Timeline for Normal subgroup of prime index
Current License: CC BY-SA 3.0
20 events
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| Feb 12, 2022 at 15:45 | history | rollback | Arturo Magidin |
Rollback to Revision 3
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| Feb 12, 2022 at 14:59 | history | edited | Shaun♦ | CC BY-SA 4.0 |
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| Jan 11, 2022 at 13:03 | comment | added | William | Frankly I was impressed by the result because really, what other theorems do we have that determine a subgroup $H$ is normal, just from it's cardinality? I know you could sometimes show the subgroup is unique using Sylow, and hence it's normal but that's only sometimes when the subgroup is maximal $p$-subgroup or you could use the counting statement where you have $|H \cap K|$ copies of elements in $|H| \cdot |K|$ . Are there other theorems to conclude a subgroup is normal from its cardinality? | |
| Jan 10, 2022 at 21:53 | comment | added | Arturo Magidin | @Wlliam: Yes, your reading is correct; no, it is not particularly "powerful", because the condition (i) only applies to a single value for each finite group; and (ii) "most" of the time you don't have such subgroups. The idea of using the action on cosets to get a morphism to a finite group is really the take-away you want to take from this result. | |
| Jan 10, 2022 at 21:41 | comment | added | William | Just to be sure if I understand "$p$ is the smallest prime" part. Suppose $H≤G$ and $|G|/|H| = 5$ so this would mean $H$ is a normal subgroup of $G$ if $|G| =5^2 \cdot 7$ or $5^{100} \cdot 19^{51}$ but for $G'$ where $|G'| = 3 \cdot 5^3 $ this test theorem would be inconclusive. So $H$ may or may not be normal in $G'$. Am I getting this right? Looks like a powerful theorem. The only tradeoff I see is that for $p=2$, it works even when $G$ is infinite but this theorem only works when the group is finite. | |
| Aug 22, 2021 at 14:15 | comment | added | Hetong Xu | ... And the same trick is used in classifying groups of order six: math.stackexchange.com/questions/496096 | |
| Aug 22, 2021 at 13:28 | comment | added | Hetong Xu | The standard trick in this solution was isolated in this post: math.stackexchange.com/questions/88719 , where a "a little bit" stronger result was proved. | |
| May 31, 2017 at 5:23 | history | edited | Error 404 | CC BY-SA 3.0 |
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| Nov 1, 2016 at 1:05 | history | edited | Alex Ortiz | CC BY-SA 3.0 |
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| Dec 6, 2015 at 2:46 | comment | added | user169852 | @sequence: If $p-1$ is composite, then it can be expressed as a product of prime numbers, each of which will be smaller than $p-1$, then the argument in my previous comment applies. | |
| Dec 6, 2015 at 2:33 | comment | added | sequence | @Bungo: why does $|G/K|$ have to be of prime order? Can't $|G/K|=p(p-1)$? | |
| Dec 4, 2015 at 19:07 | comment | added | user169852 | @sequence: $|G/K|$ has $p$ as a prime factor since $|G/H| = p$ divides $|G/K|$. Also, $|G/K|$ divides $p!$, which does not have $p^2$ as a factor, so $p^2$ is not a factor of $|G/K|$. No prime smaller than $p$ divides $|G/K|$ because no such prime divides $|G|$. No prime larger than $p$ divides $|G/K|$ because no such prime divides $p!$. Conclusion: $|G/K|$ must be exactly $p$. | |
| Nov 27, 2015 at 6:04 | comment | added | sequence | How does it follow that if $p$ is the smallest prime dividing $|G|$ then $|G/K|=p$? | |
| Nov 4, 2015 at 20:18 | comment | added | user169852 | @hermes: If $x \in K$ then $xaH = aH$ for every coset $aH$. In particular this is true for the coset $H$ itself: $xH = H$, and so $x \in H$. It's easy to verify that $K = \cap_{g \in G}(gHg^{-1})$, which is the largest normal subgroup of $G$ which is contained in $H$. This normal subgroup $K$ is called the core of $H$ in $G$: en.wikipedia.org/wiki/Core_(group) | |
| Mar 17, 2015 at 22:30 | comment | added | SorcererofDM | @SquiresMcGee yes | |
| Mar 17, 2015 at 19:32 | comment | added | Squires McGee | What is $S_p$? The symmetric group? | |
| Apr 17, 2013 at 7:23 | comment | added | misi | Is there a simpler homomorphism that would work for the case $|G|=p^2$? | |
| Jun 28, 2012 at 17:51 | comment | added | Sigur | Thanks. The part 'it follows that $|G/K|=p$' is crucial. Now everything is done! Bye. | |
| Jun 28, 2012 at 17:49 | vote | accept | Sigur | ||
| Jun 28, 2012 at 17:01 | history | answered | Arturo Magidin | CC BY-SA 3.0 |