By Thomas C. Craven

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**Example text**

Proof. Associativity is automatic. Since there is some element a ∈ H, our hypothesis implies that e = aa−1 ∈ H. And so, for any h ∈ H, we have h−1 = eh−1 ∈ H. Finally, for closure, if a, b ∈ H, then b−1 ∈ H, hence ab = a(b−1 )−1 ∈ H. 8(2). But then aan−1 = an = e, so a−1 = an−1 ∈ H by closure. Example from page 183: H = { f ∈ S5 | f (1) = 1 } consists of all permutations fixing the element 1. Since H is finite, we only need to check that it is closed under the operation: for g, h ∈ H, g(h(1)) = g(1) = 1, so g ◦ h ∈ H.

21. Every finite group order n is isomorphic to a subgroup of Sn . In practice, it can be very useful to have a specific representation of a group as a set of permutations because it is possible to do computations in a straightforward way. But the abstract proof above is not helpful: for example the group Z10 is given as a subgroup of S10 which has 10! = 3, 628, 800 elements–a horrible thing to do to such a nice group! Exercises, pages 196–199. 3. GL(2, Z2 ) = 1 0 0 1 , 0 1 1 0 , 1 1 0 1 , 0 1 1 1 , 1 0 1 1 , 1 1 1 0 because these are the matrices of nonzero determinant.

Then b = a−1 gives a counterexample. (b) ab ∈ Z(G) implies that it commutes with a−1 , in particular. Thus, b = a−1 ab = aba−1 ; multiplying by a on the right gives ba = ab. This is typical of proving things about groups: you have to find the right element to apply things to. Experimentation becomes more important than grand ideas. Exercise 45, page 190: Zm × Zn is cyclic iff gcd(m, n) = 1. If gcd(m, n) = 1, then (1, 1) has order mn, hence Zm × Zn = (1, 1) is cyclic. Assume gcd(m, n) = d > 1 and let (a, b) ∈ Zm × Zn .