Exams as a prerequisite for a degree are useless. Exams as part of a degree must be abolished. And degrees must be abolished. Ultimately the only metrics that really matter are money and fame. See also: motivation.

The only thing exams should matter for is as a screening tool to select people with specific abilities that you care about as an employer or principal investigator. If:then exams are useless for your purposes. then might as well just go by interviews (basically what all employers do already, though not PIs). Degrees are too course grained to mean anything to anybody. Employers and PIs likely only care about very few specific subjects.

Once the question of an exam has been formulated, the usefulness of the problem is already been completely destroyed, because formulating the problem that matters is the most important part of things. And any problem with an answer, is useless to put effort into: give answers.

Furthermore, preventing people from searching for answers while answering an exam, AKA preventing "cheating", also makes absolutely no sense. In the real world, we want people to find answers as quickly as possible! We should be teaching people how to "cheat"! What we should teach them instead is what a fucking license is, and what you have to do to comply with it.

And if you must absolutely have exams, they must be open to anyone who wants to applies. Then people have to pay to take the exam, with subsidies for "official course takers", who are spending 100x more anyways due to not living with their parents.

And if you pass the exam, you pass the course, without any further time requirements.

And those exams must be applied by professional test application companies to ensure no cheating and to factor out the anti-cheat work, while still making the tests available to people anywhere.

A quote from Richard Feynman present in the book Surely You're Joking, Mr. Feynman chapter O Americano, Outra Vez!:

The only metric that matters is "to feel that you've satisfied youre curiosity". When one studies for that, it can take a lot more time to actually learn everything, because it is sometimes not as clear when you should stop. But it is the only way to go deeper.

A person's understanding is the most illiquid asset that exists, to judge that based only on standardized exams, is a certain way to fail to identify top talent.

Related: exams and homework are useless, only projects matter

By Ciro Santilli: Section "How to teach".

Others:

Vs: image: the codomain is the set that the function might reach.

The image is the exact set that it actually reaches.

E.g. the function:could have:

Note that the definition of the codomain is somewhat arbitrary, e.g. $x^2$ could as well technically have codomain:even though it will obviously never reach any value in ${\mathbb{R}}^2$.

The exact image is in general therefore harder to characterize.

The bane of multicellularity.

Obesity is an extremely serious disease that is very hard to cure, and has deep psychological implications.

As per en.wikipedia.org/w/index.php?title=Semidirect_product&oldid=1040813965#Properties, unlike the Direct product, the semidirect product of two goups is neither unique, nor does it always exist, and there is no known algorithmic way way to tell if one exists or not.

This is because reaching the "output" of the semidirect produt of two groups requires extra non-obvious information that might not exist. This is because the semi-direct product is based on the product of group subsets. So you start with two small and completely independent groups, and it is not obvious how to join them up, i.e. how to define the group operation of the product group that is compatible with that of the two smaller input groups. Contrast this with the Direct product, where the composition is simple: just use the group operation of each group on either side.

Product of group subsets

So in other words, it is not a function like the Direct product. The semidiret product is therefore more like a property of three groups.

The semidirect product is more general than the direct product of groups when thinking about the group extension problem, because with the direct product of groups, both subgroups of the larger group are necessarily also normal (trivial projection group homomorphism on either side), while for the semidirect product, only one of them does.

Conversely, en.wikipedia.org/w/index.php?title=Semidirect_product&oldid=1040813965 explains that if $G=N⋊H$, and besides the implied requirement that N is normal, H is also normal, then $G=N\times H$.

Smallest example: $D_6=C_3⋊C_2$ where $D$ is a dihedral group and $C$ are cyclic groups. $C_3$ (the rotation) is a normal subgroup of $D_6$, but $C_2$ (the flip) is not.

Note that with the Direct product instead we get $C_6$ and not $D_6$, i.e. $C_3\times C_2=C_6$ as per the direct product of two cyclic groups of coprime order is another cyclic group.

TODO:

Bibliography: math.stackexchange.com/questions/1726939/is-this-intuition-for-the-semidirect-product-of-groups-correct