By Andrew Baker

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Notice that for such a group, |G| = |c|. 22. The group (Z, +) is cyclic of infinite order with generators ±1. 23. If 0 < n ∈ N0 , then the group (Z/n, +) is cyclic of finite order n. Two generators are ±1n ∈ Z/n. More generally, tn is a generator if and only if gcd(t, n) = 1. Solution. We have that for each k ∈ Z, k = ±(1 + 1 + · · · + 1) (with ±k summands). From this we see that ±1n are obvious generators and so Z/n = ±1n . 9, there is an integer u such that ut ≡ 1. Hence 1n ∈ tn n and so Z/n = tn .

Suppose that the distinct elements of X are x1 , . . , xm where m = |X| and those of Y are y1 , . . , yn where n = |Y |. A function f : X −→ Y is determined by specifying the values of the m elements f (x1 ), . . , f (xm ) of Y . Each f (xk ) can be chosen in n ways so the total number of choices is nm . Hence |Y X | = nm . , Y = {0, 1}. The set {0, 1}X is called the power set of X, and has 2|X| elements and indeed it is often denoted 2|X| . It has another important interpretation. For any set X, we can consider the set of all its subsets P(X) = {U : U ⊆ X is a subset}.

Find the number of distinguishable ways there are to colour the edges of an equilateral triangle using four different colours, where each colour can be used on more than one edge. Solution. Let X be the set of all possible such colourings of the equilateral triangle ABC whose symmetry group is G = S3 , which we view as the permutation group of {A, B, C}; hence |G| = 6. Also |X| = 43 = 64 since each edge can be coloured in 4 ways. G acts on X in the obvious way. A pair of colourings is indistinguishable precisely if they are in the same orbit.

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