Practice Test for Chapter 7.

I.  Specific Groups.


(1)  Show that any Quotient Group of Z is finite cyclic.

First show that any subgroup of Z is cyclic.
Let H be some subgroup of Z.
Let n be the least positive integer in H.

Let m be any other integer in H.

Then by division algorithm, m = nq + r where r is either 0 or positive and
< n.

Since n is in H and m is in H, then m - nq = r is in H.

-1 < r < n.  But n is the least positive integer in H.

Therefore r = 0, and m is a multiple of n.

That is, every element in H is a multiple of n, the least positive integer
in H.

H = <n>.

Every quotient group of Z is Z/n for some integer n.

The cosets in Z/n are <n>, 1+<n>, 2+<n>,...(n-1)+<n>.

1+<n> has order n in the quotient group Z/n.  Z/n has n elements. The
coset 1+<n> generates all the cosets of Z/n.  Z/n is cyclic.

(2)  Show that any finite group is isomorphic to a subgroup of S_n.

Let G be a finite group of n elements.

Let H be the subset of S_n defined as follows:

H consists of permutations of the elements of G.

Define p_g in H where g is in G by the following.

p_g (b) = b*g for all b in G, where * is the group multiplication of G.

Consider the mapping theta: G -> H defined as follows.

theta(g) = p_

[ theta(g) ] (b) = b * g.


To show that theta is one to one, 

theta(g_1) = theta(g_2) implies p_g1 = p_g2 implies for all b in G that 

b * g1 = b * g2 implies g1 = g2.

To show that theta is onto,

Let p_b be an arbitrary element of H.

By definition of theta, theta (b) = p_b.

To show that theta preserves the group multiplication,

Theta( g_1 * g_2) = p_(g_1*g_2)

Theta( g_1 * g_2) (b) = b *( g_1 * g_2) for all b in G.

                      = ( b * g_1) * g_2 for all b in G.
                      
                      =[ theta(g_1) (b) ] * g_2  for all b in G

                      = ( [theta)g_2 ] ( [ theta(g_1) (b) ])

                     = [ theta(g_1) * theta(g_2) ](b) for all b in G.

Theta (g_1 * g_2) = theta(g_1) * theta(g_2)

(3) List all the elements of A_4 as a product of disjoint cycles, and find
all the normal subgroups of A_4.

A_4 = {(1), (12)(34), (13)(24), (14)(23), (123), (124), (132), (134),
(142), (143), (234), (243)}
                      

The only proper normal subgrop of A_4 is {(1), (12)(34), (13)(24), (14)(23)}.

To see this, we let H be a normal subgroup of A_4.  We show that if H
contains one 3 cycle, then it contains all 3 cycles of A_4.

Suppose H contains the 3 cycle (abc) which represents any 3 cycle in A_4.

Let d represent the element of {1,2,3,4} not in the cycle (abc).

Since H is normal it also contains

(ad)(abc)(ad) = (a)(bcd)=(bcd)
(bd)(abc)(bd) = (adc)
(cd)(abc)(cd) = (abd)

(abc)^-1 = (acb)
(bcd)^-1 = (bdc)
(adc)^-1 = (acd)
(abd)^-1 = (adb)

This has shown that if H is a normal subgroup that contains one 3 cycle it
contains all 3 cycles.  The 3 cycles generate all of A_4.  H = A_4.

Suppose H contains one of {(12)(34),(13)(24),(14)(23)}

Represent the element in H of this form by
(ab)(cd).

Then since H is normal, H also contains

(ac) [ (ab)(cd) ] (ac) = (ad)(bc)
(ad) [ (ab)(cd) ] (ad) = (ac)(bd)

Thus if H contains any of { (12)(34), (13)(24), (14)(23) }, it contains
all of them.

The only proper normal  subgroup of A_4 is H = {(1), (12)(34), (13)(24),
(14)(23)}.

The normal subgroups of A_4 are {(1)}, {(1), (12)(34), (13)(24),
(14)(23)},  A_4.

(4) Prove that no subgroup of order 2 in S_n (n > 2) is normal.

Let H be a subgroup in S_n of order 2.  

Then H = { (a),(ab) } where a and b are distinct elements from {1,2,...n}.

Let c be an element from {1,2,3...n} that is different from a and b.

(ac)(ab)(ac) = (a)(bc) = (bc) is not in H.  Therefore H is not normal.

No subgroup of order 2 can be normal in S_n if n > 2.

(5)  Show that up to conjugation, there is only one k-cycle in S_n for
every k = 1,2,3....n.

Let (a1a2a3...ak) and (b1b2b3...bk) represent two arbitrary k cycles in
S_n.

Then

[(a1b1)(a2b2)(a3b3)...(akbk)] (a1a2a3...ak) [(akbk)...(a3b3)(a2b2)(a1b1)]

is the conjugation of (a1a2a3...ak) that is equal to

(a1)(a2)(a3)...(ak)(b1b2b3...bk) = (b1b2b3...bk).

Thus any two k-cycles in S_n are conjugates of one another.

Up to conjugation there is only one k-cycle in S_n, for any k = 1,2,3...n.


II.  Subgroups.

(1) If H is any subgroup of G, not necessarily normal, then there is a
bijection between the set of left cosets and the set of right cosets.
Also given the element a in G, there is a bijection between aH and Ha.


Define f to map the left cosets into right cosets by

f( aH) = H a^-1.

To show that f is well defined and that f is one to one,

aH = bH if and only if

b^-1  a H = H    if and only if

b^-1 a is in H   if and only if

a^-1 b is in H  if and only if

H a^-1 b = H   if and only if

H a^-1 = H b^-1  if and only if

f(aH) = f(bH)

To show that f is onto, let Hb be a right coset.

Then the left coset [b^-1] H is mapped onto Hb by f.

f is onto.

f is a well defined bijection of left cosets onto right cosets.

To show that, given the element a in G, there is a bijection between aH
and Ha, proceed as follows:

Define r: H -> Ha by  r(h) = ha for each h in H.
Define l: H -> aH by  l(h) = ah for each h in H.

Both r and l are bijections since

r(h1) = r(h2) implies h1 a = h2 a implies h1 = h2.  r is 1-1.
l(h1) = l(h2) implies a h1 = a h2 implies h1 = h2.  l is 1-1.

Given k_r in Ha implies there exist h_r in H such that k_r = h_r a
implies k_r = r(h_r).   r is onto.

Given k_l in aH implies there exist h_l in H such that k_l = a h_l
implies k_l = l(h_l).  l is onto.

Both r and l are bijections.

Then r(l^-1) is a bijection from  aH to Ha.



(2) Show that if H is a finite nonempty subset of the group G which is
closed under the group operation, then H is a subgroup.


Let b be some non-identity element in H.
Consider the set of consecutive powers of b = { b, b^2, b^3, ...}

Since H is finite, there must be some repetition of values, say for i< j
that b^i = b^j.

Then if c is any element in H,

c b^j = c b^i

c b^(j-i) = c

b^(j-i) is the identity in H.

also since [ b ] [ b^(j-i-1] = b^(j-i) is the identity

and [ b^(j-i-1)] b = b^(j-i) is the identity,

b^(j-i-1) is the inverse of b.

H is a subgroup.


(3) If the element a in G has order n, what is the order of a^t?  Give a
formula for order(a^t) in terms of t and n.  Justify your answer.

n = order(a)

n is the least non-negative integer such that a^n = identity element.

Let k = order(a^t).

k is the least non-negative integer such that (a^t)^k = identity element.

k is the least non-negative integer such that a^( t k) = identity element.

k is the least non-negative integer such that t k  is a multiple of n.

t k =t n/gcd(t,n)

k = n/gcd(t,n)


order(a^t) = order(a) / gcd( t, order(a) )

(4)  if ab is in the center of group G, then prove that ba is also in the
center of group G.

ab commutes with every element of G.

ab commutes with a^-1.

[ ab] [a^-1] = [a^-1] [ab] = b

ab [ a^-1] a = ba

ab = ba

III. Miscellaneous

(1)  Let G be any group.  From the group axioms, show that the identity
element is unique, and that if a is in G that a^-1 is unique.

Suppose e and f are both identity elements.

Then e f = e since f is an identity element.
and 
     e f = f since e is an identity element.

Therefore, e = e f = f.

The identity element is unique.

Suppose  b and c are both inverses of the element a.

Then ba = identity and ac = identity.

and b = b ( ac ) = (ba) c = c.

The inverse of a is unique.

(2)  If a and b commute, and order(a) = p, and order(b) = q, with p and q
not necessarily distinct or prime, what is the order(ab)?

Let n = order(ab)

n is the least non-negative integer such that (ab)^n = identity.

a and b  commute.

Let e be the identity element.

(ab) ^ [ pq/gcd(p,q) ] = a^[pq/gcd(p,q)]  b^[pq/gcd(p,q)]

= [ a^p ]^(q/gcd(p,q))  [b^q]^(p/gcd(p,q)) 

=  e^(q/gcd(p,q) )  e^(p/gcd(p,q) )

= e * e

= e

n divides pq/gcd(p,q).

n = order ( <ab> )

(ab)^p = a^p b^p = b^p

(ab)^p is an element of <ab>

order(  (ab)^p ) = order( (b^p) ) = p/gcd(p,q)

p/gcd(p,q) divides n

(ab)^q is an element of <ab>

order( (ab)^q ) = order( a^q  b^q ) = order( a^q ) = q/gcd(p,q)

q/gcd(p,q) divides n

lcm( p/gcd(p,q) , q/gcd(p,q) ) divides n

pq/[ gcd(p,q) ]^2 divides n.

n divides pq/gcd(p,q).

To show that no better result is possible, look at the example of z mod
210, with p = 30 and q = 105.

   a        b       a+b      n

  49       4         53     210

   7       2         9      70

  77       8        85      42

   7       8        15      14


(3)  Produce a list of Groups such that every homomorphic image of Z_8 is
homomorphic to exactly one group on the list.

z_8 is isomorphic to Z_2 X Z_2 X Z_2

list = 
 group of single element
 Z_2
 Z_2 X Z_2
 Z_2 X Z_2 X Z_2

(4)  Give an example of a group G and a normal subgroup N in G such that N
and G/N are abelian, but G is not an abelian group.

Let G = S_3 = { (1),(12),(13),(23),(123),(132)}
G is not abelian.

Let N = {(1),(123),(132)}

N is an abelian group of order 3.

G/N = {  {  (1),(123),(132) }, { (12),(13),(23) }  } is an abelian group
of order 2.