Gomini Mistry
Velia Valle
Carlos Carrosino
Andrew Bloomstone, S.
Ben Chui
Marco Acuna
Math 455, Modern Algebra I
January 23, 2007
Professor Krebs
Homework: Section 6, pages 67-68, #46, 55
#46: Let a and b be elements of a group G. Show that if ab has finite order n, then ba also has order n.
Solution/Proof:
Definition: If a Є G is of finite order n,
then n is the smallest positive integer such that an = e
Proof:
Assume that a has order 5 and that a4b = ba4
Quimby writes: You can't assume that a has order 5.
Prove → ab = ba
Can we express a as a product of a4s?
Yes we can, (a4)2 = a8 = a
Therefore → ab = (a4)2b= a4ba4 = b(a4)2 = ba
Quimby writes: I'm confused. How did you get (a4)2b= a4ba4?
Then ab = ba
Quimby writes: Actually, ab = ba might not be true. So, time to try a new approach . . .
#55: Show that Zp has no proper nontrivial subgroups if p is a prime number.
Solution/Proof: Let H be a subgroup of Zp, where p is a prime number. We need to show that H is a trivial or an improper subgroup of ZP. In other words, need to show that H = {0} or H = Zp.
By Example 5.21, Z_p is cyclic.
(Quimby writes: Let's rephrase this last sentence as follows. "By Example 5.21, Z_p is cyclic." You don't need to go through the reasoning of Example 5.21.)
By thm. 6.6, a subgroup of a cyclic group is cyclic. Hence, H is cyclic. Then there exists some element x in Z_p such that <x> = H.
(Quimby writes: This is not quite right. The fact that H is cyclic means that there exists some element x in Z_p such that <x>=H. The way you've written it, it sounds as if <x>=H for any x in H, which is not true.)
By Thm. 6.14, then |x| =p/(gcd(p,x)), an element in Z^+.
(Quimby writes: Delete the word "the." Capitalize "Thm." Change Z+ to Z^+.)
But the gcd(p,x) = 1 or p since they are the only divisors of p.
Case 1: If gcd(p,x) = 1, then <x> = {1, 2, 3, ... , p-1, 0} = Z_p. Thus, H = Z_p.
Case 2: If gcd(p,x) = p, then <x> = <p> = {0}, the trivial subgroup of Z_p. Thus, H = {0}.
From case 1 and 2, {0} and Zp are the only subgroups of Zp. Thus, Zp has no proper nontrivial subgroups if p is a prime number. Q.E.D.
(Quimby writes: It is not true that x = p or 1. So your two cases should be Case 1: gcd(p,x)=1 and Case 2: gcd(p,x)=p. Then your argument will work just fine. This is a good first draft! Make the changes above, and you should have a good final draft.)
Homework: Section 8, 86-87, # 46, 47, 52
#46: Show that Sn is a nonabelian group for n greater than or equal to 3.
Proof: Let n be a positive integer greater than or equal to 3.
(Quimby writes: This is very good!)
Let A = {m in Z^+| 1 less than or equal to m less than or equal to n}
Define f from A to A by
f(a) = m if a=1, and
f(a) = a-1 if a>1
Define g from A to A by
g(a) = n if a=1, and
g(a) = n-a+1 if a>1
As f(a) is in Sn, and g(a) is also in Sn.
fg(3) = f (m-2) = m-3, but gf(3) = g(2) = m - 1.
Then fg is not equal to gf.
Therefore Sn is not abelian.
QED.
#47: Strengthening Exercise 46, show that is n is greater than or equal to 3, then the only element of (sigma)of Sn satisfying (sigma)(gamma) = (gamma)(sigma) for all gamma in Sn is (sigma) = i,the identity permutation.
Solution/Proof (DRAFT II):
For this proof: let (sigma) = s. Let (gamma) = g. Let i = the identity permutation.
Let n be in Z^+ such that n>=3. Suppose s is in S_n such that sg=gs for all g in G. Show s=i. Assume temporarily s does not equal i. Show there exists a g in G such that sg does not equal gs.
Since s does not equal i, there exists an x such that s(x) does not equal x.
Define g in S_n by g(y) = s(x) if y=x, g(y) = x if y=s(x), and g(y) = y otherwise.
Case 1: x does not equal s(s(x)). Define g in S_n by
(Quimby writes: Reverse order of these two lines.)
Therefore,
gs(x) = g(s(x)) = x
sg(x) = s(g(x)) = s(s(x))
Thus,
gs(x) does not equal sg(x)
Thus, gs does not equal sg.
Case 2: x equals s(s(x))
x, s(x) in {1, 2, 3, ... , n }.
There exists a b in {1, 2, ... n} such that b does not equal x and b does not equal s(x), so there are at least three elements in the set: x, s(x), and b which is not equal to x nor s(x).
Define g in S_n by g(y) = s(x) if y=x. g(y) = b if y=s(x). g(y) = x if y=b. g(y) = y otherwise.
gs(x) = g(s(x)) = b
sg(x) = s(g(x)) = s(s(x)) = x.
So, gs(x) does not equal sg(x).
So, gs does not equal sg. But this is a contradiction under the assumption s does not equal i. So, s does equal i if sg=gs.
(Quimby writes: This is very good! Let's call this a final draft, even though here's a few minor things to touch up---see comment above. By the way, I deleted my previous comments here, but you can still see them by clicking the "history" link below.)
#52: Let G be a group. Prove the permutations pa: G (mapped onto) G, where pa(x) = xa for a in G, do form a group isomorphic to G. (Note: Let rho symbol = p)
Solution/Proof:
??? Hmmm....So we have to prove that the set of permutation functions on G are isomorphic to G itself! What do you need to prove to establish an isomorphism? Ok, i more or less know what an isomorphism is, but I must look over section 8 some more to find a precise mathematical definition of an isomorphism. In the meantime, you are welcome to jump in on this.
[Quimby writes: Let G' be the set of all p_a, where a is in G. Find a group law on G' that makes G' into a group. (Tricky point: ordinary function composition might not work. I'm not sure, but you might have to define p_a * p_b as p_b composed with p_a for this to work.)]
-Ben
In accordance with our conversation with you last Thusrday, we are entering our proof for the problem on isomorphism that we handed in last week.
GROUP PROBLEM ON ISOMORPHISM
We were given the groups Q8, D8, C8, C2xC4, and C2xC2xC2, to determinbe which of these 5 groups of order 8 is isomorphic to the Mini-Sudoku.
We have proven that the Mini-Sudoku (Gmn) of order 8 has:
1 element of order 1,
2 elements of order 2, and
5 elements of order 4.
We have also shown that the D8 group has:
1 element of order 1,
2 elemnt of order 2, and
5 elements of order 4.
Lemma: If A and B are groups of order n, for n in Z, and A is isomorphic to B,
then A has exactly p1 elements of order q1,
p2 elements of order q2,
......................................
pn elements of order qn for n in Z, and
B has exactly p1 elements of order q1,
p2 elements of order q2,
......................................
pn elements of order qn for n in Z,
where q is in Z and q is a divisor of n.
As Gmn has the same number of elements of order 1, 2 and 4 as D8, by the Lemma, Gmn is isomorphic to D8. QED.
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