Abstract algebra warmup
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) {[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.
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.
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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.
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.
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!
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!
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?
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?
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
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
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
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.
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
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
Nameless Posted:
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>.
I know that it is not 30. The question asks which one cannot be the order.CoCoA,
How do you know that the order of <H,K> is 30 ??? please remember |<H,K>|>= |K|
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>.