## § 3.8 Cosets, Normal Subgroups, and Factor Groups: Solved problems

27. List the cosets of < 7 > in Z16×. Is the factor group Z16× / < 7 > cyclic?

Solution: Z16× = { 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 32 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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28. Let G = Z6 × Z4, let H = { (0,0), (0,2) }, and let K = { (0,0), (3,0) }.

(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 Z12 or Z6 × Z2. 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 Z6 × Z2.

On the other hand, (1,1) + K has order 12 in G / K, and so G/K Z12.

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29. Let the dihedral group Dn be given via generators and relations, with generators a of order n and b of order 2, satisfying ba=a-1 b.

(a) Show that bai = 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 bak = a-kb, we have

b ak+1 = b ak a = a-k b a = a-k a-1 b = a-(k+1) b.

(b) Show that any element of the form ai b has order 2.

Solution: We have (ai b)2 = ai b ai b = ai a-i b2 = a0 = e.

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

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

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

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30. Let G = D6 and let N be the subgroup < a3 > = {e, a3 } of G.

(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 ba3 = a-3b = a3b, so bxb-1 = x if x = a3. The only other element of N to check is e, and since any power of a will also commute with a3, 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, a3 }
Na = {e, a3 }a = {a, a4 }
Na2 = {e, a3 }a2 = {a2, a5 }
Nb = {e, a3 }b = {b, a3b }
Nab = {e, a3 }ab = {ab, a4b }
Na2b = {e, a3 }a2b = {a2b, a5b }

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

N = {e, a3 }
aN = a{e, a3 } = {a, a4 }
a2N = a2{e, a3 } = {a2, a5 }
bN = b{e, a3 } = {b, ba3b } = {b, ba3 }
abN = ab{e, a3 } = {ab, aba3 } = {a, aa3b } = {a, a4b }
a2bN = a2b{e, a3 } = {a2b, a2ba3 } = {a2b, a2a3b } = {a2b, a5b }

(b) Is G / N abelian?

Solution: For aN = {a, a4 } and bN = { b, a3b }, we have
(aN)(bN) = abN = {ab, a4b }, while
(bN)(aN) = baN = a5 b N = { a5 b, a2 b }. Thus
(aN)(bN) (bN)(aN), and G / N is not abelian.

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31. Let G be the dihedral group D12, and let N = { e,a3,a6,a9 }.

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

Solution: Since N = < a3 >, it is a subgroup. It is normal since
ai ( a3n ) a-i = a3n and
ai b ( a3n ) ai b = ai a-3n a-i = (a3n)-1.
(We are using the fact that bai = a-i b.)

The cosets of N are

N = { e,a3,a6,a9 },
Nb = { ab, a3b, a6b, a9b },
Na = { a, a4, a7, a10 },
Nab = { ab, a4b, a7b, a10b },
Na2 = { a2, a5, a8, a11 },
Na2b = { a2b, a5b, a8b, a11b }.

(b) You may assume that G / N is isomorphic to either Z6 or S3. Which is correct?

Solution: The factor group G / N is not abelian, since
Na Nb = Nab but Nb Na = Na2b, because
ba = a11b in Na2b. Thus G/N S3.

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32. (a) Let G be a group. For a,b in G we say that b is conjugate to a, written b ~ a, if there exists g in G such that b = gag-1. Show that ~ is an equivalence relation on G. The equivalence classes of ~ are called the conjugacy classes of G.

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.

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33. Find the conjugacy classes of D4.

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

Let D4 = { e,a,a2,a3,b,ab,a2b,a3b },
with a4 = e, b2 = e, and ba = a-1b.
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 = ai b, then
x a x-1 = ai b a a-i b = ai ai-1 b2 = a2i-1,
so this shows that a3 is the only conjugate of a (other than a itself).

The solution of an earlier problem shows that
x a2 x-1 = a2 in D4, so a2 is not conjugate to any other element.

If x = ai, then
x b x-1 = ai b a-i = ai ai b = a2i b.
If x = ai b, then
x b x-1 = (ai b) b (ai b)-1 = ai ai b = a2i b.
Thus a2 b is the only conjugate of b.

If x = ai, then
x (ab) x-1 = ai ab a-i = ai+1 ai b = a2i+1 b.
If x = ai b, then
x ab x-1 = (ai b) ab (ai b)-1 = ai a-1 ai b = a2i-1 b.
Thus a3 b is the only conjugate of ab.

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34. Let G be a group, and let N and H be subgroups of G such that N is normal in G.

(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=h1n1 and y=h2n2,
for some h1,h2 in H and some n1,n2 in N. We have

xy-1 = h1n1(h2n2)-1 = h1n1n2-1h2-1 = (h1h2-1)(h2(n1n2-1)h2-1),

and this element belongs to HN since the assumption that N is normal guarantees that
h2(n1n2-1)h2-1 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.

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