Exercise 1. That G is abelian follows from the commutativity of addition: x? For each a we need to use the methods of exercises 0. The proof for left inverse is similar. Thus it can be only 1 or 2. It inherits associativity from G.
From exercise 1. Thus p and q both divide n, which means that [p, q] divides n. This is clearly reflexive,symmetric, and easily shown to be transitive. There are two elements per equivalence class on t G : one element and its inverse. Thus t G has an even number of elements. It contains the identity, so it must contain at least one nonidentity element.
Only the identity has order 1, so this nonidentity element must have order 2. Let k be an arbitrary power of x. WLOG, assume that this element is a. From 1. So any element that is not equivalent to rk for some k is equivalent to srk for some k. Thus the order of rk s cannot be 1.
To show that this is the only element other than the identity that commutes with all elemenets of D2n , we use a very unsatisfying proof by exhaustion of cases. For multiplication, p : p is prime by the fundamental theorem of arithmetic. A more explicit example: In S6 we have 1 2 3 4 5 which has order 6.
Each distinct m-cycle can be written in m different ways. Thus for S5 we can construct elements of orders 1,2,3,4,5 trivially and 6.
The order 6 element is given by 1 2 3 4 5 Exercise 1. Thus S3 can be represented as string of 3 or fewer alternating elements a, b. Then a has no multiplicative inverse. The proof changes very little. The previous parts of this exercise show that H F is a group. Thus every nonidentity element has infinite order.
This contradicts exercise 1. Associativity and closure is inherited from the properties of function composition. This function is clearly injective and a homomorphism. From the exercise 1. The text bottom of p38 assures us that this is sufficient to guarantee an isomorphism between G and D2n. There are a lot of relations that need to be algebraically verified e.
Thus the kernel contains a nonidentity element. The element 1 2 3 replaces each 1 with 2, each 2 with 3, and each 3 with 1. We show that the action of Sn on k-element subsets of A is faithful. But these were arbitrary elements of SA , thus no two elements of SA perform the same action.
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