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I am trying to understand the proof of the following theorem.

Theorem: If $G$ is a finite abelian group and $d$ is a divisor of $|G|$, then $G$ contains a subgroup of order $d$.

Proof: Let $d$ be any divisor of $|G|$, and let $p$ be a prime divisor of $d$. From Cauchy's theorem, it follows that there is $S\leq G$ of order $p$. $S$ must be normal since $G$ is abelian, and hence $G/S$ is an abelian group of order $n/p$. By induction on $|G|$, $G/S$ has a group $H^*$ of order $d/p$. From the Correspondence Theorem, $H^*=H/S$ for some subgroup $H$ of $G$ containing $S$, and $|H|=|H^*||S|=d$.

I totally understand the big picture of this proof. However, I don't understand why $G/S$ has a group $H^*$ of order $d/p$? Based on my understanding, the inductive assumption should be that the theorem holds for any group $G$ of order not greater $n$. Therefore, by induction on $|G|$, there exists a subgroup $H^*$ of order $d$. Why there is a subgroup of order $d/p$?

Forgive me if this question is too dumb, but I really think I just miss something very obvious.

Thank you.

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  • $\begingroup$ @N.S. Are you suggesting that the "correct" proof should do induction on $d$? Actually, this proof is copied from Rotman's Advanced Modern Algebra, and I am learning it. $\endgroup$ – YYF Jun 3 '14 at 16:16
  • $\begingroup$ this is problem 4 of section 3.4 of Dummit and Foote $\endgroup$ – Jorge Fernández Hidalgo Jun 3 '14 at 16:18
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Here is the inductive statement:

$P(n)$: If $G$ is an abelian group of order $n$ and $d$ is any divisor of $n$ then $G$ contains a subgroup of order $d$.

Now, in the inductive step, you use that $G/S$ is a group with $n/p$ element. Thus, for any divisor of $n/p$ there exists a subgroup of that order.

Since this is true for any divisor, it is true for the particular divisor $d/p$.

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  • $\begingroup$ Thank you! It is an another something very obvious but I missed! $\endgroup$ – YYF Jun 3 '14 at 16:25

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