Every subgroup of a cyclic groupis cyclic. Ifgenerates the group(so that) and the order ofisso thatthen the order of any subgroup ofis a divisor ofand for each divisorofthe grouphas exactly one subgroup of order– the subgroup generated by

Proof: Letand suppose thatis a subgroup ofIfconsists of the identity element thenis cyclic, so assumeandcontains some elementthenLetbe the least positive integer such thatThe claim is thatLetbe an arbitrary member ofSincewe havefor someApply the division algorithm toandto obtain a quotientand remaindersatisfyingso thatbutandhencebutis the least power ofinand sincewe must haveso that

Conversely suppose thatis any subgroup ofWe have already shown that for someSincewe have that the order ofis a divisor ofhence the order ofis a divisor of

Now letbe any divisor ofand letso no smaller power of equalssohas orderis the only subgroup of ordersince ifis any other subgroup of orderthenwhereis the least positive integer such thatWritewithWe haveso thatThenandAlso soand