By Thomas Markwig Keilen

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**Additional resources for Algebraic Structures [Lecture notes]**

**Sample text**

N . Mathematicians are lazy people, or maybe one should better say that they are efficient. Therefore, they have thought of a way to represent a permutation in such a way that each of the numbers 1 to n has to be written at most once, not twice. For this we need the notion of a cycle, which permutes k of the numbers 1, . . , n cyclically. 4 a. Let {1, . . , n} = {a1, . . , ak} ∪ {b1, . . , bn−k}, k ≥ 2, and σ= a1 a2 . . ak−1 ak b1 . . bn−k a2 a3 . . ak a1 b1 . . bn−k ∈ Sn, then we call σ a k-cycle, and we say that it cyclically permutes the numbers a1, .

61 If G = g is a cyclic group of order |G| = n < ∞ then: U≤G ⇐⇒ ∃ m ∈ {1, . . , n} with m divides n : U = gm . For such a subgroup the following holds true: n . m In particular, G has precisely one subgroup of order d for each divisor d of n. 42. 60, since lcm(m, n) = n. Finally we n should note that with m also m runs through all divisors of n. 42. 62 Each subgroup of a cyclic group is cyclic. 63 If (G, ·) is a group and p = |G| a prime number then G is isomorphic to (Zp, +). 64 Let (G, ·) be a group, g ∈ G and n ∈ Z>0.

44. If m < 0 then −m > 0 and we get analogously lcm(m, n) lcm(−m, n) = . o g−m = −m |m| Since moreover the order of an element and its inverse coincide the claim follows also in the case m < 0. 61 If G = g is a cyclic group of order |G| = n < ∞ then: U≤G ⇐⇒ ∃ m ∈ {1, . . , n} with m divides n : U = gm . For such a subgroup the following holds true: n . m In particular, G has precisely one subgroup of order d for each divisor d of n. 42. 60, since lcm(m, n) = n. Finally we n should note that with m also m runs through all divisors of n.