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Math Help - Problems with Abstract Algebra as a freshman student

  1. #1
    Senior Member Paze's Avatar
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    Problems with Abstract Algebra as a freshman student

    Hey. I'm studying discrete mathematics and for some reason we're taking abstract algebra in our first semester (pretty sure it's because there's literally only 1 student on the 3rd year and he needs the course so they shoe-horned it onto us freshmen). Anyways, naturally, me and everyone else that have NOT been studying mathematics on a university level for the past three years are having terrible trouble grasping the subject fully. I can only speak for myself, though and my problem is that I can't seem to ''see it'' like I can see other mathematics.

    I can claw my way through problems algebraically, geometrically etc. but this idea of groups and stuff in Abstract Algebra is just completely new and I've never seen anything like it so when I'm left alone with a problem I'm just stumped...However, if I receive help, just hints even, then usually the lights go on and I can throw some sort of proof together using theorems in the book...But I won't be having anyone helping me on the finals in a month obviously...So what am I to do? How can I use this month to vastly improve my problem-solving techniques for this course? Any hints? Thanks!
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    Re: Problems with Abstract Algebra as a freshman student

    Quote Originally Posted by Paze View Post
    This idea of groups and stuff in Abstract Algebra is just completely new and I've never seen anything like it so when I'm left alone with a problem I'm just stumped...However, if I receive help, just hints even, then usually the lights go on and I can throw some sort of proof together using theorems in the book...But I won't be having anyone helping me on the finals in a month obviously...So what am I to do? How can I use this month to vastly improve my problem-solving techniques for this course? Any hints? Thanks!
    See if you can find the book ABSTRACT ALGEBRA An Active Learning Approach by Davidson & Gulick.
    That book is written for self-study in that it is meant for a Moore type class.
    Thanks from emakarov
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    Re: Problems with Abstract Algebra as a freshman student

    Quote Originally Posted by Plato View Post
    See if you can find the book ABSTRACT ALGEBRA An Active Learning Approach by Davidson & Gulick.
    That book is written for self-study in that it is meant for a Moore type class.
    I took abstract algebra from Frances Gulick many years ago. She was an excellent teacher. Her husband Denny Gulick was in the department as well and wrote the calculus book we all used for undergrad calc I-III.
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    Re: Problems with Abstract Algebra as a freshman student

    Abstract algebra is...well, "abstract" and the presentation usually doesn't allow for making it "transparent". I can't write a whole text in this post, so here's a reader's digest version:

    I'd like to talk about a few VERY SIMPLE groups. The first one I'd like to look at is: {-1,1} under ordinary multiplication. There are only 4 possible products:

    1*1 = 1
    1*-1 = -1
    -1*1 = 1
    -1*-1 = 1

    If we liked, we could replace "1" with "P" (for positive), and "-1" with "N' (for negative), giving the mantra school-children are often taught:

    Positive times Positive is Positive, Negative times Positive is Negative.....

    Note that we have a striking similarity with the rules:

    Even + Even is Even
    Odd + Even is Odd
    Even + Odd is Odd
    Odd + Odd is Even

    We can "dress this up" by writing: Even = {the set of all even integers} and Odd = {the set of all even integers, plus 1}, but this obscures the SIMPLICITY of what we have. Technically, we have this as:

    Z/2Z = {[0],[1]}, where [0] stands for ANY even integer, and [1] stands for ANY odd integer. I leave it to you to deduce how we define "plus".

    Let's talk about another group: say we have a very simple mechanism, a switch, with 2 positions: on, and off. Let's use "1" to denote: "flip the switch", and "0" to denote: "do nothing".

    Suppose by A+B we mean: do B, then do A. what is:

    1+1
    1+0
    0+1
    1+1

    under these circumstances?

    Let's talk about a more "grown-up" example. Suppose we have a set with just 2 things in it. We'll call this set S = {a,b}. Consider all 1-1 onto functions f: S-->S. There are only two of these:

    f1(a) = a
    f1(b) = b

    f2(a) = b
    f2(b) = a

    consider all possible compositions of f1,f2. What are your conclusions? Which one is the most likely candidate to "be an identity", and why is it obvious "each f" has an inverse (function)?

    Each of these examples is the "same" in some way, this "sameness" is the underlying idea for the concept of ISOMORPHISM (we just change the names, the rules remain).

    The last example is THE MOST IMPORTANT. EVERY group arises in this way.

    One group you have been using for a very long time is the integers (under addition). You have used "group properties" every time you solved an equation, like:

    3 + x = 5. The "usual way this is done":

    (-3) + (3 + x) = -3 + 5 (adding -3 to both sides)
    (-3 + 3) + x = -3 + 5 (associative law)
    0 + x = -3 + 5 (additive inverse law)
    x = -3 + 5 (identity law)
    x = 2 (evaluating -3 + 5)

    In "abstract form" in a general group, this would be:

    ax = b
    a-1(ax) = a-1b
    (a-1a)x = a-1b
    ex = a-1b
    x = a-1b (and then one would "look up" what a-1b is in a "multiplication table" or "Cayley table").

    This is the same thing you have been doing for YEARS, just "in disguise".

    Let me show you how the additive group of integers is the SAME THING as a group of 1-1 onto functions on "some set".

    Which set shall we use? Why, the integers itself!

    Let us identify every integer n, with the function fn:k --> k+n

    So for example, we identify the number 4 with the function +4 (or f(k) = k+4, if you prefer).

    Consider what we get when we compose +4 with +7:

    k-->k+7--->(k+7)+4 = k+(7+4) = k+(4+7) = k+13. We get the function +13.

    So, our identity FUNCTION: f(k) = k, for all k, is the same thing as: +0.

    The inverse function of +n, is -n, that is: (+n)(-n)(k) = (k+(-n))+n = k+(-n+n) = k+0 = k = (+0)(k).

    That is, we can regard an integer n as an INSTRUCTION to "add n to whatever we see". This instruction is "reversible" (we can always SUBTRACT n), so we have an "inverse function".

    The identity function (+0) "does nothing" (which is basically what identities do, a whole lotta nothin').

    This is essentially what groups ARE: "reversible transformations of a set". Often, the sets involved are SMALL (a finite group may only have a few elements in it, like 4, or 8).

    Now, a lot of "abstract concepts" will come up in the course of what you cover, but try not to get "confused", these ideas are actually SIMPLE, it's just the same kind of algebra you've always done with "fewer rules" (and as a bonus, you usually don't have to deal with weird things like "square roots" or "logarithms").
    Thanks from Paze
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  5. #5
    Senior Member Paze's Avatar
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    Re: Problems with Abstract Algebra as a freshman student

    Thanks a lot guys. I will keep these things in mind!
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