Quimby writes:  This is excellent!  You've got all the main concepts, and your function is indeed an isomorphism.  My comments below mostly concern how to write all of this.  The only part I want you to rewrite is Step 3---you need to prove a couple of lemmas.

 

(Please click "edit page" or find lulu's files for better view.)

 

Find the group of order 8 that is isomorphic to H.

 

H = { s, t, u, v, w, x, y, z }

 

where

 

s = 1 2 3 4

1 2 3 4

 

t = 1 2 3 4

2 1 3 4

 

u = 1 2 3 4

1 2 4 3

 

v = 1 2 3 4

2 1 4 3

 

w = 1 2 3 4

3 4 1 2

 

x = 1 2 3 4

4 3 1 2

 

y = 1 2 3 4

3 4 2 1

 

z = 1 2 3 4

4 3 2 1

 

 

Find the group of order 8 that H is isomorphic to.

 

After a series permutation multiplication (for detail see attached file: isoExercise), here is the result:

 

 

s s t u v w x y z

t t s v u u z w x

u u v s t x w z y

v v u t s z y x w

w w x y z s t u v

x x w z w u v s t

y y z w x t s v u

z z y x w v u t s

 

 

 

From observation, we conclude that

1. H is non-abelian since wx=/xw.

2. H is non-cyclic since non of the elements in H generates H.

 

For the same reason, Q8 and D8 are also non-abelian and non-cyclic.

For the rest of tables, we found that: C8 is abelian and cyclic, C2 * C4 is abelian but not cyclic, and C2*C2*C2 is abelian but not cyclic. Thus we are left with Q8 and D8 to consider. We eliminate Q8 for the following reason: there are two elements in Q8 that multiplies itself twice to be the identity element p, while in H, six such elements exist.

 

Quimby writes: This reasoning is perfect.  In step 3, take what you've written here and make it into a formal proof.  Fr example, to show that H is not isomorphic to Q_8, you might prove the following lemma: If A is a group containing exactly two elements of order 2, and A is isomorphic to B, then B contains exactly two elements of order 2.

 

Claim: H is isomorphic to Table D8.

 

Quimby writes: Delete the word "Table."

 

Proof:

 

Let f: H map onto D8 be defined by the following: s=a, x=b, y=d, v=c, t=h, u=f,w=g,z=i

for s,t,u,v,w,x,y,z in H, and a,b,c,d,f,g,h,I in D8. Show f is isomorphic.

 

Quimby writes: Rephrase this as "Let f: H -> D_8 be defined by the following: f(s)=a, f(x)=b, f(y)=d, f(v)=c, f(t)=h, f(u)=f, f(w)=g, and f(z)=i. Show f is an isomorphism."  Excellent!  You found a function that works.

 

First, f is 1-1 and onto by the way it’s defined.

 

Quimby writes: Change "by the way it’s defined" to "by inspection."

 

Second, show that f is homomorphic, that is, show that for every x, y in H,

f(x)f(y) = f(xy).

 

Quimby writes: Change "homomorphic, that is" to "a homomorphism.  That is" 

 

Comparing the two tables, we found:

 

f(ss) = f(s) = a = f(s)f(s) = aa

f(st) = f(t) = h = f(s)f(t) = ah

…..

f(ts) = f(t) = h = f(t)f(s) = ha

f(tu) = f(v) = c = f(t)f(u) = hf

f(tw) =f(y) = d = f(t)f(w) = hg

f(tx) = f(z) = i = f(t)f(x) = hb

f(ty) = f(w) = g = f(t)f(y) =hd

f(tz) = f(x) = b = f(t)f(z) =hi

.....

 

f(wt) = f(x) = b = f(w)f(t) = gh

......

 

So f is homomorphic.

 

Quimby writes: Change "homomorphic." to "a homomorphism."

 

Therefore H is isomorphic to D8.

 

Q.E.D.

 

 

Sec 6 #55 Third Draft

 

Let H be a subgroup of Zp, where p is a prime number. We’ll show that H is not a proper nontrivial subgroup of Zp. First of all, Zp is a cyclic group generated by 1. Second, H is a cyclic group by theorem 6.6: a subgroup of a cyclic group is cyclic. Let b be a generator of H. Since b in Zp, b = s1 (1+1+1…+1 s times). By theorem 6.14, H contains p/d elements, where d=gcd(p,s). Since p is prime, d can only be 1 or p. Case 1: d = 1, p/d = p/1 = p. Since H has p elements, H is Zp itself, or the improper subgroup of Zp. Case 2: d = p, p/p = 1. Since H has 1 element, which is e, H the trivial subgroup of Zp. Therefore, if p is a prime number, then Zp has no proper nontrivial subgroup. Q.E.D.

 

Quimby writes: Excellent! This is a final draft! I deleted your earlier versions of this problem; you can view them by clicking the "history" link below.

 

Number 46 section 8:

Show that Sn is a nonabelian group for n>=3.

Solution:

i

so is nonabelian.

Therefore, it is a non-abelian, so for is a nonabelian group for .

 

 

Quimby writes: You have shown that S_n is nonabelian if n=3 or n=4. What about n=5, 6, 7, 8, . . . ? Here's a hint: Let f such that f(1)=2, f(2)=1, and f(x)=x if x>2. (f switches 1 and 2.) This f makes sense for any n greater than or equal to 3. Now find g such that fg does not equal gf.

 

 

Number 46 Section 6:

Let a,b be elements of a group G. Show that if ab has finite order n, then ba also has order n.

 

Solution:

 

Let n = ab.

 

 

 

(Quimby writes: There were some copy-and-paste issues here. I know from Kevin's e-mail that ab with a box around it means ab.)

 

 

Then (ab)n = e.

 

 

 

(Quimby writes: Write "(ab)^n" for ab to the nth power.)

 

Multiplying this equation by b on the left and a on the right gives (ba)(n+1) = b(ab)na = bea = ba. Cancellation of ba from this gives (ba)n = e and so ba <= n = ab.

 

 

Interchanging a and b in the above argument yields ab <= ba. So, in fact, ab = ba.

 

 

(Quimby writes: I like the use of "interchanging"! Good argument! This is a final draft!)

 

 

 

#52 section 8

Let G be a group. Prove that the permutations ra : G -> G, where ra(x) = xa for a ÃŽ G and x ÃŽ G, do form a group isomorphic to G.

 

 

Quimby writes: Please change "ÃŽ" to "ÃŽ"

 

 

Proof:

Let G be a group.

Let G' = {ra | a ÃŽ G}.

 

 

 

 

 

Let j = G -> G'.

So j(a) = ra.

 

 

Quimby writes: Rewrite this as "Define j = G -> G' by j(a) = ra."

 

Let x, y ÃŽ G. j(xy) = rxy = rxry = r(x)r(y). And by Lemma 8.15 this group is isomorphic to G.

 

 

Quimby writes: Using Lemma 8.15 is a good idea. But one of the hypotheses of Lemma 8.15 is that the function is one-to-one. So in order to use Lemma 8.15, you must first show that j is one-to-one. You're on the right track---now you've got some details to fill in.