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Abstract algebra question

by Benzoate
Tags: abstract, algebra
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Benzoate
#1
Aug24-07, 01:27 PM
P: 569
1. The problem statement, all variables and given/known data

Show that {1,2,3} under multiplication modulo 4 is not a group but that {1,2,3,4} under multiplication modulo 5 is a group

2. Relevant equations

a mod n=r ;a=qn + r

3. The attempt at a solution

I'm going to assume when the problem says modulo 4, the problem is read as: a mod n=4 . I don't want you tutors to do the problem for me, I just want you to help me where I should start because I have no idea how set {1,2,3,4} is related to modulo 4 . I know a for modulo four is equivalent to: U(4)={1,2,3,4} I have no idea how to proved that U(4) ={1,2,3,4} is not a group under multiplication.
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Dick
#2
Aug24-07, 01:34 PM
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Thanks
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In the multiplicative 'group' mod 4 {1,2,3}, calculate 2*2. Does 2 have an inverse? What prevents similar similar problems in the mod 5 group?
matt grime
#3
Aug24-07, 01:36 PM
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P: 9,398
You take the symbols 1,2,3 and multiply them together as if they were integers (2*3=6) and then reduce the answer mod 4, so 2*3=6 which is congruent o 2 mod 4. Clearly this is not a group.

Note that what you talk about, {1,2,3.,4} NOT being a group under mult. mod 4, has no relation to the question which asks you to show that {1,2,3} with mult. mod 4 is NOT a group and that {1,2,3,4} with mult. mod 5 IS a group.

matt grime
#4
Aug24-07, 01:37 PM
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Abstract algebra question

Quote Quote by Dick View Post
In the multiplicative 'group' mod 4 {1,2,3}, calculate 2*2.

Yes, this is not 1,2,3 mod 4 so the set under the operation is not even closed

Does 2 have an inverse?
is immaterial since does not even satisfy closure.
ChaoticOrder
#5
Aug24-07, 02:37 PM
P: 11
This is the brute force method, at will take some time.

Make a group table. Recall that for a group, each element in the group must appear only once in each row and in each column. If you see any element appearing more than once then you know the structure you have is not that of a mathematical group.

There are more elegant ways to solve this, but when it comes to small finite groups, its fun to see the entire structure under the binary operation all at once. Cheers.
frog22
#6
Feb26-09, 05:41 PM
P: 1
1. The problem statement, all variables and given/known data

Show that {1,2,3} under multiplication modulo 4 is not a group but that {1,2,3,4} under multiplication modulo 5 is a group


Alright, take, for example, Set {1,2,3,4,5} for modulo 6 and set {1,2,3,4,5,6} modulo 7.
A group meets 4 basic requirements...
1. Closed
2. Associative
3. Has an identity
4. It has an inverse

Looking at the first set....Prove that it is closed.

To prove that this is closed, you must multiply each member in the set together one by one then divide it by the modulo (6). The remainder must be a number in the set. If not, the set is not closed.
Example:
1*2=2 2/6= 0.3 r 2 The remainder (2) is in the set, repeat this
3*4=7 7/6= 1 r 1 The remainder (1) is in the set, repeat this
4*5=20 20/6= 3 r 2 (2) is in the set
2*3=6 6/6= 1 r 0 (0) is not in the set

because (0) is not in the set, it is not closed and is not a group.

Set #2.

It is closed and can be proven by the previous step.

Associative. I have been told, at this time in my studies, to assume this is
associative.

Indentity. I'm still trying to figure this one out, sorry..

Inverse. It's not 100% clear, but this is what I know...
To find the inverse you take your starting number in the set, find
another number in the set that, when multiplied together then
divided by the modelo (7) has a remainder of 1.

example:

1 - the inverse is 1 (I had to push the I believe button)
2 - 2*4=8 8/7= 1 r 1. Then, the inverse of 2 is 4.
3 - 3*5=15 15/7= 2 r 1. 3 is 5.
4 - above is shown 4 is 2.
5 - above is shown 5 is 3.
6 - the inverse is 6 (I believe button again)

Anyway, I hope this helps. If anyone can help fill in the gaps (for my own personal benefit) I would greatly appreciate it.


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