Timeline for Almost squared finite groups
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Feb 23, 2021 at 8:51 | comment | added | Taras Banakh | I can also prove that $G$ is even and $a^2=b^2$ for two distinct $a,b\in A$, then the center of $G$ has even index in $G$. All non-commutative examples (i.e., $D_8$, $S_4$ and $GL(2,3)$ of almost squared groups show that exacty this happens: any almost square root contains two distinct elements $a,b$ with $a^2=b^2=1$. | |
| Feb 23, 2021 at 7:48 | comment | added | Taras Banakh | Observe that for $f=(|A|\pm 1)\sum_{a\in A}a$ and $e=\sum_{g\in G}g$ we have $(f-e)^2=(|A|\pm 1)^2x$. Maybe this equality can be used to show that $x$ showld have small 2-power order? Maybe this recent result (arxiv.org/abs/2101.02586) about matrices could be of help? | |
| Feb 22, 2021 at 23:10 | comment | added | Taras Banakh | Unfortunately, this my argument does not work. It only proves what I already wrote earlier: if $x$ has order $2^n$, then $|G|$ is divisible by $2^{n+1}$. No so much. | |
| Feb 22, 2021 at 21:26 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Deleted incorrect arguments
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| Feb 22, 2021 at 16:58 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Partially answered Question 5.
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| Feb 22, 2021 at 16:48 | comment | added | Taras Banakh | I am trying to determine the smallest odd cardinality of a group for which it cannot be immediately shown that it is not almost squared. It seems that all groups of odd cardinality $3<4n^2-1<5^2\cdot 3^3$ are not almost squared. So, finding such a group by brute force in GAP is a hopeless task. I do not know whether there is an almost squared group aboung groups of order $5^2\cdot 3^3$. | |
| Feb 22, 2021 at 11:44 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Reduced to the case exceptional element $x$ has order $2$.
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| Feb 21, 2021 at 23:32 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
typo
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| Feb 21, 2021 at 23:26 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Clarified and expanded earlier arguments.,
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| Feb 21, 2021 at 22:11 | comment | added | Taras Banakh | Therefore $x\in D$ and since elements of $D$ do not commute, we conclude that $|D|=1$ and $H=\{1,x\}$. Therefore, if $(D^+)^2=H^+-x$, then $x\in D$, $x^2=1$ and $C_G(x)=\{1,x\}$. | |
| Feb 21, 2021 at 22:06 | comment | added | Taras Banakh | I noticed (even on the answers to my preceding "squared" question) that there is a kind of telepathy when different mathematicians on huge distances arrive to the same conclusions in the same time. By the way, it is not the end of the story. If $(D^+)^2=H^+-x$, then $x\in D$ and $H=C_G(x)$ is a 2-element group. Assuming that $x\notin D$, we can find an element $u\in D$ with $u^2=1$. Since $u$ and $x$ commute and $x^2=1$, they generate a 4-element subgroup of $H$, which implies that $|H|$ is divisible by 4, which is not true. Therefore $x\in D$. | |
| Feb 21, 2021 at 21:57 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Went further with the case $x \neq 1.$
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| Feb 21, 2021 at 21:45 | comment | added | Taras Banakh | If $(D^+)^2=H^+-x$, then $x^2=1$. Indeed, in this case $|H|=|D|^2+1$ and $|H|$ is even since $x\in H$ has 2-power order. Then $|D|=2m+1$ for some $m$ and hence $|H|=|D|^2+1=4(m^2+m)+2$ is not divisible by 4. So, the order of $x$ should be 2. | |
| Feb 21, 2021 at 17:28 | comment | added | Taras Banakh | If $x=ab$ for some $a,b\in A\setminus C_G(x)$, then $xax^{-1}=x^{-1}ax$ and $xbx^{-1}=x^{-1}bx$ and hence $a,b\in C_G(x^2)$. Maybe this can be used somehow? For example that $A\cap C_G(x^2)$ is an almost square root of the subgroup $C_G(x^2)$? | |
| Feb 21, 2021 at 16:46 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
further comments
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| Feb 21, 2021 at 16:32 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
typos corrected
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| Feb 21, 2021 at 16:08 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Corrected by removing some text
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| Feb 21, 2021 at 15:59 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
added more remarks
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| Feb 21, 2021 at 13:13 | history | answered | Geoff Robinson | CC BY-SA 4.0 |