Question #238358

Determine the elements of order 15 of U(225)


1
Expert's answer
2021-09-21T02:54:39-0400

We can write U225U15×U15U_{225}\backsimeq U_{15}\times U_{15} [By chinese remainder theorem]

Hence any element of U15U_{15}  has orders 1,2 or 4. Since  U15=φ(15)=8|U_{15}|=φ(15)=8

there is no element of U15U_{15}  that generates it, hence U15U_{15}  is not a cyclic group.

So, we can write:

U(225)=U(152)=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 threesubgroups of order 2 and two of order 4.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.


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