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?

Re: Subgroup isomorphic to a direct sum.

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

$\displaystyle G \cong \mathbb{Z}_{p_1^{k_1}} \oplus \cdots \oplus \mathbb{Z}_{p_r^{k_r}}$, for primes p_{1},...,p_{k}.

since G is assumed non-cyclic, we cannot have gcd(p_{1},...,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 $\displaystyle \mathbb{Z}_{p_1^{k_1}} \oplus \mathbb{Z}_{p_2^{k_2}}$ has a subgroup isomorphic to$\displaystyle \mathbb{Z}_{p_1} \oplus \mathbb{Z}_{p_2}$

(since each $\displaystyle \mathbb{Z}_{p_i^{k_i}}$ is cyclic and thus has a cyclic subgroup of order p_{i}).

but p_{1} = p_{2}, so setting p = p_{1} = p_{2}:

$\displaystyle \mathbb{Z}_{p} \oplus \mathbb{Z}_{p} \oplus \{0\} \oplus \cdots \oplus \{0\} \cong \mathbb{Z}_{p} \oplus \mathbb{Z}_{p}$

is isomorphic to a subgroup of G.

Re: Subgroup isomorphic to a direct sum.

I just can't thank enough. Orz