By Thomas Markwig Keilen

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

**Sample text**

It thus remains to show that α injective. For this let g, h ∈ G/ Ker(α) with α(g) = α g = α h = α(h), then eH = α(g) −1 α(h) = α g−1 α(h) = α g−1h . e. g−1h ∈ Ker(α) and hence g = h. This shows that α is injective. 51 Consider the groups (Z, +) of integers with addition and (C \ {0}, ·) of complex numbers with multiplication. From the lecture Grundlagen der Mathematik it is known that i·π α : Z −→ C \ {0} : z → e 2 ·z is a group homomorphism since the power law e i·π ·(z+z′ ) 2 =e i·π ·z 2 ·e i·π ′ ·z 2 holds true.

In particular each subgroup of Zn is cyclic. 41 it suffices to find the subgroups U of Z with nZ ⊆ U. 39 such a subgroup has the form U = mZ for an integer m ≥ 0. e. m must lie between 1 and n and m is a divisor of n. We next want to compute the order of an element m ∈ Zn for positive integers m and n. For this we introduce the following notation. 43 For two integers a, b ∈ Z let lcm(a, b) := min{z > 0 | a and b divides z}, 0, if a, b = 0, if a = 0 or b = 0. 4. 44 Let m, n ∈ Z>0. Then o m = lcm(m, n) m is the order of m ∈ Zn.

19 Note again that the group operation in (Z, +) is the addition. The condition “g−1 h ∈ U” translates thus to “−g + h ∈ U”. And since the addition is commutative we usually prefer to write “h − g ∈ U”. 45 however, we know that 0 < j − i < n is not a multiple of n. This shows that i and j do not coincide. 7 From now on we will usually write Zn instead of Z/nZ. Moreover, we sometimes write an instead of a for a coset in Zn if we want to indicate modulo which integer we are currently working. Since the set Zn is so important for our lecture we will introduce some common notions.

### Algebraic Structures [Lecture notes] by Thomas Markwig Keilen

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