Ok so there has been a lot of clutter recently with like 12 threads of people posting their homework problems. No need for that.

I am a moderator on another site where we get 1000 Math/physics hw posts per day

Our homework section is pretty big on that website, and there are some basic rules that i think should be followed here.

1. Dont berate people
- this should be obvious. You were once at their level too.
2. In order for us to help you, you must post a reasonable attempt at a solution.
- We are not here to do your homework for you. We already passed the class in question
3. When you post a question, to make it easy put in into the following form:

1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

i URGE posters not to help people unless they post an attempt at a solution.

Mods feel free to add anything else.

Bring on the first questions.

Surely it's many times more useful to hint at a solution, or comment on someone's proposed solution, than it is to just give the solution to them? It seems like nothing will be learnt if you just hand the answer to someone.

Woot! I start three math courses in about a month. Thanks for starting this thread!

first time in ~18 years that i won't be in school, maybe i'll use this thread to see if i retained any of the **** i learned there

ok this is Abstract algebra

Show by means of an example that it is possible for the quadratic equation
x^2=e to have more than two solutionsin some group G with identity e.

Now I have been thinking about this on my own all day and trying to figure it out for myself and i came up with two ideas but havent figured it out

so a hint in the direction would be great

1.im thinking it has to do with a group in the complex or matrices

2. or I define x to be an element of a group s.t. a quadratic as well.

EX: (x^2)^2 giving four roots possibly.

but the point i think here is e is always e.

btw it will be much harder to search for individual questions in one subfolder. but i like your format

you can do it with four elements, so just write out the multiplication tables for all of the groups with four elements (how many of them are there?)

Suppose you have a rectangle. What symmetries are possible with this object? Well, we can leave it alone, rotate it through 180 degrees, reflect it vertically, or reflect it horizontally. Draw a rectangle and convince yourself these are all different maneuvers.

Each of these symmetries is such that if you repeat it twice, you return the rectangle to it's original position, i.e. every symmetry x satisfies x^2 = e. The symmetries of any object always make a group under the obvious operation of composition.

The examples given so far all work.

However, it's also possible to come up with a solution using the group of invertible 2 by 2 matrices over a given field. This is what you seemed to be getting at with the idea above.

Okay we have a bunch of examples (actually my example is the same as blah_blahs but whatever).

The way I would think about it would be 'can I think of some nice easy examples of elements that have order two'.

Well multiplication by 1 and -1 has order two, since -1*-1 = 1, but that doesn't help much as it only gives us two elements - though in fact {1,-1} X {1,-1} gives a group with the property you require, the same group me and blah_blah mentioned above.

Think about geometry, the most obvious elements with order two are reflections about some line of symmetry. If I reflect something, and then reflect it again in the same line, I'm back where I started.

Consider an equilateral triangle. I can reflect it in any line which bisects a side and bisects the angle at the opposite vertex (draw a picture). So, there are three reflections here as well as the identity element. However, these do not make a group in themselves as the composition of two reflections through different lines is not a reflection. We need to add the other two symmetries, which are a clockwise rotation of 120 degrees, and a clockwise rotation of 240 degrees. This gives us a group of six elements, four of which have order 2 (the other two have order three).

The symmetry group of any regular polygon will have more than two elements of order two, as a square has 4 reflections, a pentagon 5, etc.

Do not know if you have looked at permutations yet, but if so another obvious example of an element of order 2 is a transposition. Roughly speaking, the Symmetric group on n elements, Sn, is the group of all possible permutations of those elements, with the operation 'composition'. A transposition is a permutation which just swaps two elements, i.e. for the group S3:

Our set is {1,2,3}.
The transpositions are {2,1,3} (swapping 1,2), {1,3,2} (swapping 2,3), and {3,2,1} (swapping 1,3).
The other permutations move all elements {2,3,1}, and {3,1,2}

For n = 3 or more, there will be more than one possible transposition.

In the specific case of n=3, this is just the symmetry group of the triangle again, but this is not true for larger n.

Good thinking Rosie imo. It says in the sticky about no h/w threads offering money for a solution. Maybe this part should be amended.

It seems to be seasonal anyway. Like high volume around exams and beginning of term. Do you get that on the site you mod?

So is this valid for my algebra question above.

by example show that x^2=e has more than 2 solutions

let x = {x^2|x is an element of C)

then

(1^2)^2=e
(-1^2)^2=e
(-i^2)^2=e
(i^2)^2=e

where a*e=e*a=a (identity element)
what do you think is this legit?

Your terminology is nonsensical, I'm afraid.

let x = {x^2|x is an element of C)

This means nothing. I assume you mean something like

let S = {x^2|x is an element of C).

Note that this set is exactly the same as the set C, since every element of C is the square of some other element of C.

The set C under multiplication does not have more than two elements satisfying x^2 = e, the only ones which do are 1 and -1.

You misunderstood. Anybody can just post a problem and we can just post an answer and they wont learn anything.

The point is that they make an attempt at a solution, and we tell them where they went wrong and put them on the right track.

Subgroup Theorem:
A subset of a group G is a subgroup of G if and only if
1.H is closed under binary operation of G
2.the identity element e of G is in H
3.for all a is an element of H it is true that a^-1 is an element of H also

This is what my book says and then gives examples and discussion.

I am wondering whose theorem this is? When I search subgroup theorem many come up. I am looking for a proof of the subgroup theorem. I handed in my homework, and my teacher said I needed to prove for each question each individual category.

So for example we know e for the matrices of invertible matrices exists. But I am not allowed to use axioms for proofs.

So I am looking for the subgroup theorem proof, he did it on the board really quick at the end of class so I couldnt get it all down. The identity proof was pretty lengthy and I did confirm that I had to prove it for all identities.

A subgroup of G is defined as a subset which under the group-action of G, forms a group.
The three points you list are just three of the group axioms, so there is nothing to prove.
What you have written is more like a definition than a theorem.

What does your book say is the definition of a subgroup, then? The above looks to me like a standard definition and not a theorem.

BTW this is not homework, I am just trying to get an understanding


So for example if I want to show that do this one:

determine whether the given set of invertible nxn matrices with real number entries is a subgroup of GL(n,r)

the given set is nxn matrice with determinant 1 or -1

For property 3

We know there exists an identity matrix such that a*a^-1=a^-1*a=a

So would I then go on to show:

[ab][10]=[ab]
[cd][01] [cd]

[10][ab]=[ab]
[01][cd] [cd]

Im just trying to get an appropriate format. In my mind I understand how to verify these conditions now. Now I need to figure out out to express them in appropriate mathematical terms.

sorry that was for the identity not the inversse but same questions apply

Let the given set be H.

Is H closed under multiplication? Yes, because the determinant of the product of two matrices is equal to the product of their two determinants, so this will be +/- 1.

Is the identity element in H? The identity element of GL(n,r) has determinant 1, so it is in H.

Does every element of H have an inverse in H? A square matrix is invertible if it does not have determinant zero, so all elements of H have an inverse in G. For any element h of H, we have h * h^-1 = e, and so |h| * |h^-1| = |e|, so |h^-1| = 1 / |h|. Thus h^-1 has determinant 1 (if h has determinant 1), or -1 (if h has determinant -1). Thus h^-1 is in H as well, and we have shown that H forms a subgroup.

This was given as theorem 5.14

The definition was given as

5.4 definition

If a subset H of a group G is closed under binary operation of G and if H with the indeuced operation from G is itdelf a group, then H is a subgroup of G


The both say the same thing obviously. But they do seperate the 2 as given

I total agree with you. I am going to submit this to my proffesor to get his exact comments.

I'll be a little more general here. If you just think of a group as some generators and some relations, then:

G_n : < a_1, a_2, ..., a_n | (a_1)^2 = (a_2)^2 = ... = (a_n)^2 = 1 >

is an infinite multiplicative group with at least n+1 elements satisfying x^2 = 1.

Real Analysis

Given

a_n, l, and e

we know |a_n-l|<e for convergent sequences

So my question is, to find the right e do we solve the inequality |a_n-l|<e
or a_n<e?

also im a little confused on when to flip the inequality. I know we flip it when we divide by a negative

Typically to prove convergence of a sequence we solve for the N (in terms of e), such that |a_n - L| < e for all n >= N.