## 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 for all dividing .

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

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

Now suppose that every group of width has the property that if divides , then there exists a subgroup with . Let be a group of width ; say . Let . If , then is a subgroup of of order . Suppose and without loss of generality, say divides . By Cauchy’s Theorem, there exists an element of order , and is normal in since is abelian. Now has width and order . By the induction hypothesis, there is a subgroup of of order , and so has order as desired.

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## Comments

How does d=m imply G is order d?

Good question.

I just noticed that this proof is broken for another reason: the line “if then or ” 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.

Thanks!

Why does the image of ‘pm’ seem ? lol

And it may be an error that is a subgroup of order