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MATH 410 - ABSTRACT DISCUSSIONS - WEEK 8

CAN OZAN OGUZ

1. Theorems In theory, there are three main . They all follow from the first isomorphism theorem. Let’s try to develop some intuition about these theorems and see how to apply them. We already say the first isomorphism theorem in the 6th discussion: First Isomorphism Theorem: Let Φ : G → H be a group . Then G/ker(Φ) ' Im(Φ) If you just consider the map Φ, it doesn’t have to be injective or surjective. To ensure surjectivity, we restrict the target to the of Φ. Injectivity is a bigger problem. Recall that a homomorphism is injective if and only if its is trivial. But it may be the case that kerΦ contains many elements. Our strategy is to consider another group where ker(Φ) becomes a single element. This is exactly the group G/ker(Φ).

ker(Φ) G H

1G 1H Im(Φ) g1 h1 g 2 g3

g4

g5

G/ker(Φ) 1G/ker(Φ) Second Isomorphism Theorem: Let K,H ≤ G. Under some assumtions, we have HK/H ' K/(K ∩ H) The assumptions we need are to make sure we have a group on the right hand side, and on the left hand side. For example for two H,K, their product HK is not a group in general. We need one of them to be normal. Also we will quotient out by H to get a group, hence H should be normal in G. You should check if this guarantees K ∩ H is normal in K.

Date: October 12th, 2017. 1 2 CAN OZAN OGUZ

Let’s just focus on the right hand side of the equation first: HK/H. We multiply K by H, and then divide by H. If we were doing this with , they would just cancel each other. With sets and groups, it is a bit different. The reason is that intersection of H and K doesn’t contribute to the product HK. For example, if we take H = K, then HK = KK = K, and we didn’t gain anything by multiplication. However we will divide by H regardless of how big the product became. If we had H ∩ K = 1, then second isomorphism theorem tells us HK/H ' K , so multiplication and cancel out. But with a non-trivial intersection, we need to quotient K by K ∩ H. Below is a diagrammatic picture of what is going on. Remember H is normal of G, so I will draw it as a vertical rectangle all the way to the left. It will partition G into its cosets. However K doesn’t need to be normal. So I will draw it as a horizontal rectanlge. It has to contain 1G. H aH bH cH G H ∩ K h0 = k00 1G k1 k2 K

h1 h1k1

h2 h2k2

HK In the above picture, you can see that H partitions HK into cosets just as H ∩K partitions K into cosets. This is exactly the content of the second isomorphism theorem. Multiplication of cosets of H in HK has the same multiplication structure as cosets of H ∩ K in K. Example Let G = Z, H = 6Z, K = 8Z. Since G is abelian, H is a . In general, K doesn’t need to be normal in G to apply the second isomorphism theorem. In our case, K is also normal since every subgroup of an is normal. Let’s start by understanding HK and H ∩ K:

HK = 6Z8Z = {a + b : a ∈ 6Z, b ∈ 8Z} = {6k + 8` : k, ` ∈ Z} = {gcd(6, 8)n : n ∈ Z} = gcd(6, 8)Z since we saw in chapter one that the linear combinations of two integers is multiples of their gcd. Also

H ∩ K = 6Z ∩ 8Z = lcm(6, 8)Z Therefore by second isomorphism theorem, we get an isomorphism gcd(6, 8)Z/6Z ' 8Z/lcm(6, 8)Z MATH 410 - DISCUSSIONS - WEEK 8 3

The two groups above are isomorphic, in particular have the same order:

6 lcm(6, 8) = gcd(6, 8) 8

or

gcd(6, 8)lcm(6, 8) = 6 × 8

Of course this formula works for any pair of positive integers. Third Isomorphism Theorem Let H,K ≤ G. If we assume the above ex- pressions are groups, then . G/K H/K ' G/H

It is easy to see that we need H and K to be normal subgroups, and moreover we need K ≤ H. Check that these assumptions are enough, that is with these assumptions, K is normal in H and H/K is normal in G/K. The of third isomorphism theorem is simpler. If we quotient out the nominator and denominator with K, they cancel. Exercise: Try to understand the above statement by drawing a diagram for G. Remember, H should partition G, and K should partition both H and G.

Example Let G = R, H = Q and K = Z. By third isomorphism theorem, we have

. R/Z Q/Z ' R/Q

We saw earlier that R/Z gives us a circle via the homomorphism : R → R2, f(θ) = (cos(2πθ), sin(2πθ)). So we have

1. S Q/Z ' R/Q

We don’t have an understanding of the groups Q/Z and R/Q yet, however if we understand one of them, third isomorphism theorem is going to let us understand the under. Also notice that the group Q/Z is a normal subgroup of S1. Correspondance Theorem: All isomorphism theorems require taking a quo- tient by a normal subgroup. Taking the quotient results in a smaller group. So these isomorphism theorems relate larger groups to smaller groups via . Then we should think about the following question: Question: By studying G/H, how much information do we get about G? Suppose we have a chain of subgroups of G: 4 CAN OZAN OGUZ

G G/H ≤

K1 ≤

K2 ≤

K3 ≤

H 1G/H ≤

K4 The image of H under the quotient map is going to be the identity element of the . Therefore you can easily tell that there isn’t enough left for the image of K4 under the quotient map. But for K3 the situation is different. It’s image will be K3/H. This is a subgroup of G/H. Similarly K1/H and K2/H will be subgroups of G/H. They will have the same inclusion : G G/H ≤ ≤

K1 K1/H ≤ ≤

K2 K2/H ≤ ≤

K3 K3/H ≤ ≤

H 1G/H ≤

K4 This is the first part of the correspondance theorem. We have an inclusion preserving between the sets

bijection {Subgroups of G contatining H} ←−−−−→{Subgroups of G/H}

Furthermore, Ki is normal in G if and only if Ki/H is normal in G. Moreover we have [G : Ki] = [G/H : Ki/H]. So we can study subgroups of G bu studying subgroup of G/H. Example Let G = GLn(R), the group of n by n invertible matrices. Recall that ∗ the determinant map was a surjective from GLn(R) to R , with kernel the special linear group SLn(R). Then by first isomorphism theorem, we get ∗ GLn(R)/SLn(R) ' R and R∗ is an abelian group. Therefore all of its subgroups are normal. This tells us an important fact about subgroups of GLn(R) containing SLn(R): They are normal in GLn(R) by the second part of correspondance theorem. MATH 410 - ABSTRACT ALGEBRA DISCUSSIONS - WEEK 8 5

2. Direct Products of Groups Now we will focus our attention to an important construction in : of two groups. This is going to be a very valuable tool to obtain new groups from old ones, providing us many new examples to work on. The construction is simple and relies on the of sets. Let (G, ∗) and (H, #) be two groups. Then G × H = {(g, h): g ∈ G, h ∈ H} is the cartesian product the sets G and H. We can turn G × H into a group via:

(g1, h1).G×H (g2, h2) := (g1 ∗ g2, h1#h2) Remark: G is not a subgroup of G × H since it is not a of it. Elements of G × H are pairs, and elements of G are not. However, the group G is isomorphic to the group G × {1H }, which is a subgroup of G × H. So we will say G × H has a subgroup isomorphic to G. Example Z/3Z × Z/5Z = {a(mod 3, b(mod 5) : a, b ∈ Z}' Z/15Z The last isomorphism comes from Chienese Remainder theorem: there is a unique integer 15 which is a modulo 3 and b modulo 5. More generally we have if (m, n) = 1, then Z/nZ × Z/mZ ' Z/mnZ. Example Z/2Z × Z/2Z 6' Z/4Z. This is easy to see by comparing elements of order 4 on both sides. Z/4Z is a generated by an element of order 4, however all elements of Z/2Z × Z/2Z have order two. In general, when m and n are not coprime, we have Z/nZ × Z/mZ 6' Z/nmZ An important result in group theory is the structure theorem for all finite abelian groups. You don’t need to know this result for now, but I am writing it down to show importance of direct products: r1 r2 rk Theorem Any finite abelian group is of the form Z/p1 Z × Z/p2 Z × · · · × Z/pk Z where are distinc primes, ri are positive integers. We can obtain many new groups from the old ones using the direct product construction. For example we now have S3×Z, S3×Z/nZ, S2×S3, S2×S3×S4...etc. I suggest you to try to use the correspondance theorem to understand subgroups of some of these groups. It is a good practice.