Lagrange's Theorem -- Abstract Algebra 10

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Опубликовано:

7 апр 2023

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Комментарии : 10

@Zebinify Год назад

There might be a typo at 46:03? should it be (h^k)*g = g*h^(P-k)?

@lexinwonderland5741 Год назад

Great lecture!! I'm particularly glad you mentioned a matrix group at the end to connect with the seemingly abstract finite groups, it really helps tie it together and feel like "math" to a newcomer. I really wish you'd pointed out the (most?) canonical isomorphism between GL2(Z2) and D3, especially noting that [0,1;1,0] is very similar to the identity but a "flip"... hmm, makes you think of the element s in a dihedral group...

@matheusjahnke8643 Год назад

I thought of an bijection between a subset of GL2(Z2) and S2... for every permutation f of Zn.... there is a matrix M with 0's and 1's(exactly N ones) such that M[1;2;3;...;n] = f([1;2;3;...;n])

@dewookus Год назад

This video and the entire course are fantastic. One question though: Looking at the board at time 46:53, I wonder if one could just go directly from the observation that G is the disjoint union of H and gH to the line where the elements of G are listed. We know the h^k's are all distinct since the order of h is p, and we can see the g(h^k)'s are distiinct by the cancellation law for groups. That gives us all of G, as desired. Thanks, Dave

@FelipeMontealegreS 9 месяцев назад

American hero.

@matheusjahnke8643 Год назад

at 4:50 fi(h)=gi * h=left[gi](h) for all h in H except we know left[gi] is a member of SG(he showed that before*)... so it is bijective at G.... if you create an fi with an adequately restrained domain and codomain.... you create another bijective function fi. *before = ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-qEvL1F8leRQ.html

@aashsyed1277 Год назад

19:34 we will look at in a previous video ?????

@visheshlonial9624 Год назад

hello can anyone please help me understand how to prove the result at 45:30. I can't seem to figure it out myself

@jaa2174 11 месяцев назад

hg has order two, so hg is equal to its inverse: hg = (hg)^-1 = g^-1 * h^-1. Moreover, g also has order 2, so g = g^-1 and h has order p so h^p = e => h^(-1) = h^(p)*h^(-1) = h^(p-1). Combining both equations, hg = gh^(p-1), which should remind you of Dp (rs = sr^(n-1))

@visheshlonial9624 11 месяцев назад

@@jaa2174 Thank you