Originally Posted by

**WannaBe** Well, I'm trying to solve some old exams in abstract algebra, and I need your help in the following:

**Question 1:**

Let $\displaystyle U(Z/15Z) $ be the multiplicative group of the invertible elemenets in $\displaystyle Z/15Z $ . Write $\displaystyle U(Z/15Z) $ as a product of cyclic groups.

**My try:**

It's easy to show that x is in U iff gcd(x,15)=1... Hence:

U={1,2,8,4,7,13,11,14} ... If we'll take the element 2 to be a generator we'll get the cyclic group {1,2,4,8} ... If we'll take <7> we'll get {1,7,4,13} etc... I can't figure out how to write U as a product of cyclic groups when the intersection of each two cyclic subgroups of U is "bigger" than {1}...