernel is {z ∈ C | |z| = 1} = S . The First Isomorphism theorem guarantees that × 1 ∼ C /S = R>0.

G. In S3, consider the subgroup K generated by (12).

(1) Write out the elements of S3, partitioned into right K-cosets in an organized way. (2) Write out the elements of S3, partitioned into left K-cosets in an organized way. Use this to explain why h(1 2)i is not normal. (3) Explore what goes wrong if we try to define the operation ? on left cosets.

(1) There are three: K = {e, (1 2)}, K(2 3) = {(2 3), (1 2 3)} and K(1 3) = {(1 3), (1 3 2)}. (2) Also three: K = {e, (1 2)}, (2 3)K = {(2 3), (1 3 2)} and (1 3)K = {(1 3), (1 2 3)}. Since K(1 3) 6= (1 3)K, the subgroup K is not normal. (3) Since (2 3)K = (1 3 2)K, we should have (2 3)K?(1 3)K = (1 3 2)K?(1 3)K. But according to our formula, (2 3)K? (1 3) = (2 3 1)K whereas (1 3 2)K? (1 3)K = (1 2)K = K. It is not true that (2 3 1)K = eK, so the operation is not well-defined. This is just one of many ways to see that the operation is not well-defined.

I.PRODUCTSAND QUOTIENT GROUPS: Let K and H be arbitrary groups and let G = K × H. (1) Find a natural homomorphism G → H whose kernel K0 is isomorphic to K. (2) Use the first isomorphism theorem to prove that G/K0 ∼= H. (3) Prove the K-cosets are all of the form K × h for h ∈ H. (4) Without using the First Isomorphism theorem, prove directly that G/K0 is isomorphic to H.

(1) Consider the map π : K × H → H sending (k, h) 7→ h. This is easily checked to be a group homomorphism: 0 π((k1, h1)(k2, h2)) = π(k1k2, h1h2) = h1h2 = π(k1, h1)π(k2, h2). The kernel is K = {(k, eH ) | k ∈ K}, 0 and the map K → K sending k 7→ (k, eH ) is obviously a group homomorphism. (2) The map in (1) is clearly surjective, and its kernel is K0, so by the first isomorphism theorem, G/K0 =∼ H. (3) K0 is a kernel, so normal, so we don’t distinguish between left and right cosets. Every coset (k, h)K0 is 0 0 the same as (eK , h)K since (k, h) = (eK , h)(k, eH ) ∈ (eK , h)K , and we know two cosets are either the 0 same or completely disjoint. But then the elements of (eK , h)K can be written {(eK , h)(k, eH ) | k ∈ K} = {(k, h) | k ∈ K} = K × h. 0 0 (4) The map H → G/K sending h 7→ K × h = (eK , h)K is obviously surjective, and it respects the group product. So we need only check that it is injective. For this, we compute the kernel is eH .