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Group of order 56 is not simple
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- เผยแพร่เมื่อ 7 ก.พ. 2021
- Telegram link
. Upasana Pahuja Taneja
Mathematics for graduation, post graduation, NET, iit jam, Gate
t.me/upasanataneja
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
#Groupoforder56isnotsimple
open set
• Open sets
supremum and infimum
• Supremum and infimum r...
upper bound and lower bound
• Upper bound and lower ...
Group definition • Definition of a group ...
Cyclic group
• Group theory part 6-Cy...
Subgroups
• Group theory part 3-Su...
One step subgroup test
• Group theory part 4-on...
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
• Permutation Groups || ...
Function domain and Range
• Function domain and r...
concept of 1-1 onto and bijective mapping
• one one onto function
Difference between Range and codomain
• Difference between Ran...
Relation function domain codomain range and image
• Relation: Image, Range...
HOMOMORPHISM
• Group theory part 8-Gr...
Isomorphism
• Group theory part 9-Gr...
normal subgroups
• Normal Subgroups
first theorem of homomorphism
• Group theory part 11- ...
cosets
• Group theory part 12-C...
Quotient groups
• Group theory part 13- ...
sylow p subgroup
• Sylow P subgroup
sylow theorems
• Group theory part 14- ...
Beautifully explained, Ma'am, thank you so much
Thank you❤🌹 so much mam
1lk
Beautifully explained
Good presentation my friend
thank you!
Very nice explanation mam 😊
Muchas gracias. Saludos desde México.
Thank you mam for this presentation, it is so helpful.
Many Thanks....please refer to other Videos of modern Algebra and real analysis...
Very nice explanation mam❤
Mam pls make a video on group of order 30 .
What will be the cases?
Ma'am if last condition was hold then will what??
Thank you mam
Mam isme jo 7 or 8 order ka group h Usme common elements bi to ho sakte h
Mam,why are you take such as n1,n7 .which method ,please explain
Thank You so much ma'am!!😭
You are so welcome!
Keep watching
Mam Kya sirf identity common ha or koi elements common nahi Ho sakta.
thanks mam
Most welcome 😊
Thanks mam 🙏🙏🙏
Thank you
Very usefull
Thank you...😊
Thank you so much ma'am
Thank you 😊
@@dr.upasanapahujataneja1707 Ma'am is it possible to personally contact with you ??
Mail me at upasna.pahuja@gmail.com
@@dr.upasanapahujataneja1707 okk thank you ma'am
Mam there can b more elements common in two Sylow p subgroups...why are we taking only identity element in intersection?
They Will not have any element common other than identity as they are groups of prime order...
@@dr.upasanapahujataneja1707 by this you are claiming 8 is a prime number, again, “2-SSG” is a group with order being powers of 2, in this video every 2-SSG has order 8, by no ways you can restrict it to just a cyclic subgroup or order 8.
Ma'am why there is only identity element common between two p-ssg?
Thanks for asking the question😊
Ans- As every group has Only one identity element and these are Subgroups of this group so the identity element is common.
@@dr.upasanapahujataneja1707 the question here is not why identity is a common element between two p-ssg, but the “only common”.
Because she made a mistake on assuming all “p-SSG” has prime order, forgetting the p here indicates the group has order of a power of p, not that group has prime order.
Please make a video on "group of order 120 is not simple"
Hii..have u ans for tiz question?? Explain it..
Is (1, 2) a simple group ?
Write me on mathsclasses87@gmail.com
Mam ssg ki full form btana
Sylow subgroup
Mam, here in 2-SSG order is 2
1 is identity common in all H1,.....H7 and rest contains one more element in total 1+7 = 8 (you took 48 here)
And 7 SSG is 48
In total giving 56 which divides 56
Am I right ????
The order of 2-ssg is not 2. It is 2^3 = 8
Mam in case of group of order 30,
n2 = 1,3,5,15
n3 = 1,10
n5= 1,6
Now how to proceed further?
What will be the different cases?
This prove fails at 7:49 here the number of “2-SSG” is 7 and number of elements in every “2-SSG” is 8, we know nothing about these subgroups of order 8 and you just assumes they are all isomorphic to cyclic group of order 8 and trivially intersects. THIS IS SIMPLY FALSE. The same problem arises multiple times in your videos, and you have refused to address this issue.
Will definitely look into this and will revert asap
English please? Can't understand Indian
Please stop giving such wrong solutions...