Every subgroup of a cyclic group is cyclic.

[@devikasahare2386]

Thank you sir excellent 👍❤

[@Abhaybhadouria1670]

Found this after being disappointed by gazendra purohit sir ❤‍🩹

[@rashidulislam4555]

Bhai tu 5mahine pehse se padai kar raha tha😮kal exm hein Aaj Sikh Raha hoon😭😭

[@Radheradhe23468]

Sir book m lengthy tha boht apne easy way m smjha diya ♥️

[@Semicolon-l8f]

Are aap yha bhi

[@nityamsharma9485]

Me to ratunga ise, kuch smjh nai aa rha

[@Yeahmansingh]

thanks you made it very easy to understand

[@eshaparvez4972]

Well explained Sir!!

[@sandeepkaur2567]

Thank you sir.😊 you explained very well .

[@Subho_0513]

Dua koro sir
Mera paper thik se jaye

[@Renu-us6fs]

Sir pdf iss chapter k chaiye

[@jasrotia7350]

Thanku . Nd ur writing too good sir ...

[@user-yr9dy7oj1p]

Sir .. kindly guide me to answer the below question. Let G be a finite group. Then show that o(G) = o(Z(G)) + ∑[G:N(a)]>1[G : N[a]]

[@JasveerSingh-kh9qd]

Very hardworking sir 👨‍🏫

[@Rahul12345-p]

Thank you dada,khub help holo

[@Renu-us6fs]

Excellent sir

[@mathematics9334]

Sir ,
b€H and a^k€H
=> b.( a^k)€H. (By closure property)
How can we say that
b.( a^k) ^(-q)€H.

[@MathsICU]

We know that if a belongs to H (a€ H) then every integral power of a is also in H. (a^k€ H for every k in the set of integers) , H being a subgroup.
Thus a^k € H then every integral power of a^k is also in H.
(a^k)^(-q) € H

[@mathematics9334]

@@MathsICU Ok sir , now I got it.
Thank you so much . Now it is clear.

[@sanamsanam7529]

Sir is theorem ka hum use kaha per karta hai

[@MathsICU]

For example Z5 is a cyclic group with respect to addition modulo 5
Therefore every subgroup of Z5 is also cyclic
You do not need to check their cyclicness.

[@Subho_0513]

Sir kal mera group theory ka exam he .

[@MSDian-w3]

Trivial and improper case kaha hai,, University ka koi acche teacher khata check karenge toh negative marks milenge, pure maths ka proof eise likhenge toh

[@sachin1bollywood]

New Subscriber Sir ☺️

[@MathsICU]

Thanks for subbing!

[@crazyaman6450]

Itna lambaa

[@rutujapande2020]

Thank you sir

[@harshitapandey8342]

Thank you

[@Subho_0513]

Love From Purulia (west bengal)

[@MathsICU]

Best wishes 👍

[@Subho_0513]

@@MathsICU thank you Sir

[@awaisshoukat150]

Good

[@Anujkumar-we3ky]

good

[@sm-qh2zp]

H =({i^-³, i², i³,i⁴},×)
Here in i²,i³,i⁴→2,3,4 are positive so take it as it is
In i-³ → -3 is negative taki inverse of i-³ i.e. i³.
Now the obtained element ao far is in some positive power of i and belongs to H (i.e. {i³,i²,i³,i⁴} but are not forming group

[@inzanevlog1816]

your handpen 😧

[@birupakshchoudhury7057]

lovely proof

[@varunchoudhary8149]

.

[@gourabbiswas2002]

Thank you sir

[@sudhasistla60]

Thank you sir