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 = .
Every quotient group of Z is Z/n for some integer n.
The cosets in Z/n are , 1+, 2+,...(n-1)+.
1+ has order n in the quotient group Z/n. Z/n has n elements. The
coset 1+ 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)^p = a^p b^p = b^p
(ab)^p is an element of
order( (ab)^p ) = order( (b^p) ) = p/gcd(p,q)
p/gcd(p,q) divides n
(ab)^q is an element of
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.