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Math 481, Abstract Algebra, Winter 2004 Problem Set 6, due Friday, February 13 6.31 Show that the mapping a → log10 a is an isomorphism from R+ under multiplication to R under addition. Let’s call the map in question φ. Now φ is obviously well defined (there is no ambiguity about its definition), and it is not difficult to see that it is a homomorphism if we remember a little fact from calculus 142 (that log10 (ab) = log10 (a) + log10 (b) for all a, b ∈ R+ ). So for a, b ∈ R+ we have φ(ab) = log10 (ab) = log10 (a) + log10 (b) = φ(a) + φ(b) as required. Now if a ∈ R then 10a ∈ R+ and φ(10a ) = log10 (10a ) = a, so φ is onto, and if log10 (a) = log10 (b), then elog10 (a) = elog10 (b) , that is a = b, so the map is injective. We conclude that φ is an isomorphism. 6.35 Let a belong to a group G and let |a| be finite. Let φa be the automorphism of G given by φa (x) = axa−1 . Show that |φa | divides |a|. Exhibit an element a from a group for which 1 < |φa | < |a|. To show that |φa | divides |a| (here |φa | denotes the order of φa as an element of Aut(G)), it is enough to demonstrate that (φa )|a| = id where id is the identity automorphism (this by corollary 2 of theorem 4.1). But note that (φa )|a| (x) = (φa )|a|−1 (axa−1 ) = (φa )|a|−2 (a2 xa−2 ) = · · · = a|a| xa−|a| = x for all x ∈ G so that (φa )|a| = id as required. Note that in an Abelian group G, φa (x) = axa−1 = aa−1 x = x for all x ∈ G (that is, |φa | = 1). Thus to find an example of the type required we must look to non-Abelian groups—but it can’t be S3 , because this group only has elements of prime order. So let’s try D4 . We’ll take a to be the rotation by 90 degrees, and thus |a| = 4. We claim that |φa | = 2. We have that (φa )2 (x) = a2 xa−2 where a2 = a−2 is rotation by 180 degrees,and it is easy to check (by looking at a multiplication table for D4 ) that a2 xa2 = x for all x ∈ D4 . 6.42 Prove that Q, the group of rational numbers under addition is not isomorphic to a proper subgroup of itself. This is not difficult to do after establishingthe fact that if φ is an isomorphism from Q under addition to a subgroup of itself, then φ ab = ab φ(1). Indeed, with this observation we note that if φ(1) = dc , then for all ab ∈ Q, φ ac−1 db−1 = ac−1 c a ac−1 φ(1) = = . bd−1 bd−1 d b Thus φ is onto Q, and thus the image of φ is not proper. So suppose that φ is an isomorphism of Q with some subgroup of Q. Then a 1 1 = φ 1 + · · · + φ 1 = aφ 1 = φ + · · · + φ b b b} b b b | {z | {z } a times for all a b a times ∈ Q. But we can also say that b 1 = bφ φ(1) = φ b b and thus So we conclude that φ 1 φ(1) φ = . b b a b = ab φ(1) as required. 7.15 Suppose that |G| = pq, where p and q are prime. Prove that every proper subgroup of G is cyclic. By Lagrange’s theorem, any subgroup of G must have order dividing |G| = pq. The order of any proper subgroup of G must therefore be a proper divisor of pq, either p, q, or 1. A subgroup of order 1 is obviously cyclic—for subgroups of order p or q, we resort to corollary 3 of Lagrange’s theorem (that any group of prime order is cyclic). 7.21 Suppose that G is an Abelian group with an odd number of elements. Show that the product of all of the elements of G is the identity. By Lagrange’s theorem, no element of G can have order two (two doesn’t divide an odd number) so no element is its own inverse. Inverses are also unique and thus writing |G| = 2n + 1 we can choose distinct elements g1 , . . . , gn ∈ G such that G = {e, g1 , g1 −1 , . . . , gn , gn −1 } (the implication being that {e, g1 , g1 −1 , . . . , gn , gn −1 } are the distinct elements of G). Now it is clear that the product of all the elements of G gives the identity, that is, that eg1 g1 −1 g2 g2 −1 · · · gn gn −1 = e. 7.34 Prove that a group of order 12 must have an element of order 2. Let G be a group of order 12. By Lagrange’s theorem, an element in G can have order 1, 2, 3, 4, 6, or 12. Note that only the identity has order 1. Now, if G contains an element g of order 6 12 (that is, if G is cyclic), if n then |g | = 2 (we use the fact—shown many times in class—that |a| = n and k|n, then a k = k). Similarly, if G contains an element g of order 6, then |g 3 | = 2, and if G contains an element g of order 4, then |g 2 | = 2. So we conclude that G contains an element of order 2 unless all element of G (except the identity) have order 3. This gives us exactly 11 elements of order 3. But this cannot occur, because elements of order 3 in a group occur in pairs. That is, for each g ∈ G\{e}, we have g −1 ∈ G\{e}, and g −1 is distinct and unique (but we cannot pair off eleven elements). We conclude that G must have an element of order 2 as required. 7.46b Calculate the order of the group of rotations of a regular octahedron (a solid with eight congruent equilateral triangles as faces). We will use the fact that if G is a finite group of permutations on a set S, then |G| = |stabG (i)||orbG (i)| for any i ∈ S. For us, S consists of the faces of the regular octahedron, and choosing one of the faces to be the “top” face, call it i, we see that |stabg (i)| = 3 (the only rotations which fix the top face are rotations through that face, and there are three rotations of an equilateral triangle). On the other hand, |orbG (i)| = 8 (clearly we can reorient the regular octahedron so that our original face lies where any of the other faces originally abode), so that |G| = 24.