Question

Question #d4cb7

Answer 1

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.