## Abstract algebra warmup

Forum for the GRE subject test in mathematics.
lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am

### Abstract algebra warmup

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) {, }
b) {, , , , , , }
c) {, , , }

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
Posts: 11
Joined: Sun Jul 20, 2008 12:11 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 *= is not in the subset.

JcraigMSU
Posts: 13
Joined: Wed Aug 06, 2008 9:25 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

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 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.

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 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
Posts: 13
Joined: Wed Aug 06, 2008 9:25 am
4. a)5 b)6...Am I missing something here??

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
Posts: 13
Joined: Wed Aug 06, 2008 9:25 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
Posts: 13
Joined: Wed Aug 06, 2008 9:25 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)

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am
Very good job, JCraig. I like your proof for problem 5.

Problem 6. Consider two groups:
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
Posts: 1
Joined: Thu Aug 14, 2008 2:12 pm

### Problem 6

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

### Group theory

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 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.

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am

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
Posts: 128
Joined: Sun Aug 31, 2008 4:42 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
Posts: 42
Joined: Wed Sep 03, 2008 5:39 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
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
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 )
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
Posts: 42
Joined: Wed Sep 03, 2008 5:39 pm
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
Oh God, so silly am I You are right, I missed understood the problem. the order of |<H,K>| NOT the order of subgroup of <H,K>