# Subgroup isomorphic to a direct sum.

• Nov 12th 2012, 10:30 PM
Rita
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 Zp+Zp.

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?
• Nov 13th 2012, 09:09 AM
Deveno
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 p1,...,pk.

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

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 pi).

but p1 = p2, so setting p = p1 = p2:

\$\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.
• Nov 14th 2012, 05:19 AM
Rita
Re: Subgroup isomorphic to a direct sum.
I just can't thank enough. Orz