if then there's nothing to prove. so let and suppose that the claim is true for all groups of order with let G be a group of order and let we want

to show that G has a subgroup of order it's trivial for so we'll assume that we have to consider two cases:

Case 1: G is abelian. in this case choose with since G is abelian, < x > is normal in G. now G/< x > is a group of order so by induction hypothesis it has a subgroup H/<x>

of order therefore: and we're done in this case.

Case 2: G is non-abelian. let Z(G) be the center of G and then: if then by induction hypothesis Z(G) has a subgroup of order and we're done. if then

consider G/Z(G), which is a group of order by induction G/Z(G) has a subgroup K/Z(G) of order hence: