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Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
Skewbrow:
I'll be teaching freshman abstract algebra this coming Spring, and Ankh's sheet could well come from my course. Which questions are giving you a headache?
ankhtahr:
This is the math I need to do for Computer Science here. This is linear algebra. I also have advanced mathematics/analysis.
Currenly I'm still struggling with some of the definitions, but 1 c) and 4 are probably the most difficult.
I'm working a little bit ahead right now, This sheet is only due on tuesday, but I'd like to maybe sleep next week.
Skewbrow:
--- Quote from: Ankhtahr's Teacher ---Assignment 1: Let G and H be groups with respective neutral elements e G and e H
Let f be a homomorphism of groups from G to H. Show that:
c) Let p: G -> G/ ker( f ) be the canonical projection mapping the element g to its coset g ker(f) (sorry, I don't know
what "coset" is in German). Then there exists a unique homomorphism f from G/ ker( f ) to H such that
f = f o p (using underscore instead of the overline bar, sorry). Furthermore, f is injective and its image
is equal to that of f.
--- End quote ---
Here we have no choice in constructing the homomorphism f. If g ker(f) is an arbitrary element of the quotient group G/ ker( f ), we have no choice. Because f = f o p, and g ker(f)=p(g), we must define f(g ker(f))=f(g). Can you follow this?
This means that the uniqueness part will be clear, once we get existence. But there are several catches before we are there. First and foremost you should check that f is well defined (=wohldefiniert?). This is because distinct elements of the group G may give rise to the same coset. So we need to clear the following obstacle. If g ker(f) = g' ker(f) for two different elements g and g' of the group G, then we must be sure that f(g ker(f))= f (g'ker(f)). In other words, we need to check that f(g)=f(g'). Can you do that? Hint: here it is absolutely crucial that we are talking about cosets of ker(f).
Ok, if that is clear, then the rest is relatively easy. Namely you need to check that f is a homomorphism of groups. This is not very difficult, because you only need to recall what the group operation of G/ ker(f) is, and then utilize the fact that f was known to be a homomorphism of groups.
This exercise is designed to make you once more think through the meaning of the relevant definitions.
Skewbrow:
--- Quote from: Ankhtahr's teacher ---Assignment 4: Let M be a set and P(M) its power set. For all elements A, B of P(M) define the
symmetric difference (Can't draw the Delta symbol here. I'm using '+' instead, because this will become
the addition of the ring you are creating. Similarly 'n' denotes intersection when standing alone).
A + B := (A U B) \ (A n B).
Show that (P(M),+,n) is a ring. You are allowed the assume the associativity of the symmetric difference.
--- End quote ---
Ok. So the first task is to show that (P(M),+) is a commutative group. Before we start let me recap. The set A+B consists of those elements of M that belong to either A or B but not to both. This will be another subset of M, i.e. an element of P(M), so '+' is an operation on P(M). We were allowed to assume associativity, so let's skip that. A neutral element? Can you think of a subset S of M with the property that no matter which set A is, then S+A=A. Hint: S should not contain any elements of A, because those would be excluded from S+A.
Proving the existence of inverse is dependant on knowing what the neutral element is, so I stop here, and let you sweat a bit. :evil:
Edit: I am trying to guess a helpful level of chatter and hints here. If it is Greek to you, say so, and I will make another attempt. Getting used to the level of abstractness here requires a fair amount of work, and takes a little while. For example, I would guess that may be top 10% of the students in my course would make any serious headway in your Exercise #4. A vast majority of students here are unprepared to fathom that they can do addition and multiplication on sets. German high school may be a bit better in that regard. The exercise is not very difficult, but you have to absorb the rules of the game of definitions and axioms to stand a chance.
Pilchard123:
(click to show/hide)An abelian grape
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