abstract algebra

HW 3

February 26, 2018

Due Friday, March 9. Be sure to justify all of your answers to receive full credit.

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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