Group of order 36 is not simple
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- เผยแพร่เมื่อ 9 ก.ย. 2024
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. Upasana Pahuja Taneja
Mathematics for graduation, post graduation, NET, iit jam, Gate
t.me/upasanata...
In this video you Will how to prove a group is not simple .
you Will learn about counting of p-ssg.
I have tried my best to clear concept for you. If you like then please like share and subscribe my channel.
Many Thanks for watching
#Groupoforder36isnotsimple
open set
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supremum and infimum
• Supremum and infimum r...
upper bound and lower bound
• Upper bound and lower ...
Group definition • Definition of a group ...
Cyclic group
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Subgroups
• Group theory part 3-Su...
One step subgroup test
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Two step subgroup test
• Group theory part 5-Tw...
Order of a group and order of an element
• Group theory part 2-Or...
Permutation Group
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Function domain and Range
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concept of 1-1 onto and bijective mapping
• one one onto function
Difference between Range and codomain
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Relation function domain codomain range and image
• Relation: Image, Range...
HOMOMORPHISM
• Group theory part 8-Gr...
Isomorphism
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normal subgroups
• Normal Subgroups
first theorem of homomorphism
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cosets
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Quotient groups
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sylow p subgroup
• Sylow P subgroup
sylow theorems
• Group theory part 14- ...
group of order 56 is not simple
• Group of order 56 is n...
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This presentation is Very useful to understand this simple group concept .Thank you mam
Mam first Time I am watching this video I don't understood but 2nd time see this video super mam
result:Group of order 36 is not simple is correct.
The prove is wrong at 04:56 by saying there are 32 (distinct) non identity elements in K1,K2,K3,K4 combined is asuming K1,2,3,4 to be trivially intersected. Which might not be true. Ki intersect Kj can be 3.
TQ soo much mam .... really kl mra exam h or muje 1 unit s kush ni aata tha .....apki 3-4vedio dekh kr lg ra ki ab ...1 unit bi attempt ho jayegi...... specially yh bale questions... 👀❤️
suppose that g is a group of order p^nm , where p is prime and p doesn't divide m. show that the number of sylow p-subgroups divides m
Thanku ma'am
Mam maza aa gye very nice 👍 explain big helpful your vide
Thanks mam
Many Thanks 🙏
Tq so much your teaching very very super
Many Thanks 🙏
I can't tell how much proud I am feeling...just superbbb maam
Many Thanks...it really feels great😍🤗
4:46 NEIN NEIN NEIN. Thanks! you helped me a lot
Thanks
Thankyou ma'am
Keep watching
Important sharing mam 👍👍👍
Tq mam
5 like 👍👍
Super excellent upload shai....
2-SSG and 3-SSG don't have prime orders. Therefore we can't expect to have intersection trivial. How did you conclude that elements are greater than the order of G ?
Please watch this...In this i have explained in detail...hopefully you Will get it
th-cam.com/video/oI2zcwkgVxg/w-d-xo.html
@@dr.upasanapahujataneja1707 I watched your other video and I see the same issue in that video as well. For example, when you are considering last case: n2 = 7 and n7 = 8. Note that for 2--SSG order is not prime. therefore intersection might not be trivial. So you can't make an assumptions that total elements (identity common) are 1+49. This is only possible if intersection is trivial. Which is not the case with 2-ssg (Since order of 2-SSG is not prime). I think one can look at this way to simplify. Consider first n7 = 8, since order of 7-SSG is prime, therefore you have 48 elements. So 56 - 48 = 8 elements left. Since |G| = 56, this will only be possible with one unique 2-ssg. So we end up showing that there is a unique 2-ssg. Hence G is not simple even in the last case.
@@harrypotter-ku4lf for this group it is better to think about the size of intersection of two syl-p subgroups.
Tyy
If Order of a group is p^2 q^2 where both pq are prime, then it is so simple to call it not simple.
Thanks a lot....
Thanks for watching....please share n subscribe
Awesome mamm
Thank you....
This works only if the only common element of H1, H2, H3, respectively K1 K2 K2 K4 is 1. How do you know that?
Because identity element in a group is unique and identity of group and subgroup is same...
@@dr.upasanapahujataneja1707 I think she is asking about the possibility that there may be some non-identity element g belong to H1 and K1 at the same time (here H1 and K1 are taken for the explanation without loss of generality).
However, the answer is negative since the order of H1 and that of K1 are coprime to each other (4 and 9). To be more clearly, if there is some element g belong to H1 and K1 at the same time, we can make a cyclic group A generated by g, and note that A is also a subgroup of H1 and K1 simultaneously. By Lagrange's theorem, the order of A, o(A), must divide o(H1) and o(K1), but since gcd(o(H1), o(K1))=1, o(A)=1, and g must be the identity.
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Group of order 36 will necessarily have a normal subgroup of order 4 ??
Mam mujhe yeah galat lag raha app kese yeah bol sakte hen ki
Wo sare group unique hai ?????????????????
Well explained
Ma'am , if H1 and H2 have common elements other than identity, then how can we approach ?
They Will not have any element common other than identity as they are groups of prime order...
Any two groups of prime order are identical If they have any common element other than identity
@@dr.upasanapahujataneja1707 the fact is they do not have prime order, by claiming H1 or H2 have prime order you are saying 4 is a prime number.
Mam, how to find number ring related videos upload mam
Mam non abelian group of order 36 ka ek Example pls ap de sakte hoo plss
Mam please do the Vedio on show that any group of order 5 square seven square is abelian
Thanks 😊🙏
I have many Videos on group of order p square, q square and pq is abelian....
Will definitely upload the same
Ma'am I need guidence about CSIR net mathematics...so how can I text you personally??
Mail me at mathsclasses87@gmail.com
@@dr.upasanapahujataneja1707 thank you ma'am
Meaning of ssg??
Sylow subgroup
@@dr.upasanapahujataneja1707 thank you
You have did wrong process ma'am. Sorry to say but the subgroups may contains common non identity elements . So this counting is false .
This proof is wrong.
2-SSG and 3-SSG do not have a trivial intersection given their orders aren't prime. Therefore, you saying that the number of elements are greater than order of G overlooks the fact that you have overcounted the overlapping terms.
th-cam.com/video/1kjSNUTydmk/w-d-xo.html
Please watch this video
I clarified here
@@dr.upasanapahujataneja1707 Yes, thank you. The video you sent went with the correct approach, yet this video doesn't follow that approach. The proof is still incorrect.