Thread: Subgroup isomorphic to a direct sum.

One question from textbook:
Show that a finite abelian group that is not cyclic contains a subgroup which is isomorphic to the direct sum Z_p+Z_p.

Can anyone prove it?

Is it somehow related to a corollary saying that if G is a finite abelian group of order n, then G has a subgroup of order m for every positive integer m that divides n?

Any thought on it?

by the structure theorem for finitely-generated abelian groups we may write:

, for primes p₁,...,p_k.

since G is assumed non-cyclic, we cannot have gcd(p₁,...,p_r) = 1 (by the chinese remainder theorem), so for some i ≠ j, p_i = p_j.

without loss of generality (we may re-arrange the direct sum as we like) take i = 1, j = 2.

now has a subgroup isomorphic to

(since each is cyclic and thus has a cyclic subgroup of order p_i).

but p₁ = p₂, so setting p = p₁ = p₂:



is isomorphic to a subgroup of G.

I just can't thank enough. Orz