*Solution:*
**Z**_{16}^{×} = { 1,3,5,7,9,11,13,15 }.

Here are the cosets of the subgroup < 7 > :

< 7 > = { 1,7 }

3 < 7 > = { 3,5 }

9 < 7 > = { 9,15 }

11 < 7 > = { 11,13 }

Since 3^{2} is not in < 7 >,
the coset 3 < 7 > does not have order 2,
so it must have order 4, showing that the factor group is cyclic.

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(a) List all cosets of H; list all cosets of K.

*Solution:*
The cosets of H = { (0,0), (0,2) } are

(0,0) + H = { (0,0), (0,2) }

(1,0) + H = { (1,0), (1,2) }

(2,0) + H = { (2,0), (2,2) }

(3,0) + H = { (3,0), (3,2) }

(4,0) + H = { (4,0), (4,2) }

(5,0) + H = { (5,0), (5,2) }

(0,1) + H = { (0,1), (0,3) }

(1,1) + H = { (1,1), (1,3) }

(2,1) + H = { (2,1), (2,3) }

(3,1) + H = { (3,1), (3,3) }

(4,1) + H = { (4,1), (4,3) }

(5,1) + H = { (5,1), (5,3) }

The cosets of K = { (0,0), (3,0) } are

(0,0) + K = { (0,0), (3,0) }

(0,1) + K = { (0,1), (3,1) }

(0,2) + K = { (0,2), (3,2) }

(0,3) + K = { (0,3), (3,3) }

(1,0) + K = { (1,0), (4,0) }

(1,1) + K = { (1,1), (4,1) }

(1,2) + K = { (1,2), (4,2) }

(1,3) + K = { (1,3), (4,3) }

(2,0) + K = { (2,0), (5,0) }

(2,1) + K = { (2,1), (5,1) }

(2,2) + K = { (2,2), (5,2) }

(2,3) + K = { (2,3), (5,3) }

(b) You may assume that any abelian group of order 12 is
isomorphic to either **Z**_{12} or
**Z**_{6} × **Z**_{2}.
Which answer is correct for G / H? For G / K?

*Solution:*
Adding an element of G to itself 6 times
yields a 0 in the first component
and either 0 or 2 in the second component,
producing an element in H.
Thus the order of an element in G / H is at most 6,
and so G/H
**Z**_{6} × **Z**_{2}.

On the other hand, (1,1) + K has order 12 in G / K,
and so G/K **Z**_{12}.

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(a) Show that ba^{i} = a^{-i} b for all i with 1 i < n.

*Solution:*
The identity holds for all positive integers i,
and can be proved inductively:

assuming ba^{k} = a^{-k}b, we have

b a^{k+1}
= b a^{k} a
= a^{-k} b a
= a^{-k} a^{-1} b
= a^{-(k+1)} b.

(b) Show that any element of the form a^{i} b has order 2.

*Solution:*
We have (a^{i} b)^{2}
= a^{i} b a^{i} b
= a^{i} a^{-i} b^{2}
= a^{0} = e.

(c) List all left cosets and all right cosets of < b >

*Solution:*
The left cosets of < b >
have the form a^{i} < b > = { a^{i}, a^{i} b },
for 0 i < n.

The right cosets of < b >
have the form < b > a^{i} = { a^{i}, a^{-i} b },
for 0 i < n.

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(a) Show that N is a normal subgroup of G.

*Solution:*
To use Definition 3.7.5,
we need to show that gxg^{-1} belongs to N
whenever x is in N and g is an element of G.
Problem 29 (a)
implies that ba^{3} = a^{-3}b = a^{3}b,
so bxb^{-1} = x if x = a^{3}.
The only other element of N to check is e,
and since any power of a will also commute with a^{3},
we see that gxg^{-1} does belong to N
whenever x is in N and g is an element of G.

Another method of sulution would be to
check that the left cosets are the same as the right cosets.
These are the right cosets of N.

N = {e, a^{3} }

Na = {e, a^{3} }a = {a, a^{4} }

Na^{2} = {e, a^{3} }a^{2}
= {a^{2}, a^{5} }

Nb = {e, a^{3} }b = {b, a^{3}b }

Nab = {e, a^{3} }ab = {ab, a^{4}b }

Na^{2}b = {e, a^{3} }a^{2}b
= {a^{2}b, a^{5}b }

To find the left cosets, we need to use the formula
ba^{i} = a^{-i} b
from problem 29 (a).
Here are the left cosets of N in G.

N = {e, a^{3} }

aN = a{e, a^{3} } = {a, a^{4} }

a^{2}N = a^{2}{e, a^{3} }
= {a^{2}, a^{5} }

bN = b{e, a^{3} } = {b, ba^{3}b }
= {b, ba^{3} }

abN = ab{e, a^{3} } = {ab, aba^{3} }
= {a, aa^{3}b } = {a, a^{4}b }

a^{2}bN = a^{2}b{e, a^{3} }
= {a^{2}b, a^{2}ba^{3} }
= {a^{2}b, a^{2}a^{3}b }
= {a^{2}b, a^{5}b }

(b) Is G / N abelian?

*Solution:*
For aN = {a, a^{4} } and bN = { b, a^{3}b },
we have

(aN)(bN) = abN = {ab, a^{4}b }, while

(bN)(aN) = baN = a^{5} b N = { a^{5} b, a^{2} b }.
Thus

(aN)(bN) (bN)(aN),
and G / N is not abelian.

Next problem | Next solution | Up | Table of Contents

(a) Prove that N is a normal subgroup of G, and list all cosets of N.

*Solution:*
Since N = < a^{3} >, it is a subgroup.
It is normal since

a^{i} ( a^{3n} ) a^{-i} = a^{3n} and

a^{i} b ( a^{3n} ) a^{i} b
= a^{i} a^{-3n} a^{-i}
= (a^{3n})^{-1}.

(We are using the fact that ba^{i} = a^{-i} b.)

The cosets of N are

N = { e,a^{3},a^{6},a^{9} },

Nb = { ab, a^{3}b, a^{6}b, a^{9}b },

Na = { a, a^{4}, a^{7}, a^{10} },

Nab = { ab, a^{4}b, a^{7}b, a^{10}b },

Na^{2} = { a^{2}, a^{5}, a^{8}, a^{11} },

Na^{2}b = { a^{2}b, a^{5}b, a^{8}b, a^{11}b }.

(b) You may assume that G / N is isomorphic to either
**Z**_{6} or S_{3}.
Which is correct?

*Solution:*
The factor group G / N is not abelian, since

Na Nb = Nab but Nb Na = Na^{2}b, because

ba = a^{11}b in Na^{2}b.
Thus G/N S_{3}.

Next problem | Next solution | Up | Table of Contents

*Solution:*
We have a **~** a since we can use g = e.

If b **~** a, then b = gag^{-1} for some g in G, and so

a = g^{-1} b g = g^{-1} b (g^{-1})^{-1},
which shows that a **~** b.

If c **~** b and b **~** a, then

c = g b g^{-1} and b = h a h^{-1} for some g,h in G,
so

c = g (h a h^{-1}) g^{-1} = (gh) a (gh)^{-1},
which shows that c **~** a.

Thus **~** is an equivalence relation.

(b) Show that a subgroup N of G is normal in G if and only if N is a union of conjugacy classes.

*Solution:*
The subgroup N is normal in G if and only if
a in N implies gag^{-1} in G, for all g in G.
Thus N is normal if and only if whenever it contains an element a
it also contains the conjugacy class of a.
Another way to say this is that N is a union of conjugacy classes.

Next problem | Next solution | Up | Table of Contents

*Solution:*
The notion of a conjugacy class was defined
in Problem 32.

Let D_{4} =
{ e,a,a^{2},a^{3},b,ab,a^{2}b,a^{3}b },

with a^{4} = e, b^{2} = e, and ba = a^{-1}b.

Since xex^{-1} = e, the only element conjugate to e is e itself.

If x is any power of a, then x commutes with a,
and so

xax^{-1} = a.
If x = a^{i} b, then

x a x^{-1}
= a^{i} b a a^{-i} b
= a^{i} a^{i-1} b^{2}
= a^{2i-1},

so this shows that a^{3} is the only conjugate of a
(other than a itself).

The solution of an earlier problem shows that

x a^{2} x^{-1} = a^{2} in D_{4},
so a^{2} is not conjugate to any other element.

If x = a^{i}, then

x b x^{-1}
= a^{i} b a^{-i}
= a^{i} a^{i} b
= a^{2i} b.

If x = a^{i} b, then

x b x^{-1}
= (a^{i} b) b (a^{i} b)^{-1}
= a^{i} a^{i} b
= a^{2i} b.

Thus a^{2} b is the only conjugate of b.

If x = a^{i}, then

x (ab) x^{-1}
= a^{i} ab a^{-i}
= a^{i+1} a^{i} b
= a^{2i+1} b.

If x = a^{i} b, then

x ab x^{-1}
= (a^{i} b) ab (a^{i} b)^{-1}
= a^{i} a^{-1} a^{i} b
= a^{2i-1} b.

Thus a^{3} b is the only conjugate of ab.

Next problem | Next solution | Up | Table of Contents

(a) Prove that HN is a subgroup of G.

*Solution:*
See Proposition 3.3.2.
It is clear that

e = e · e belongs to the set HN,
so HN is nonempty.

Suppose that x and y belong to HN.
Then

x=h_{1}n_{1} and y=h_{2}n_{2},

for some h_{1},h_{2} in H
and some n_{1},n_{2} in N. We have

xy^{-1}
= h_{1}n_{1}(h_{2}n_{2})^{-1}
= h_{1}n_{1}n_{2}^{-1}h_{2}^{-1}
= (h_{1}h_{2}^{-1})(h_{2}(n_{1}n_{2}^{-1})h_{2}^{-1}),

and this element belongs to HN since the assumption that N is normal
guarantees that

h_{2}(n_{1}n_{2}^{-1})h_{2}^{-1}
belongs to N.

(b) Prove that N is a normal subgroup of HN.

*Solution:*
Since N is normal in G,
it is normal in the subgroup HN, which contains it.

(c) Prove that if H N = { e }, then HN / N is isomorphic to H.

*Solution:*
Define µ : H -> HN / N
by µ (x) = xN for all x in H.

(Defining a function from HN/N into H is more complicated.)

Then µ (xy) = xyN = xNyN =
µ (x) µ (y) for all x,y in H.

Any coset of N in HN has the form
hnN for some h in H and some n in N.

But then hnN = hN = µ (h),
and so this shows that µ is onto.

Finally, µ is one-to-one since if h in H belongs
to the kernel of µ
then

hN = µ (h) = N,
and so h is in N.

By assumption, H N = { e },
and so h = e.

Forward to §4.1 | Forward to Chapter 3 Review | Back to §3.7 problems | Up | Table of Contents

Forward to §4.1 | Forward to Chapter 3 Review | Back to §3.7 problems | Up | Table of Contents