Timeline for Groups (not necessarily finite) with a given number of maximal subgroups
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Dec 2 at 21:10 | history | became hot network question | |||
| Dec 2 at 13:00 | comment | added | Carl-Fredrik Nyberg Brodda | @Gro-Tsen I suppose it is worth mentioning that if $G$ is finitely generated then every proper subgroup is indeed contained in a maximal subgroup (proved using Zorn’s Lemma as usual). | |
| Dec 2 at 12:10 | comment | added | YCor | @Gro-Tsen the Prüfer group $C_{p^\infty}$ (increasing union of cyclic groups $C_{p^n}$) has no maximal proper subgroup. Not so magic. | |
| Dec 2 at 12:01 | history | edited | semisimpleton | CC BY-SA 4.0 |
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| Dec 2 at 10:58 | comment | added | Gro-Tsen | @YCor “Not every proper subgroup is contained in a maximal one.” 😱 HOW? WHY? WHAT IS THIS MAGIC? 😵 I feel like I've just been told that Santa Claus doesn't exist. (And yes, indeed, when you think of it, it's obvious, but it makes you reconsider why maximal ideals exist.) This is one for the big list. | |
| Dec 2 at 10:27 | history | edited | semisimpleton | CC BY-SA 4.0 |
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| Dec 2 at 10:24 | comment | added | semisimpleton | Oh, I see! It's strange that Zorn's Lemma can be used to prove that "every proper ideal is contained in a maximal ideal", but it can't be used to prove that "every proper subgroup is contained a maximal subgroup". | |
| Dec 2 at 9:04 | comment | added | YCor | The group $C_{p^\infty}\times C_6$ has exactly 2 maximal proper subgroups but is not cyclic. (In your reasoning "any element in $G-M_1\cup M_2$ generates the whole group" is not true — indeed not every proper subgroup is contained in a maximal one.) | |
| Dec 2 at 7:55 | comment | added | YCor | The group $C_{p^\infty}\times C_6$ has exactly 2 maximal subgroups but is not cyclic. (In your reasoning "any element in $G-M_1\cup M_2$ generates a cyclic subgroup" is not true.) | |
| Dec 2 at 5:17 | comment | added | Carl-Fredrik Nyberg Brodda | (My comment should obviously read "any non-cyclic group") | |
| Dec 2 at 4:04 | answer | added | Jeremy Rouse | timeline score: 6 | |
| Dec 2 at 3:48 | comment | added | Carl-Fredrik Nyberg Brodda | Experimenting with GAP suggests that any group with exactly $3$ maximal subgroups is a $2$-group (and there are many such $2$-groups). Is there some simple argument for why this is true (or a counterexample)? | |
| Dec 2 at 3:28 | history | asked | semisimpleton | CC BY-SA 4.0 |