prove directly without using sylow first thrm or couchy . that an abelian group G with order n=p1 to the power

Question

alpha 1 * p2 to the power alpha 2 ............pn to the power alpha n has asubgroup of order p1p2p3............pnwhen pi not equal pj , for every i not equal j ?

Answer 1

This follows immediately from the converse of Lagrange's theorem for abelian groups, since p1p2p3...pn divides |G|.

Answer 2

I'd look for an element with each of the orders p1, p2, ..., then multiply all those together to get an element of order p1*p2* ..., and then look at the (cyclic) subgroup it generates.

I think that gets it done for you.