MATH 3005 Homework Solution Han-Bom Moon

Homework 10 Solution
Chapter 9.

1. Let H = {(1), (12)}. Is H normal in S3 ?

(13)H = {(13)(1), (13)(12)} = {(13), (123)}

H(13) = {(1)(13), (12)(13)} = {(13), (132)}

Because (13)H 6= H(13), H is not a normal subgroup of S3 .

8. Viewing h3i and h12i as subgroups of Z, prove that h3i/h12i is isomorphic to Z4 .

Similarly, prove that h8i/h48i is isomorphic to Z6 . Generalize to arbitrary integers
k and n.
I will prove the general formula: For any positive integers k and n, two groups
hki/hkni and Zn are isomorphic.
Sol 1. Because hki is cyclic, all elements in hki is of the form mk for m Z. So all
elements in hki/hkni is of the form mk + hkni = m(k + hkni). Therefore hki/hkni
is cyclic and it is generated by k +hkni. So it suffices to check the order of k +hkni.
Note that n(k + hkni) = nk + hkni = hkni and for 0 < m < n, m(k + hkni) 6= hkni.
Therefore |k + hkni| = n and hki/hkni = Zn .
Sol 2. Note that an element in hki/hkni is of the form mk + hkni for some m Z.
Define a map : hki/hkni Zn as (mk + hkni) = m mod n.
Step 0. is well-defined.
If m1 k + hkni = m2 k + hkni, then m2 k + m1 k + hkni = hkni. So (m1 m2 )k =
m2 k + m1 k hkni. Therefore m1 m2 is a multiple of n so m1 mod n =
m2 mod n. This implies that (m1 k + hkni) = (m2 k + hkni).
Step 1. is one-to-one.
If (m1 k + hkni) = (m2 k + hkni), then m1 mod n = m2 mod n. Therefore
n|m1 m2 and nk|m1 km2 k. So m1 km2 k hkni, and m1 km2 k+hnki = hnki.
So m1 k + hnki = m2 k + hnki.
Step 2. is onto.
Obviously, for m Zn , (mk + hkni) = m mod n = m.
Step 3. has the operation preserving property.

(m1 k + hkni)(m2 k + hkni) = m1 mod n + m2 mod n = m1 + m2 mod n

= ((m1 + m2 )k + hkni)
= ((m1 k + hkni) + (m2 k + hkni)).

Therefore is an isomorphism.

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MATH 3005 Homework Solution Han-Bom Moon

9. Prove that if H has index 2 in G, then H is normal in G.

Because H has index 2, there are exactly two left cosets (say {H, aH}) and two
right cosets ({H, Hb}). Note that the disjoint union of H and aH is G. Also the
disjoint union of H and Hb is G. So aH = G H = Hb.
If x H, then xH = H = Hx. If x / H, then xH = G H = Hx. So in any cases,
the left coset is equal to the right coset. Therefore H is normal.

11. Let G = Z4 U (4), H = h(2, 3)i, and K = h(2, 1)i. Show that G/H is not isomor-
phic to G/K. (This shows that H K does not imply that G/H G/K.)
In G/H, ((1, 3)H)2 = (1 + 1, 32 )H = (2, 1)H, ((1, 3)H)3 = (1 + 1 + 1, 33 )H =
(3, 3)H, ((1, 3)H)4 = (1 + 1 + 1 + 1, 34 )H = (0, 1)H = H (Note that (0, 1) is the
identity of G.). So |(1, 3)H| = 4 in G/H.
On the other hand, in G/K, for any element (a, b)H, ((a, b)H)2 = (a + a, b2 )H =
(2a, b2 )H. But in U (4) = {1, 3}, all squares are 1, so (2a, b2 )H = (2a, 1)H =
(2, 1)a H = H because (2, 1)a h(2, 1)i = H. Therefore |(a, b)H| 2.
So G/H 6 G/K.

12. Prove that a factor group of a cyclic group is cyclic.

Let G = hai and H / G. Then any element in G/H is of the form ak H, which is
(aH)k for some k Z. Therefore G/H = haHi and it is cyclic.

14. What is the order of the element 14 + h8i in the factor group Z24 /h8i?

14 + h8i = 6 + h8i

2 (6 + h8i) = 12 + h8i = 4 + h8i

3 (6 + h8i) = 18 + h8i = 2 + h8i
4 (6 + h8i) = 24 + h8i = h8i
So |14 + h8i| = 4.
Can you generalize it? Answer: In Zn , |a + hbi| = lcm(a, gcd(n, b))/a.

21. Prove that an Abelian group of order 33 is cyclic.

Let G be an Abelian group of order 33. There is a G with |a| = 11 and b G
with |b| = 3.
Sol 1. Because G is Abelian, (ab)i = ai bi . The order |ab| is one of 1, 3, 11, 33.
If |ab| = 1, then ab = e and b = a1 . So we obtain a contradiction |a| = |b|. If
|ab| = 3, then e = (ab)3 = a3 b3 = a3 . This is impossible because |a| = 11. If
|ab| = 11, then e = (ab)11 = a11 b11 = b11 . From b9 = e, we have b2 = e, which is
impossible too. Therefore |ab| = 33 and G = habi.
Sol 2. Let H = hai and K = hbi. H K = {e} because |H K| is a common divisor
of |H| = 11 and |K| = 3. Because |HK| = |H||K|/|H K| = 11 3/1 = 33 = |G|,
G = HK. Since G is Abelian, both H and K are normal subgroups. Therefore

G = H K H K Z11 Z3 Z33 .

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MATH 3005 Homework Solution Han-Bom Moon

22. Determine the order of (Z Z)/h(2, 2)i. Is the group cyclic?

For (1, 0) Z Z, m(1, 0) = (m, 0) / h(2, 2)i for all m > 0. This implies that
m((1, 0) + h(2, 2)i) 6= h(2, 2)i for any m > 0. So |(1, 0) + h(2, 2)i| = , and
|(Z Z)/h(2, 2)i| = .

2((1, 1) + h(2, 2)i) = (2, 2) + h(2, 2)i = h(2, 2)i

So |(1, 1) + h(2, 2)i| = 2. On the infinite cyclic group Z, except the identity, there
is no element with finite order. Therefore (Z Z)/h(2, 2)i is not cyclic.

24. The group (Z4 Z12 )/h(2, 2)i is isomorphic to one of Z8 , Z4 Z2 , or Z2 Z2 Z2 .

Determine which one by elimination.
Note that h(2, 2)i = {(0, 0), (2, 2), (0, 4), (2, 6), (0, 8), (2, 10)}. For (0, 1) Z4 Z12 ,
k(0, 1) = (0, k) h(2, 2)i only if 4|k. So |(0, 1)+h(2, 2)i| = 4. So it is not isomorphic
to Z2 Z2 Z2 where all nonidentity elements have order 2.
Furthermore, an element (a, b) Z4 Z12 has order lcm(|a|, |b|). Because both
|a| and |b| are divisors of 12, |(a, b)| = lcm(|a|, |b|) is a divisor of 12. Note that
12((a, b) + Z4 Z12 ) = 12(a, b) + Z4 Z12 = Z4 Z12 . So |(a, b) + Z4 Z12 |
is a divisor of 12 and there is no order 8 element. Therefore given group is not
isomorphic to Z8 .
So it is isomorphic to Z4 Z2 .

32. Prove that D4 cannot be expressed as an internal direct product of two proper
subgroups.
If D4 = H K, then because |D4 | = 8, |H| = 4 and |K| = 2 (or vice versa.). Then
K Z2 , and H Z4 or Z2 Z2 . In particular, both K and H are Abelian groups.
Because
D4 = H K H K,
D4 must be Abelian, too. But D4 is not, so it is not an internal direct product.

34. In Z, let H = h5i and K = h7i. Prove that Z = HK. Does Z = H K?

Note that Z is an additive group. So HK = {a + b | a H, b K}.
Because gcd(5, 7) = 1, there are two integers x and y such that 5x + 7y = 1. So for
any m Z, m = 5mx + 7my h5ih7i = HK. Therefore Z = HK.
But H K = hlcm(5, 7)i = h35i. So Z 6= H K.

39. If H is a normal subgroup of a group G, prove that C(H), the centralizer of H in

G, is a normal subgroup of G.
We will show that xC(H)x1 C(H) for all x G. Let a xC(H)x1 . Then
a = xyx1 for some y C(H). We need to show that ah = ha, or aha1 = h for
all h H.
aha1 = (xyx1 )h(xyx1 )1 = xyx1 hxy 1 x1 .
Because H / G, x1 hx x1 Hx H. So yx1 hx = x1 hxy. Thus we have
xyx1 hxy 1 x = xx1 hxyy 1 x1 = h. Therefore aha1 = h and a C(H).

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MATH 3005 Homework Solution Han-Bom Moon

44. Observe from the table for A4 given in Table 5.1 on page 111 that the subgroup
given in Example 9 of this chapter is the only subgroup of A4 of order 4. Why does
this imply that this subgroup must be normal in A4 ? Generalize this to arbitrary
finite groups.
Generally, if a finite group G has only one subgroup of H of fixed order k, then
H / G.
Indeed, for any x G, xHx1 G and |xHx1 | = |H| = k. From the assump-
tion, xHx1 = H. This implies that H / G.

51. Let N be a normal subgroup of G and let H be a subgroup of G. If N is a subgroup

of H, prove that H/N is a normal subgroup of G/N if and only if H is a normal
subgroup of G.
Suppose that H is a normal subgroup of G. Then for any a G, aHa1 H.
So for any hN H/N , aN hN (aN )1 = aha1 N H/N . In other words,
aN (H/N )(aN )1 H/N for any aN G/N . Therefore H/N / G/N .
Conversely, suppose that H/N /G/N . Then for any aN G/N , aN (H/N )(aN )1
H/N . So aha1 N = aN hN (aN )1 H/N for all h H. Therefore there ex-
ists h0 H such that aha1 N = h0 N . So aha1 = h0 n for some n N . Then
aha1 = h0 n H, because N H. This implies that aHa1 H. So H / G.

58. If N and M are normal subgroups of G, prove that N M is also a normal subgroup
of G.
In general, N M is not a subgroup! But if (at least one of) N and M are normal,
we can prove that N M is also a subgroup. e = ee N M , so N M is nonempty.
Take n1 m1 , n2 m2 N M . Then (n1 m1 )(n2 m2 )1 = n1 m1 m1 1
2 n2 . Because M
is normal, there is m3 M such that m1 m1 1 1 1 1
2 n2 = n2 m3 . So n1 m1 m2 n2 =
n1 n1
2 m3 N M and N M G.

Because N and M are normal subgroups, aN a1 N and aM a1 M for any

a G. Now aN M a1 = aN a1 aM a1 N M . Therefore N M / G also.