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Math Help - Subgroup isomorphic to a direct sum.

  1. #1
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    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?
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  2. #2
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    Re: Subgroup isomorphic to a direct sum.

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

     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 \mathbb{Z}_{p_1^{k_1}} \oplus \mathbb{Z}_{p_2^{k_2}} has a subgroup isomorphic to \mathbb{Z}_{p_1} \oplus \mathbb{Z}_{p_2}

    (since each \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:

    \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.
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  3. #3
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    Re: Subgroup isomorphic to a direct sum.

    I just can't thank enough. Orz
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