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I'm solving Exercise 2.5.26 from Topics in Algebra by Herstein. Could you please verify if my attempt is fine or contains logical mistakes?

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

My attempt: First, we prove a simple lemma

Lemma: Let $G$ be an abelian group and $H,K$ its subgroups such that $o(H)= m$, $o(K|)= n$, and $H \cap K = \{1\}$. Then the $o(H \lor K) = mn$ where $H \lor K$ is the smallest subgroup containing both $H$ and $K$.

Proof: Because $G$ is abelian, $H \lor K =\{hk \mid h \in H \text{ and } k \in K\}$. Let $a,b \in H$ and $x,y \in K$ such that $ax = by$. Then $a^{-1}b = xy^{-1} \in H \cap K$ and thus $a^{-1}b =1$. Consequently, $a=b$ and $x=y$. The result then follows.

Come back to our general case, let $\operatorname{lcm}(m,n) = p_1^{n_1} \cdots p_k^{n_k}$ where $p_1,\ldots,p_k$ are prime numbers. For each $i$, we pick $C_i$ from the set $\{C \in \{H, K\} \mid p_i^{n_i} \text{ divides } o(C)\}$. By below Sylow theorem for abelian groups (in which I use first isomorphism theorem to give a simple proof here),

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 $p^n$.

we have $G$ has a subgroup $G_i$ with $o(G_i) = p_i^{n_i}$ for each $i$. Notice that $G_i \cap G_j = \{1\}$ for all $i \neq j$. Then the result follows by applying our Lemma repeatedly.

Update: Thanks to @aschepler for pointing out my typo. The correct version should be

we have $\color{blue}{C_i}$ has a subgroup $G_i$ with $o(G_i) = p_i^{n_i}$ for each $i$. Notice that $G_i \cap G_j = \{1\}$ for all $i \neq j$. Then the result follows by applying our Lemma repeatedly.

Akira
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    (+1) Your attempt is correct as far as I can see. – Arnaud Mortier Jul 22 '20 at 14:10
  • Is $C_i$ an element or a subgroup? "$C \in {H,K}$" doesn't really make since either way. "$C_i \in H \cup K$" would name an element, but it's not generally true an element of the needed order exists. Probably simplest to say $C_i$ is a subgroup of either $H$ or $K$ with $o(C) = p_i^{n_i}$, which exists by the quoted theorem of Sylow. – aschepler Jul 22 '20 at 14:29
  • @aschepler, it's a typo. Please see my edit. – Akira Jul 22 '20 at 14:34
  • Oh, I see. You really do mean that $C_i=H$ or $C_i=K$. – aschepler Jul 22 '20 at 15:15

1 Answers1

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$\DeclareMathOperator{\lcm}{lcm}$One possibility is to use the formula $$ o(H K) = \frac{o(H) \cdot o(K)}{o(H \cap K)}, $$ then note that since $H \cap K \le H, K$, by Lagrange $$ o(H \cap K) \mid o(H), o(K), $$ so that $$ o(H \cap K) \mid \gcd(o(H), o(K)), $$ and finally recall that $\gcd(x, y) \cdot \lcm(x, y) = x y$.

With this you obtain that $G$ has a subgroup of order a multiple of the lcm, so you will have to know that the converse of Lagrange's theorem holds in a finite abelian subgroup.

Andreas Caranti
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