|
|
Jeremy Gibbs
Brad Robbins
Heather Rust
Abstract Algebra Project
1. Prove that C(Sn)={i} for n³ 3.
In any permutation group with n symbols, any permutation other than the identity will have at least 2 symbols. We can find an element in the same Sn which does not commute with it. Therefore no element except the identity can be in the center if n³ 3.
( q )( f ) q f ¹ f q
q cannot be in C(Sn)
Formal Proof
x = {1, 2, . . . n}
Let d Î Sn and d ¹ i, n ³ 3.
We can find l Î x, such that d (l) = m.
Let k, l, m, be in x and all different.
Take a
= 
a d (l) = a (d (l)) = a (m) = m
d a (l) = d (a (l)) = d (k) ¹ m because d is not 1-1.
Therefore, a d ¹ d a .
As a result, d Ï Sn.
i ¹ d
=> d Ï C(Sn)
and i Î C(Sn)
Therefore, C(Sn) = {i}.
2. Let H be a subgroup of a group G, and x,yÎ G. Prove that there is a 1-1 correspondence between xH and yH.
In order to show 1 - 1, we must demonstrate that xH = yH => f(x) = f(y) => x = y in the mapping from G -> G', xy Î G.
First, we need to establish f(e) with e being the identity in G. Then, f(e) = e'. With H as a subgroup of G, then with any element x of G combined with an element h of H, it will remain in G (as demonstrated in a previous proof). The same will result for yh. Thus, we can say that xH Î G, yH Î G, therefore, xH = yH.
As a result, f(x) Î G', as is f(y). Therefore, f(x) = f(y) => x = y.
3. Consider the statement: Let G be a finite group, and o(G)=n. If mï n, then G has a subgroup of order m where m is a positive integer. Show that this statement is not true in A4.
A4 is normal
A4 is not abelian
Let G=A4, o(G)=12 (or n=12) by Lagrange’s Thm
m could be 1,2,3,4,6,12. We must be able to find subgroups with these orders.
o(H)=1 H={i}
o(H)=2 H={i,s 2}
o(H)=3 H={i,t 1,t 2}
o(H)=4 H={i,s 2,s 5,s 8}
o(H)=6 H={i, s 2,s 5,s 8,t 1,t 2,t 7, …}
cannot find an H such that o(H)=6, so the statement it not true. Q.E.D.
4. Show that the statement in #3 is true for cyclic groups.
[a]={e,a,a2,a3,…an-1} o[a]=n
Let [b] be a subgroup, then o[b]=m.
By Lagrange’s Thm, because mï n, then n=mr.
o[a]=n { e,a,a2,a3,…an-1}
o[b]=2 {e,bn/2}
o[b]=3 {e,bn/3,b2n/3}
o[bn/m]=m {e,…,b(n-2)/m,b(n-1)/m,bn/m}
5. Explore the validity of the statement in #3 for abelian groups.
If T, prove it. If F, give counter example.
The statement is true because Klein’s is an example.
Klein o(K)=4, order is 4 so n=4 and m=1,2,4
o(H)=1 {e}
o(H)=2 {e,a}
o(H)=4 {e,a,b,c}
so the statement is true
PROOF: Let G be a noncyclic abelian group with o(G)=n. Let H be a subgroup of G with o(H)=m. Then mï n Î N.
6. Let f: G® G’ be a homomorphism, prove that f is 1-1 Û Kerf={e}, where e is the identity in G.
G G’
· e’
=>
(Prove: f is 1-1 means kerf = {e} of G -> G')
With e being the identity in G, since f is 1-1, e maps to e' in G'. All other entries go to other elements. For example, element x of g maps to x'. As a result of being 1 - 1, then only e can map to e'. As a result, the kerf = {e}.
<=
(Prove: if kerf = {e}, then f is 1-1)
If kerf = {e}, then e -> e'. Since only e maps to e' and no other elements do, then f must be 1-1.
7. Let f: G® G’ be a homomorphism, and H be a subgroup of G. Prove that f(H)={y=f(x) for some x in H} is a subgroup of G’.
G G’
H ¬ f(H)
We must show closure and inverse for f to be a subgroup.
Closure argument
u, v Î f(H)
f is a homomorphism
H is a subgroup of G
Then, we can find x, y Î H such that f(x) = u, f(y) = v. Since H is a subgroup, xy Î H.
Therefore, f(xy) = f(x)f(y) = uv Î H.
Inverse argument
If w Î f(H), we can find z Î H such that f(z) = w.
Notice w-1 = (f(z))-1 = f(x-1)f(z-1)(f(H)). z-1 Î H. QED
8. Let f: G® G’ be a homomorphism, and H be a subgroup of G’. Prove that f-1(H’)={xÎ G: f(x)Î H’} is a subgroup of G.
G G’
f-1(H’) ® ¬ H’
We must show closure and inverse for f-1 to be a subgroup.
Closure argument
u,v Î f-1 (H')
- f is homomorphism => f-1 will also be homomorphism
- H' is a subgroup of G
Then, we can find x, y, Î H' such that f(x) = u, f(y) = v.
Since H' is a subgroup, xy Î H'.
Therefore, f(xy) = f(x)f(y) = uv Î H'.
Inverse argument
If w Î f(H'), we can find z Î H' such that f(z) = w.
Notice w-1 = (f(z))-1 = f(z-1) Î f(H). QED
H is a subgroup of G.
9. Show that the analogue of #7 is not true for a normal subgroup H. Find a condition under which an analogue of #7 is true for a normal subgroup H.
(If H is normal in G, then H' will not necessarily be normal in G'. Show this is true.)
We must show closure and inverse.
Closure argument
u, v Î f(H)
- f is a homomorphism
- H is a normal subgroup of G
Then, we can find a,b Î H' such that f(a) = u, f(b) = v. Since H is a subgroup, xy Î H. Therefore, f(xy) = f(x)f(y) = uv Î H.
Inverse argument
If w Î f(H), then we can find z Î H such that f(z) = w. Notice w-1 = (f(z))-1 = f(z-1)(f(H)).
Failure of Normal argument
Given g-1hg Î H, for all g Î G, for all h Î H. To show that H is normal in G, i.e. gH = Hg, for all g Î G.
Take gh Î gH
We know g-1h-1g Î H-1
g-1h-1g Î H-1
For this to be normal, then g-1h-1g = h’ which will not necessarily be true.
Conditional Agreement
For this to be true, g’s inverse must not be contained in H-1 if g is not.
An infinite field has infinite entries.
A characteristic of a field takes the smallest integer of the field and takes it to 0, the additive identity.
The first example of a field that is infinite, yet contains a finite characteristic is Z3 {x} = a + bx. This creates several lines, six total, with infinite points on them. The characteristic, as defined above, is three.
Another example of a field that is infinite, yet has a finite characteristic is again found in Z3 and is x2 + y2 = c2. With c in Z3 this creates a series of 2 circles with infinite points. The characteristic, as defined above, is three.
Several more examples of such fields could be generated such as these. The form of Z can be changed as well as how many unknown elements, especially in the second example, will determine several more examples of the same type.