How would you prove that a finite group of composite order contains a proper non-trivial subgroup?

Answer 1

I can give an outline of the proof.

Step 1. If G is a finite group that includes an element whose order is equal to the order of the group, then G is cyclic.

Step 2. Let G be a finite group of composite order. Then either G is cyclic, or G is not cyclic.

Step 3. Suppose that G is not cyclic. Then G does not have an element whose order is equal to the order of the group. (by step 1) So, every element of G has an order that is less than the order of G. Let z be any element of G, other than the identity element. Then there is a cyclic subgroup of G, which is generated from z. The subgroup consists of the elements { e, z, z^2, ... z^(n-1) }. (where e is the identity element; and n is the order of z)

Step 4. Suppose that G is cyclic. Its order is a composite number, by step 2. Let f be a factor of the order of G, such that (f > 1) and (f < order(G)). Then G contains a cyclic subgroup whose order is f.

In each case (step 3 and step 4), the group G contains a proper, non-trivial subgroup. End of proof.

Answer 2

enable's be extra direct that naming Sylow and Cauchy right here. enable g = order (G). certainly you already be attentive to: For any element A of G, A generates a subgroup of order order(a). So all you choose is: Lemma: There exists a factor A of G whose order is neither a million nor g. properly, %. an A that's no longer equivalent to the id. If order(A) isn't equivalent to g, you're completed. If order(A)=g, enable B=A^p, the place p is a non-trivial portion of g. Then order(B) = g/p. QED!!!