I think a bit of the argument is missing at 14:20 - just because G has a normal subgroup of order 3 (say A) and a subgroup of order 2 (say B) doesn't immediately tell us that G is a semidirect product of A and B; we need to show that A and B have trivial intersection (which is true because their orders are coprime) and that any element of G can be written as a product of an element of A with an element of B (that is AB = G, and this is true because A being normal means that AB is a subgroup of G, and AB has A and B as subgroups, so its order must be a multiple of both 3 and 2, and hence is 6, so AB = G)
re trivial intersection: I believe the trivial intersection part is implied because the subgroups have very small order (order 2 and order 3) so it is trivial to see that they have a trivial intersection.
I do not understand either what is happening at 2:40. And I think that is the punchline in understanding what a semi-direct product is and what makes it so hard to understand for some people..
As other comment mentioned, professor forgot to mention B is a normal subgroup so the conjugate of b1 by a2 is again an element of the normal subgroup B. This argument also shows that the possible way a group can decompose into a normal subgroup and a subgroup, or conversely two groups G, H can form a group so that G is normal subgroup, is totally determined by the action discussed right here. If the action is trivial (i.e. the action does nothing), then the semidirect product is just direct product. If the action is nontrivial, then we end up with a group that is not a direct product and elements in G, H does not commute with each other (in the result group).
I really appreciate posting these lectures online for free. Maybe it is just me but I did not understand at 12:54 why if all elements have order 2 then the order of the group must be 2^N for some N. This is really unnecessarily long and probably wrong understanding... (based on the material presented so far, not using other resources or theorems not presented like Cauchys theorem) 1) If all elements have order 2 then G is abelian because (ab)(ab)^-1 = 1 so (ab)(ab) = 1 so aba = b so ab = ba. 2) Since G is abelian, the set {1, a} for each a in G forms a normal subgroup of G, and the quotient group G / {1, a} has order |G|/|{1,a} = |G|/2 3) The elements of the quotient group G / {1, a} are the cosets (left or right, doesn't matter since it is abelian) g {1, a}. There are |G| not necessarily distinct products of each element of G with {1, a}.... ie [ {1,a} , {g1, g1a} , {g2, g2a } ... {g_(n-1), g(n-1)a} ] . 4) We know that there are exactly |G|/2 distinct elements (cosets) in the products. The only way this happens is if for any particular g in G, it has some other g' in G with the same product, ie. g{1,a} = g'{1,a}. But since g{1,a} = {g, ga} and g'{1,a} = {g', g'a}, either g=g' or g=g'a. Since g != g' by assumption, g=g'a. So for any g in G, you can find another g' in G that gives you the same set by multiplying by a: g = g'a => ga = g' . 5) The quotient group's members are cosets of the form {g,ga} and the quotient group's operation is defined by {g1,g1a} * {g2,g2a} = {g1g2, g1g2a}. If we can show that the quotient group's members all have order 2, then it follows that you can apply the logic from steps 1-4 as much as necessary and that the order of G must be a power of 2. 6) The quotient group's members all have order 2 trivially. x{g, ga} = {1, a} if and only if x is of the form {1, b}...
I believe it was presented in a previous lecture; essentially, G is abelian so can be seen as a vector space over F_2, which necessarily has cardinal 2^n for some n.
