## Algebra and number theory, U Glasgow notes - download pdf or read online

By Baker.

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Solution. Setting σ = (i1 i2 · · · ir ), we have σ k (1) = hence |σ| r. As ik = 1 for 1 < k ik+1 if k < r, i1 if k = r, r, r is the smallest such power which is ι, hence |σ| = r. So for example, |(1 2)| = 2, |(1 2 3)| = 3 and |(1 2 3 4)| = 4. But notice that the product (1 2)(3 4 5) satisfies ((1 2)(3 4 5))2 = (1 2)(3 4 5)(1 2)(3 4 5) = (3 5 4), hence |(1 2)(3 4 5)| = 6. On the other hand, the product (1 2)(2 3 4) satisfies ( (1 2)(2 3 4) )2 = (1 2)(2 3 4)(1 2)(2 3 4) = (1 3)(2 4) so |(1 2)(3 4 5)| = |(1 3)(2 4)| = 2.

If w ∈ W , then by definition of W we must have w ∈ / g(w) = W , which is impossible. On the other hand, if w ∈ / W , then w ∈ g(w) = W and again this is impossible. But then w cannot be in W or the complement X − W , contradicting the fact that every element of X has to be in one or other of these subsets since X = W ∪ (X − W ). Thus no such surjection can exist. When X is finite, this result is not surprising since 2n > n for n ∈ N0 . For X an infinite set, it leads to the idea that there are different ‘sizes’ of infinity.

Then G is cyclic and so abelian. Proof. Let θ(d) denote the number of elements in G of order d. 25, θ(d) = 0 unless d divides |G|. Since G= {g ∈ G : |g| = d}, d||G| we have |G| = θ(d). 24(d), we also have ϕ(d). |G| = d||G| Combining these we obtain θ(d). 2) d||G| d||G| We will show that for each divisor d of |G|, θ(d) ϕ(d). For each such d of |G|, we have θ(d) 0. If θ(d) = 0 then θ(d) < ϕ(d), since the latter is positive. So suppose that θ(d) > 0, hence there is an element a ∈ G of order d. In fact, the distinct powers ι = a0 , a, a2 , .

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### Algebra and number theory, U Glasgow notes by Baker.

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