Finite Dimensional Vector Spaces IIT JAM Question | Mathematics CUET (PG) 2025 Question | Lec-8

[@iitjam_mathematics]

➡ 𝐅𝐑𝐄𝐄 𝐋𝐞𝐜𝐭𝐮𝐫𝐞𝐬, 𝐩𝐫𝐚𝐜𝐭𝐢𝐜𝐞 𝐪𝐮𝐞𝐬𝐭𝐢𝐨𝐧𝐬 & 𝐦𝐨𝐫𝐞: bit.ly/41buOC0 💰𝐏𝐮𝐫𝐜𝐡𝐚𝐬𝐞 𝐋𝐢𝐯𝐞 𝐂𝐨𝐮𝐫𝐬𝐞 𝐇𝐞𝐫𝐞 𝐚𝐭 𝐁𝐢𝐠 𝐃𝐢𝐬𝐜𝐨𝐮𝐧𝐭𝐞𝐝 𝐏𝐫𝐢𝐜𝐞 :- bit.ly/4dZ7UD9 💰𝐏𝐮𝐫𝐜𝐡𝐚𝐬𝐞 𝐑𝐞𝐜𝐨𝐫𝐝𝐞𝐝 𝐂𝐨𝐮𝐫𝐬𝐞 𝐇𝐞𝐫𝐞 𝐚𝐭 𝐁𝐢𝐠 𝐃𝐢𝐬𝐜𝐨𝐮𝐧𝐭𝐞𝐝 𝐏𝐫𝐢𝐜𝐞 :- surl.li/dsabzx 📺 𝐘𝐨𝐮𝐓𝐮𝐛𝐞 𝐂𝐡𝐚𝐧𝐧𝐞𝐥:: kzbin.info/door/g7ESZu0ESsWIHAwysBetbQ 📲 𝐓𝐞𝐥𝐞𝐠𝐫𝐚𝐦 𝐆𝐫𝐨𝐮𝐩: t.me/IITJAM_NBHM_MScEntranceMathDSCN 📸 𝐈𝐧𝐬𝐭𝐚𝐠𝐫𝐚𝐦 𝐏𝐚𝐠𝐞 : instagram.com/ifas_maths__iit_jam_cuet_nbhm/ 📘 𝐅𝐚𝐜𝐞𝐛𝐨𝐨𝐤 𝐏𝐚𝐠𝐞 : facebook.com/MScEntranceMaths/

[@hrishutiwari4929]

Consider the action of S4 on Z [ x1,x2,x3,x4] given by σ.p(x1,x2,x3,x4) = p( xσ(1), xσ(2), xσ(3), xσ(4) ) for σ belongs to S4.Let H ⊆ S4 denote the cyclic subgroup generated by (1423). Then the cardinality of orbit OH(x1x3+x2x4) of H on the polynomial x1x3 + x2x4 is

[@Nainagour-l1h]

Since H is generated by (1423) the orbit x1x3+x2x4 under the action of H is the set {(1423)^n(x1x2+x2x4):n belong z} Calculate the value of (1423)^4(x2x3+x2x4) for n=0,1,2,3,4 (1423)^0(x1x3+x2x4)=(x1x3+x2x4) (1423)^1(" "")=(x4x1+x3x2) (1423)^2( """)=(x2x4+x1x3) (1423)^3( """")=(x3x2+x4x1) (1423)^4( """)=(x1x3+x2x4) The cardinality of the orbit oH (x1x3+x2x4) of H on the polynomial x1x3+x2x4 is 4 Answer 4

[@hrishutiwari4929]

Let < (0,2) > denote the subgroup generated by (0,2) in Z4×Z8. Then the order of (3,1) +< (0,2) > in the quotient group Z4× Z8 / < (0,2) > is

[@hrishutiwari4929]

Sir next lecture mein ye explain kr dijiyega

[@Nainagour-l1h]

Sir homework ka answer option d is correct thank you sir radhe radhe 🙏

[@hitendrakumar2070]

Option d will be the absolutely correct answer

[@mansigupta9604]

D is the correct answer

[@hariomgupta9116]

H.W ka D correct hoga

[@as-bj4hq]

Option d correct

[@abhijitdas2976]

8:55 sir our question is G is not mentioned is abelian, cyclic, so kase HK is group

[@gauravkumaragnihotri6472]

Group was cyclic.

[@abhijitdas2976]

Option d