By Jürgen Bierbrauer

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There must be a rather large kernel. An element g ∈ S4 belongs to Ker(π) if it fixes a, b, c. We have g ∈ Ker(π) if and only if the image of any pair is either the same pair or its complement. We see that transpositions and 3cycles do not have that property. The three permutations that are products of two disjoint transpositions do have the property. 13 Proposition. Let π be the permutation representation of S4 on the pairs of complementary pairs. Then Ker(π) = V = {e, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}.

4. 2. 1. Determine the number of generators of the cyclic group of order 3125. 2. Let g have order 165. Determine the order of g 48 . 3. 3 is elementaryabelian. 4. Prove the following: if g ∈ G, o(g) = n and h ∈ H, o(h) = m, then the element (g, h) ∈ G × H has order lcm(n, m). 5. 4. 38 CHAPTER 5. 1, factor groups. Let H < G. The idea is to define a group where multiplication is like in G but elements of H are treated as if they were the neutral element. As we talk about groups in general (not just abelian groups) we use right cosets, keeping in mind that we could have used left cosets just as well.

A mapping f : G −→ H is a group isomorphism if the following hold: • f is a bijective mapping (one-to-one and onto), and • For every g ∈ G we have f (g −1) = f (g)−1, and • For every g1 , g2 ∈ G we have f (g1 g2 ) = f (g1 )f (g2 ). If there is an isomorphism between G and H, we call G and H isomorphic groups and write G ∼ = H. 33 34 CHAPTER 4. GENERATORS AND ISOMORPHISMS We can see the application of the isomorphism f as a renaming of the elements of G such that the multiplication remains unchanged.

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