Group Theory 9, Subgroups

Group Theory 9, Subgroups

[LadislauFernandes]

tks for watching my videos. Of course I can explain.
1. in the step before you get a^2 (b^2)^-1 right? so if you look at the set H x^2 =e, so a^2=e & b^2 =e, so you get e.e^-1
2. funny thing: doesn't matter what element "e" actually is. Any g times its inverse must be e.
Hope this helps, feel free to ask any question...

[xoppa]

5:38 both groups are under multiplication

[nikhilkulhari]

Thank you very much for these awesome videos which helped me a lot in understanding the important concepts of abstract algebra, once again thank you very much . Sir are you professor/Teacher  at which college or university ?

[dannysmith]

at 4.36 how diid you get that e.e^-1 = e, could you explain this please