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In my attempt to prove

Let $G$ be an abelian group and $H,K$ its subgroups such that $o(H) = m$ and $o(K) = n$. Then $G$ has a subgroup of order $\operatorname{lcm}(m,n)$.

, the crucial argument is

Sylow theorem for abelian groups: Let $G$ be an abelian group, $p$ a prime number, and $n$ a natural number. If $o(G)$ is divisible by $p^n$, then $G$ has a subgroup of order $p^n$.

IMHO, my proof is quite different from those I've seen around. Could you please verify if my attempt is fine or contains logical mistakes?

My attempt:

First, we need an auxiliary result

Cauchy's theorem for abelian group (CTFAG): Let $G$ be an abelian group. If the order of $G$ is divisible by a prime number $p$, then $G$ has an element of order $p$.

Then we define recursively a finite sequence of group epimorphisms $(\phi_k)_{1 \le k \le n}$ as follows. Let $G_0 := G$. By CTFAG, there exists $x_k \in G_k$ such that $o(x_k) = p$. Let $G_{k+1} =G_k / \langle x_k \rangle$. Because $G_k$ is abelian, the map $\phi_{k+1}: G_k \to G_{k+1}, \quad y \mapsto y \langle x_k \rangle$ is a group epimorphism. Moreover, $G_{k+1}$ is also abelian and $o(G_{k+1}) = o(G_{k})/p$.

It follows that $o(G_n) = o(G) / p^n$ and that $\phi := \phi_n \circ \cdots \circ \phi_1$ is a group epimorphism from $G$ to $G_n$. By first isomorphism theorem, we have $G / \operatorname{ker} \phi \cong G_n$ and thus $o(G / \operatorname{ker} \phi ) = o(G_n)$. Consequently, $o(G) = o(G_n) o(\operatorname{ker} \phi) = o(G) o(\operatorname{ker} \phi ) / p^n$. Finally, we have $o(\operatorname{ker} \phi ) = p^n$.

Shaun
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Akira
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  • I don't think this is hard due to the fundamental theorem of finite abelian groups. The Sylow theorems are mainly "interesting" for the non-abelian case. – Randall Jul 22 '20 at 12:55
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    @Randall My problem is that my textbook Algebra by Saunders MacLane and Garrett Birkhoff have not covered the fundamental theorem of finite abelian groups. My little proud is that the above proof is of my own and quite different than the other ones ^^ – Akira Jul 22 '20 at 12:57
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    This is a good proof in the abelian case (absent the classification theorem). I think an inductive proof of this kind is common. For example: https://math.stackexchange.com/questions/2068970/can-sylows-theorem-for-abelian-group-be-generalized – halrankard Jul 22 '20 at 13:02
  • It can be done slightly more easily: since you have an element of order $p$, you have a $p$-subgroup $P$ of $G$. Thus by induction $G/P$ has a Sylow $p$-subgroup. Take the preimage of this, which must be a Sylow $p$-subgroup of $G$. – David A. Craven Jul 24 '20 at 09:58

1 Answers1

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Here is @halrankard's comment that answers my question. I post it here to remove this question from unanswered list. All credits are given to @halrankard.

This is a good proof in the abelian case (absent the classification theorem). I think an inductive proof of this kind is common. For example: Can Sylow's Theorem for Abelian group be generalized?

Akira
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