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

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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(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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(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 ba^k = a^-kb, 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 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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(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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(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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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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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 a^i-1 b² = a^2i-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 = a^2i-1 b.
Thus a³ b is the only conjugate of ab.

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    (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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