Abstract algebra warmup

Forum for the GRE subject test in mathematics.
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lime
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Abstract algebra warmup

Postby lime » Mon Aug 04, 2008 5:10 am

Hi guys, here are some interesting abstract algebra problems which could be good warmup before our autumn preparing run for gre subject math.

Problem 1. Is the set S = {[a]: [a] in Zm, gdc(a,m)=1 } is a group under operation of multiplication?

Problem 2. Are non-zero residue classes modulo p form a group with respect to multiplication?

Problem 3. Which of the following subsets of Z13 is a group with respect to multiplication?
a) {[1], [12]}
b) {[1], [2], [4], [6], [8], [10], [12]}
c) {[1], [5], [8], [12]}

P.S. As you solve these I would add more.
Last edited by lime on Sat Aug 09, 2008 7:28 am, edited 1 time in total.

mathsubboy
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Postby mathsubboy » Tue Aug 05, 2008 10:06 pm

1. YES, it is a group

2. Not necessarily!
non-zero residue classes modulo p form a group with respect to multiplication if and only if p is a prime.

Consider Z3 and Z4 for example.

3. a) c)YES

b)NO. since [2]*[8]=[3] is not in the subset.

JcraigMSU
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Joined: Wed Aug 06, 2008 9:25 am

Postby JcraigMSU » Thu Aug 07, 2008 10:37 am

1) How did you show inverses in this set? I was playing with the linear combonation theorem to solve this thoroughly, but am having trouble with that step.

Got 2 & 3

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lime
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Postby lime » Sat Aug 09, 2008 7:24 am

mathsubboy, well done!

JcraigMSU, Good question. I was confused with inverse here as well.

Discussed set S is a subset of Ym, set of non-zero resudue classes modulo m, which is a group according to the problem 2. Let "a" be an element of S. Since it is also element of Ym, it has unique inverse (-a) in Ym. We should show that (-a) is also in S.

(*) Apparently, gcd(bc,m)= 1 => gcd(b,m)=1 and gcd(c,m)=1 (you can prove it).

Since
a*(-a) =~ 1 (mod m) (here =~ means "congruent to")
=> gcd(a*(-a), m) = 1
But then, according to (*) gcd(-a,m)=1 and hence -a is also in S.

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lime
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Postby lime » Sat Aug 09, 2008 7:25 am

Problem 4. What is the characteristic of the ring?
a) Z5
b) Z6

Problem 5. Can the identity and unity of the ring coincide?

JcraigMSU
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Joined: Wed Aug 06, 2008 9:25 am

Postby JcraigMSU » Mon Aug 11, 2008 9:56 am

4. a)5 b)6...Am I missing something here??

5. Are you asking can the additive identity = multiplicative identity?
If so, only trivially.
Assume 1 = 0 in a ring R. Take r in R:
-->r = r * 1 = r * 0 = 0 because 0 absorbs multiplicatively.
-->r = 0 = 1 --> ring with one element, 0.

JcraigMSU
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Postby JcraigMSU » Mon Aug 11, 2008 10:13 am

Also, Lime, I am not quite following your argument for 1.
If we accept that the sets in 2 are only groups when p is prime (which is true Y4 for example is not closed), we can't assume that there exists inverses for each element in the Ym set when m is not a prime. Infact for problem one, if m is a prime, S = Ym.

So, in that manner I don't see how we can assume (-a) is in Ym, thereby befuddling the proof provided.

Through relatively simple examples, it's clear that when m is non-prime, S = {units of Zm} which is a group. However, it does not suffice for the proof.

I'll keep tryin though!

JcraigMSU
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Postby JcraigMSU » Mon Aug 11, 2008 10:36 am

erm, I got it. I must have been in a fog on Friday to not see this haha.
Linear combination --> exists integers, u, v such that

au + mv = 1.
mv = 1 - au
definition of division --> m/(1 - au)
definition congruence --> 1 =~ au (mod m)
--> u = (-a)

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lime
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Postby lime » Wed Aug 13, 2008 10:22 am

Very good job, JCraig. I like your proof for problem 5.

Problem 6. Consider two groups:
G1=Z4, additive group and
G2, the multiplicative group of the non-zero elements of Z5 (check Problem 2).
It follows readily that the mapping G1->G2:
0->1
1->3
2->4
3->2
is an isomorphism. Do other isomorphisms G1->G2 exist?

yael
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Joined: Thu Aug 14, 2008 2:12 pm

Problem 6

Postby yael » Thu Aug 14, 2008 2:23 pm

Hi,

Since the above mentioned map is an isomorphism, you can see that since 1 and 3 are generators of Z4, 3 and 2 are generators of Z5*.

A homomorphism of a cyclic group is an isomorphism iff a generator of one group is mapped to a generator of the other group.

Thus the only 2 isomorphisms between Z4 and Z5* are:

1. the one mentioned above--which maps 1 in Z4 to 3 in Z5*)

2. the one that maps 1 in Z4 to 2 in Z5*, i.e.
0->1
1->2
2->4
3->3

Nameless
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Joined: Sun Aug 31, 2008 4:42 pm

Group theory

Postby Nameless » Mon Sep 01, 2008 9:22 am

You guys are awesome, by the way, I do have another question :

G is finite group and H, K are subgroups of G, of order 12 and 30 respectively. ( it means that |H|=12, |K|=30- where |H| means the order of H ). Which of the following can not be the order of subgroup of group generated by H and K

A. 30
B. 60
D. 120
E. infinite countable

we know that |<H,G>| = (|H||G|) /( |H intersection G| )
where <H,G> is the group generated by H and G
since |H|=3.2.2 and | G| =2.3.5

so if L=H intersection G then |L| is in {1,2, 3, 6 }

and I get stuck :D


For me, it seems true that the answer is E but the official answer is A
if some know, please explain . Thanks so much
Last edited by Nameless on Thu Sep 04, 2008 1:07 am, edited 1 time in total.

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lime
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Postby lime » Mon Sep 01, 2008 10:40 am

The answer is indeed A.

H*K can be infinite because G is not necessarily commutative.

You can come to solution by yourself.
Hint: Use second Sylow's theorem.

Nameless
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Joined: Sun Aug 31, 2008 4:42 pm

Postby Nameless » Mon Sep 01, 2008 10:12 pm

Lime,
You are right, there exists a group G with a, b whose orders are finite but the order of a*b is infinite so as you said, the order of H*K may be infinite so how can we apply Sylow theorem in this situation ? please explain ?

CoCoA
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Postby CoCoA » Wed Sep 03, 2008 5:52 pm

Can be solved simpler with Lagrange.

<H,K> may not be finite - but IF it is - then both H and K are subgroups of <H,K>. Since |H|=12 does not divide 30, then 30 cannot be the order of <H,K>, so (a) is correct.

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

Postby Nameless » Wed Sep 03, 2008 8:13 pm

CoCoA,

How do you know that the order of <H,K> is 30 ??? please remember |<H,K>|>= |K|

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

Postby Nameless » Wed Sep 03, 2008 11:20 pm

Okie,
Now let's discuss question #66 - GR0568 - available online on ets site :

R is a ring with identity, U is additive subgroup of R then U is called a right ideal of R id : ru in U for all r in R and u in U. If R has exactly two right ideals then Which of following must be true :
1. R is commutative
2. R is division ring ( if not know, please use google :D)
3. R is infinite

A. 1 only
B. 2 only
C. 3 only
D. 1 & 2 only
E. 1, 2, and 3


Solution :

1 . Not necessary - since we look at Halmiton Quaternion then R has exactly 2 ideals but R is not commutative
3. Let R= Zp where p is prime number then R is a finite field

(2) since if we build up a homomorphism f: R---->rR where r is not zero then r must be invertible for all r not zero so R is a division ring



Question GR0568:

Problem 60 :

Let G be the five-star picture , then the group of symmetries of regular pentagon of G is isomorphic to :

1. S_5
2. A _5
3. Cylic of order 5
4. Cylic of order 10
5. dihedral group of order 10

CoCoA
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Joined: Wed Sep 03, 2008 5:39 pm

Postby CoCoA » Thu Sep 04, 2008 1:05 am

Nameless Posted:
CoCoA,

How do you know that the order of <H,K> is 30 ??? please remember |<H,K>|>= |K|

I know that it is not 30. The question asks which one cannot be the order.

I know this because not only is |<H,K>| >= |K|, but also |<H,K>| is divisible by |K| (if it is finite), by Lagrange, because K is a subgroup of <H,K>.

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

Postby Nameless » Thu Sep 04, 2008 1:13 am

Oh God, so silly am I :D
You are right, I missed understood the problem. the order of |<H,K>| NOT the order of subgroup of <H,K>




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