Answer to Question #238358 in Abstract Algebra for adu

Question

Determine the elements of order 15 of U(225)

Answer 1

We can write $U_{225}\backsimeq U_{15}\times U_{15}$ [By chinese remainder theorem]

Hence any element of $U_{15}$  has orders 1,2 or 4. Since  $|U_{15}|=φ(15)=8$

there is no element of $U_{15}$  that generates it, hence $U_{15}$  is not a cyclic group.

So, we can write:

$U(225)=U(15^2 )=U(15)=1,2,4,7,8,11,13,14.\\ The \ cyclic \ subgroups \ generated \ by \ these \ elements \ are:\\ <2>=<8>=1,2,4,8,\\ <4>=1,4,\\ <7>=1,4,7,13,\\ <11>=1,11(note \ that \ 11=−4),\\ <14>=1,14(note \ that \ 14=−1),\\ <1>=1.\\ Observe \ that \ U(15) \ cannot \ be \ cyclic \ since \ it \ has \ three \\ subgroups \ of \ order \ 2 \ and \ two \ of \ order \ 4.$