Also since G is a subgroup of Sym X we 234 CHAPTER 4 Abstract Algebra see that

# Also since g is a subgroup of sym x we 234 chapter 4

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8. Also since G is a subgroup of Sym( X ), we
234 CHAPTER 4 Abstract Algebra see that | G | 24. But there are plenty of permutations of the vertices 1, 2, 3, and 4 which are not graph automorphisms (find one!), and so | G | < 24. On the other hand, note that the powers e, σ, σ 2 , σ 3 give four au- tomorphisms of the graph, and the elements τ, στ, σ 2 τ, σ 3 τ give four more. Furthermore, since τστ = σ 3 we can show that the set { e, σ, σ 2 , σ 3 , τ, στ, σ 2 τ, σ 3 τ } is closed under multiplication and hence is a subgroup of G . Therefore 8 | G | and so it follows that | G | = 8 and the above set is all of G : G = { e, σ, σ 2 , σ 3 , τ, στ, σ 2 τ, σ 3 τ } . Below is the multiplication table for G (you can fill in the missing elements). Notice that G has quite a few subgroups—you should be able to find them all (Exercise 3). e σ σ 2 σ 3 τ στ σ 2 τ σ 3 τ e e σ σ 2 σ 3 τ στ σ 2 τ σ 3 τ σ σ σ 2 σ 3 e σ 2 σ 2 σ 3 e σ σ 3 σ 3 e σ σ 3 τ τ στ στ σ 2 τ σ 3 τ Exercises 1. Use the corollary on page 233 to give another proof of Fermat’s Little Theorem; see page 86. 2. Suppose that G is a finite group of prime order p . Prove that G must be cyclic. 3. Refer to the multiplication table above for the group G of symme- tries of the square and list all of the subgroups.
SECTION 4.2 Basics of Group Theory 235 4. Let G be a group, let H be a subgroup, and recall the equivalence relation ( mod H ) defined by g g 0 ( mod H ) g - 1 g 0 H. The equivalence classes in G relative to this equivalence relation are called the (left) cosets of H in G . Are the cosets also sub- groups of G ? Why or why not? 5. Let G be the group of Exercise 3 and let K be the cyclic subgroup generated by στ . Compute the left cosets of K in G . 6. Let G be the group of Exercise 3 and let L be the subgroup { e, τ, σ 2 , σ 2 τ } . Compute the left cosets of L in G . 7. Here we shall give yet another proof of the infinitude of primes. Define, for each prime p the corresponding Mersenne number by setting M p = 2 p - 1 (these are often primes themselves). Assume by contradiction that there are only finitely many primes and let p be the largest prime. Let q be a prime divisor of M p = 2 p - 1. Then we have, in the multiplicative group Z * q of nonzero integers modulo q , that 2 p 1(mod q ). This says, by exercise 2 on page 227 that p is the order of 2 in the group Z * q . Apply Lagrange’s theorem to obtain p | ( q - 1), proving in particular that q is a larger prime than p , a contradiction. 4.2.8 Homomorphisms and isomorphisms What is the difference between the additive group ( Z 6 , +) and the mul- tiplicative group ( Z * 7 , · )? After all, they are both cyclic: ( Z 6 , +) has generator 1 (actially, [1]), and ( Z * 7 , · ) has generater 3 ([3]). So wouldn’t it be more sensible to regard these two groups as algebraically the same, the only differences being purely cosmetic? Indeed, doesn’t any cyclic group of order 6 look like { e, x, x 2 , x 3 , x 4 , x 5 , x 6 } ?
236 CHAPTER 4 Abstract Algebra Here’s a much less obvious example. Consider the two infinite groups ( R , +) and ( R + , · ). At first blush these would seem quite different.

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