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Let be $(G,\ast,e)$ a group so that let be $\mathfrak X$ a collection of subgroup of $(G,\ast,e)$.

If $\mathfrak X$ is an inclusive chain then it is not hard to show that $\bigcup\mathfrak X$ is a subgroup: indeed, if $p$ and $q$ are in $\bigcup\mathfrak X$ then there exists $H$ and $K$ such that $p$ is $H$ and $q$ is in $K$ so that if the proposition $$ (H\subseteq K)\vee(K\subseteq H) $$ holds then $p\ast q$ is in $H$ or in $K$ and so in $\bigcup\mathfrak X$; moreover, $e$ is element of $\bigcap\mathfrak X$ and so of $\bigcup\mathfrak X$; finally, if $x$ is in $\mathfrak X$ then there exsists $X$ in $\mathfrak X$ such that $x$ is in $X$ so that $x^{-1}$ is in $X$ and so in $\bigcup\mathfrak X$.

Now if $H$ and $K$ are subgroup such that $H\cup K$ is a subgroup then (see here for details) it is not hard to prove that $H$ is contained in $K$ or $K$ in $H$: however assuming that $\bigcup\mathfrak X$ is a subgroup of $(G,\ast,e)$ then I am not able to prove or disprove it is an inclusive chain so that I thought to put here a specific question where I ask for it. So could someone help me, please?

Antonio Maria Di Mauro
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  • Take $\mathfrak X$ the collection of all subgroups, and then see https://math.stackexchange.com/questions/3761213/if-collection-of-subgroups-of-g-form-a-chain-then-g-is-cyclic – anankElpis Apr 27 '25 at 10:15
  • Let $H,K \le G$ then the union is subgroup of G iff they lie each other. Now, it is clear that the union of all subgroups is subgroup iff $\mathfrak X$ is a chain. – nozalp10 Apr 27 '25 at 12:44

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The claim is false. Let $C_p$ be the cyclic group of order $p$ for a prime $p$. Let $G=C_p\times C_p$ and let $\Pi$ be the collection of subgroups of $G$ of order $p$. Since any nonidentity element of $G$ has order $p$, we have $G=\bigcup \Pi$. But any two members of $\Pi$ intersect trivially, so $\Pi$ is not a chain of subgroups under inclusion.

Anonymous IV
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