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Abstract Algebra
1. In this problem we will show that every group of order 15 is cyclic. Let G be a group of order 15.
The Sylow theorems (which we have not covered), say that there exist elements g of order 5 and h of order 3 in the group G.
(a) By considering the left cosets of 〈g〉, show that gh has the form hjgb, where 0 ≤ j ≤ 2 and 0 ≤ b ≤ 4.
(b) Show that gh = hgb, for some value of b which is not divisible by 5.
Hint: We may rule out j = 2 by showing that in that case gh has even order in G.
(c) Use induction to show that (gh)k = hkgb k+bk−1+...b. Conclude that the order of gh is
divisible by 3, but is not 3.
(d) Conclude that gh has order 15 and hence G is cyclic.
2. Saracino: 7.10, 8.11, 8.18, 9.6 (see pg 75 for notation), 9.18, 10.8, 10.15, 11.4, 11.16, 11.18.
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