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Group Isomorphisms - Some Intuition

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Group Isomorphisms - Some Intuition Group Isomorphisms - Some Intuition The idea of an \isomorphism" is central to algebra. It's our version of equality - two objects are considered isomorphic if they are essentially the same. Before studying the technical definitions, consider a (slightly mathematical) example from language. We'll define a few finite sets: 1. f I, II, III, IV, V, VI g 2. f uno, dos, tres, cuatro, cinco, seis g 3. f 一, 二, 三, 四, 五, m g 4. f one, two, three, four, five, six g 5. f 1, 2, 3, 4, 5, 6 g 6. f un, deux, trois, quatre, cinq, six g 7. f I, II, III, IV, V, VI g Even if you don't recognize all of them, hopefully at least one or two sets look familiar. Based on that, you might make a conjecture about a set you don't recognize. They are, in fact, all the same - the first element of each set is used to represent the idea of a single entity, or the first object in an arrangement. The second element of each set represents a pair, and so on. The fact that each set has the same number of elements is not the important part; rather, it's that we can replace one set with another, without changing the inherent meaning. Regardless of the language or symbols we use for them, numbers are the same - the number systems we use are isomorphic, and the map between them is the isomorphism (in this case, a multi-language dictionary would make a good map). There is only one set of numbers, we just have many names for them. Many groups we've already seen are isomorphic to each other. This is a stronger property than simply counting the number of elements in the group. If two groups G and H are isomorphic, then for every element of G there is an element of H that \behaves" the same way: it will have the same order, among many other properties. In addition, the groups themselves will have similar properties, even though their elements and operations are different. Even when groups appear to be similar (perhaps they have the same order, or are both infinite and Abelian, etc) they may not be isomorphic. Math 280 Fall 2015 Part 2: Groups Group Isomorphisms Definition Let φ : G ! H be a map from a group G to a group H such that 1. φ preserves the operations of the groups (that is, φ is a homomorphism) 2. φ is onto: for every h 2 H, there is some g 2 G such that φ(g) = h 3. φ is one-to-one: if g1; g2 2 G and g1 6= g2, then φ(g1) 6= φ(g2) If such a map exists, then G and H are called isomorphic, and the map φ is called an isomorphism. In this case, we write G ≈ H. Proving that groups are isomorphic requires finding a map between the groups, and showing that it is an operation-preserving bijection. Example We'll show that R, the group of real numbers under addition, is isomorphic to the group G of positive real numbers under multiplication. Define a map φ : R ! G by φ(x) = ex (or if you prefer, φ(x) = exp(x)). First note that φ is defined for any real number x, and that ex > 0 for any x, so this really is a map from R into G. Next, we show that all three properties of an isomorphism are satisfied: 1. φ preserves the operations of the groups: Let x; y 2 R. Then φ(x + y) = ex+y = exey = φ(x)φ(y) 2. φ is onto: Let g 2 G, so g is a positive real number. Let x = ln(g). Then φ(x) = ex = eln(g) = g 3. φ is one-to-one. Suppose x; y 2 R such that φ(x) = φ(y). Then ex = ey, so ln(ex) = ln(ey) and we have x = y. Example It is important to note that a map can fail to be an isomorphism, even when it is a bijective map from a group to itself. For instance, define φ : R ! R by φ(x) = x3. This is a one-to-one, onto map. However, if x; y 2 R, we have φ(x + y) = (x + y)3 6= x3 + y3 = φ(x) + φ(y) so the map does not preserve the group operation of addition. Theorem 1 Every infinite cycle group is isomorphic to Z under addition. Math 280 Fall 2015 Part 2: Groups Cayley tables can be used to decide if two groups G and H are isomorphic. If the elements in the Cayley table of G can be relabelled with the elements of H to yield the Cayley table of H (possible after rearranging rows or columns), then G and H are isomorphic. Example Compare the table of U(5) on the left to the table of Z4 on the right: ×5 1 2 3 4 +4 0 1 2 3 1 1 2 3 4 0 0 1 2 3 2 2 4 1 3 1 1 2 3 0 3 3 1 4 2 2 2 3 0 1 4 4 3 2 1 3 3 0 1 2 The operations and sets are completely different, though both groups have order 4. We'll try to find a way of relabeling the elements in the table of U(5) with those of Z4. First, the identity of U(5) is 1, while the identity of Z4 is 0, so it makes sense to relabel 1 with 0: 0 0 0 0 0 0 Next, note that the element 4 in U(5) has order 2, just like the element 2 in Z4. So, next we'll relabel 4 with 2: 0 2 0 0 2 2 0 0 2 2 2 0 Similarly, the element 3 in U(5) has order 3, just like the element 3 in Z4, so we'll keep the 3's where they are. While we're at it, we'll replace 2 in U(5) with the only remaining element of Z4, 1 (both of these have order 4): 0 1 3 2 0 0 1 3 2 1 1 2 0 3 3 3 0 2 1 2 2 3 1 0 This is still not exactly what we were looking for, but if we switch the last two rows and last two columns of the table (so the row and column labels are in order), we have So, the groups U(5) and Z4 are isomorphic. If we need the specific +4 0 1 2 3 0 0 1 2 3 1 1 2 3 0 2 2 3 0 1 3 3 0 1 2 isomorphism, it's the map that accomplishes our relabelling: φ : U(5) ! Z4 defined by φ(1) = 0; φ(2) = 1; φ(3) = 3; φ(4) = 2 This is clearly a bijective, but as an example of how the operations are preserved: φ(2 ×5 3) = φ(1) = 0 = 1 +4 3 = φ(2) +4 φ(3) Math 280 Fall 2015 Part 2: Groups The next theorem can be a very useful way of deciding if a homomorphism between groups is actually an isomorphism: we just have to find its kernel. If the kernel contains an element other than the identity, the map cannot be an isomorphism. Theorem 2 A group homomorphism is one-to-one if and only if the kernel contains only the identity element. Theorem 3 (Properties of Isomorphic Groups) Suppose G and H are isomorphic groups, with isomorphism φ : G ! H. Then 1. G and H have the same order. 2. If G and H are finite, they have the same number of elements of each order. 3. G is Abelian if and only if H is Abelian. 4. G is cyclic if and only if H is cyclic. If G = hgi, then H = hφ(g)i. 5. φ is invertible, and φ−1 : H ! G is also an isomorphism. 6. If K is a subgroup of G, then φ(K) is a subgroup of H. 7. If Z(G) is the center of G, then φ(Z(G)) is the center of H. Theorem 4 (Properties of Isomorphisms and Group Elements) Let φ : G ! H be a group isomorphism. Then 1. φ(eG) = eH , where eG; eH are the identity elements of G; H, respectively. 2. for all g 2 G, φ(gn) = (φ(g))n. 3. jgj = jφ(g)j for all g 2 G. 4. if g1; g2 2 G, then g1g2 = g2g1 if and only if φ(g1)φ(g2) = φ(g2)φ(g1). Definition An isomorphism φ : G ! G from a group G to itself is called an automorphism. Theorem 5 The set of automorphisms on a group G (that is, the set of all isomorphisms from a group to itself) forms a group under function composition, denoted Aut(G). Proof Before starting the proof, it's important to understand the group we're working with. The elements of Aut(G) are isomor- phisms (functions), not the elements of G. Furthermore, its operation is function composition, regardless of the operation on G. We need to show several things here. First, we need to prove that the composition of automorphisms leads to another automorphism; that is, that composition is a closed, binary operation on this set. Next, we have to show that the group properties hold. Let φ and be automorphisms on the group G, so both map G into itself, both preserve group operations, and both are bijections.
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