Notes on finite group theory [Lecture notes] by Peter J. Cameron

By Peter J. Cameron

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Prove that G is abelian. 4 Facts about cosets. (a) Show that, if C is a right coset of H in G, then C−1 = {c−1 : c ∈ C} is a left coset of H. Show also that the map C → C−1 is a bijection between right and left cosets. Deduce that the numbers of left and right cosets are equal. (b) Let H be a subgroup of G. Prove that a−1 Ha = {a−1 ha : h ∈ H} is also a subgroup of G. ) (c) Prove that any right coset is a left coset (of a possibly different subgroup). 5 Let H and K be subgroups of a group G. (a) Show that H ∩ K is a subgroup.

Sometimes this is written as H\G, but this is too close to the notation H \ G for set difference so I will avoid it. Sometimes it is written [G: H]. Now G acts on cos(H, G) by right multiplication. Formally, using µ(x, g) for the action of the permutation corresponding to g on the element x, the action is given by µ(Hx, g) = H(xg). Fortunately, we can write this in the briefer form (Hx)g = H(xg) without risk of too much confusion. Note that the action of G on cos(H, G) is transitive; for given any two cosets Hx and Hy, we have (Hx)(x−1 y) = Hy.

But then pv = (pu)(u−1 ) = 0 for all v ∈ Fq . Thus the additive group is elementary abelian. (b) Let k be the exponent of the multiplicative group of Fq (the smallest positive integer such that uk = 1 for all u = 0). Then k divides the order q − 1 of the multiplicative group. But the equation xk − 1 = 0 has at most k solutions. So we must have k = q − 1. Now in our proof of the Fundamental Theorem of Finite Abelian Groups, we saw that there is an element whose order is equal to the exponent. So the multiplicative group is cyclic.

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