Finite abelian groups have subgroups of every order dividing the order of the group

Use Cauchy’s Theorem and induction to prove that a finite abelian group has a subgroup of order d for all d dividing |G|.

First we make a quick definition. Let G be a finite abelian group, say with |G| = n. By the Fundamental Theorem of Arithmetic, n factors completely as a product of primes in an essentially unique way; that is, n = p_1p_2 \ldots p_k for some primes p_i, where k is unique and the p_i are unique up to a permutation. We call this k the width of G. We will proceed by induction on the width of G.

If G has width 0, then |G| = 1 so that G = 1. Now 1 itself is the only (positive) divisor of 1 and G itself is a subgroup of order 1. Thus the conclusion holds in the base case.

Now suppose that every group G of width n has the property that if d divides |G|, then there exists a subgroup H \leq G with |H| = d. Let G be a group of width n+1; say |G| = p_1 \cdots p_{n+1} = m. Let d|m. If d = 1, then 1 is a subgroup of G of order d. Suppose d > 1 and without loss of generality, say p_1 divides d. By Cauchy’s Theorem, there exists an element x \in G of order p_1, and \langle x \rangle is normal in G since G is abelian. Now G/\langle x \rangle has width n and order m/p. By the induction hypothesis, there is a subgroup H/\langle x \rangle of G/\langle x \rangle of order d/p, and so H \leq G has order d as desired.

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  • Robert Bradford  On January 17, 2011 at 5:35 pm

    How does d=m imply G is order d?

    • nbloomf  On January 17, 2011 at 7:59 pm

      Good question.

      I just noticed that this proof is broken for another reason: the line “if d|pm then d = p or d|m” is wrong.

      I’ll add this to the growing list of broken proofs that will have to be fixed later when I have more time.


  • Gobi Ree  On November 29, 2011 at 1:56 am

    Why does the image of ‘pm’ seem r? lol

  • Gobi Ree  On November 29, 2011 at 2:01 am

    And it may be an error that G \leq G is a subgroup of order d

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