Question #d4cb7

1 Answer
Aritra G.
Sep 6, 2017

Consider a group #(G,*)# which has prime order so that,

#o(G) = p# where #p# is some prime number.

Now Lagrange's theorem states that,

If #G# be an arbitrary group of finite order and #o(G)# be it's order, then an arbitrary subgroup #H# has order #o(H)# such that #o(H)# divides #o(G)#.

Now, let #a# be an arbitrary non identity element of the group.

So the cyclic subgroup #(S,*) = < a># for which #a# is the generator element must have an order such that it divides #p# (By Lagrange's theorem).

However, since #p# is prime, we must have #o(S) = o(G)#

Thus, #G = S#

And hence, #(G,*)# must be a cyclic group.

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