Abstract Algebra I, Edition: version 11 Nov 2016 by Randall R. Holmes

By Randall R. Holmes

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Therefore the proof above shows that G is abelian. Recall that the symmetric group S3 has order 3! = 6. Since Z6 also has order 6, there exists a bijection from S3 to Z6 , so one might wonder whether these two groups are actually isomorphic. 5 Example Is S3 isomorphic to Z6 ? Solution According to Exercise 3–6 the group S3 is not abelian. Since Z6 is abelian, we conclude from the preceding example that S3 is not isomorphic to Z6 . 6 Example addition). Is Q isomorphic to Z? (both viewed as groups under Solution It was pointed out in Section 2 that |Q| = |Z|, which is to say that 36 there exists a bijection from Q to Z.

Therefore, ϕ is an isomorphism and we conclude that R ∼ = R+ . 3 Example Is R isomorphic to Z? 6 that |R| = |Z|, that is, there is no bijection from R to Z. In particular, there cannot be an isomorphism from R to Z. Therefore, R is not isomorphic to Z. 35 If G ∼ = G , then G and G are indistinguishable as groups. If G has a property that can be described just using its elements and its binary operation, then G must have that same property, and vice versa. The next example illustrates this principle.

Since m was an arbitrary element of {1, 2, . . , n}, we conclude that στ = τ σ as desired. Let σ= 1 2 3 4 5 6 7 8 9 3 7 8 9 5 4 2 1 6 ∈ S9 We can write σ as a product (meaning composition) of pairwise disjoint cycles by following an algorithm as illustrated here: • Start with 1. We have σ : 1 → 3 → 8 → 1 (back to where we started). This completes the cycle (1, 3, 8). 61 • Pick the smallest number not yet appearing, namely, 2. We have σ : 2 → 7 → 2. This completes the cycle (2, 7), which we compose with the cycle above to get (1, 3, 8)(2, 7).

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