Originally Posted by

**crystalwater** Suppose that G is a group of order 143 and that H is a subgroup of G with H not equal to G. Prove H is cyclic.

Solution Attempt:

Suppose G is a group and |G| = 143.

Thus, G has an element a with order 13 and an element b with order 11. Then ab has order 143 since

(ab)^143 = (a^143)(b^143) = e. However, (ab)^13= (a^13)(b^13) = b^13 not equal to e.

Also, (ab)^11 = (a^11)(b^11) = a^11 not equal to e. Then G = <ab> is the cyclic group generated by ab.

A Theorem says that any subgroup of a cyclic group is cyclic, thus H is cyclic because G is cyclic and H is a subgroup of G.

Thanks! Any feedback would be most most appreciated.