I'm afraid the statement of the theorem: "Every finite group is isomorphic to a collection of permutations" implies that _collections_ have structure, which they don't. The correct statement should be "Every finite group is isomorphic to _a group constructed by_ a collection of permutations _and function composition_".
It should be understood that the binary operation is function composition, as discussed in the previous lecture. Similarly, we can say things like "the group of integers", without saying "under addition" each time.
@@matthewmacauley5441 While I agree that the operation is implied, the slide said _collection_, not group. Not all collections of permutations with composition qualify as groups.
This might be late, but I believe this is false. Calculating first right to left, we get the same r^2, but my rf = (15)(26)(34) and r^2f = (16)(24)(35)
So in short, we can take any order `n` and map all the possible groups in it by putting `Sₙ` at the centre and "branching" every other group as derived from that `Sₙ`? I think it could be more clear to see by looking at the multiplication table itself: Since it's been said already that every element can appear only once in each column, and all elements must appear in each column, then it's quite obvious that a column is a permutation, because it's like shuffling a deck of playing cards. I'm also curious: is there any formula for the number of all possible groups of order `n`?
Why not labeling node 1 the identity?
I'm afraid the statement of the theorem: "Every finite group is isomorphic to a collection of permutations" implies that _collections_ have structure, which they don't.
The correct statement should be "Every finite group is isomorphic to _a group constructed by_ a collection of permutations _and function composition_".
It should be understood that the binary operation is function composition, as discussed in the previous lecture. Similarly, we can say things like "the group of integers", without saying "under addition" each time.
@@matthewmacauley5441 While I agree that the operation is implied, the slide said _collection_, not group. Not all collections of permutations with composition qualify as groups.
i think at 5:05, r²=(123)(465), rf=(16)(24)(35), r²f= (15)(26)(34). you should give this example before that homework in lecture 2.3
This might be late, but I believe this is false. Calculating first right to left, we get the same r^2, but my rf = (15)(26)(34) and r^2f = (16)(24)(35)
So in short, we can take any order `n` and map all the possible groups in it by putting `Sₙ` at the centre and "branching" every other group as derived from that `Sₙ`?
I think it could be more clear to see by looking at the multiplication table itself: Since it's been said already that every element can appear only once in each column, and all elements must appear in each column, then it's quite obvious that a column is a permutation, because it's like shuffling a deck of playing cards.
I'm also curious: is there any formula for the number of all possible groups of order `n`?
The number of groups of order n (up to isomorphism) is sequence A000001 in OEIS: oeis.org/A000001
Prof. Mccauly , what symetric group is the E8 group isometric to ?
☄️
sir,u were supposed to explain the proof of cayley's theorem not proving the statement using some examples from the textbook