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1) (a) Let G = D12, H = {g ∈ D12 | g3 = R0}. Decide if H is not a

subgroup, a subgroup which is not normal or a normal subgroup

of G.

Solution

Claim: H is a normal subgroup of G. First note that all reflec-

tion in D12 have order 2 so H consists of rotations only. hence

H = {R0, R120, R120} which is a cyclic subgroup of order 3.

Next, since conjugation is an automorphism it follows that if

x ∈ D12 and g ∈ H then (xgx1)3 = xg3x1 = e. Hence

xHx1 H. Thus H CG is a normal subgroup by the normal

subgroup test.

Answer: H CG is a normal subgroup.

(b) G = S10, H = {σ ∈ S10, | such that σ can be written as

product of ≤ 3 disjoint cycles }. Decide if H is not a subgroup,

a subgroup which is not normal or a normal subgroup of G.

Solution

Let σ = (12)(34), τ = (56)(78). Then σ, and τ are in H but

στ = (12)(34)(56)(78) is a product of 4 disjoint cycles. by

uniqueness of disjoint cycle decomposition this means that στ /∈

H. Hence H is not a subgroup.

Answer: H is not a subgroup.

(c) Let G be an abelian group and let H = {g ∈ G | |g| < ∞}

Decide if H need not be a subgroup, must be a subgroup but it

need not be normal, is always a normal subgroup of G.

Solution

Let us check that H is a subgroup. let g, h ∈ H and |g| =

m, |h| = m are finite then Since H is abelian we have

(gh)mn = hmnhmn = (gn)m(hm)n = e · e = e, Hence gh has

finite order and thus gh ∈ H. This shows that H is closed

under the operation.

Next since |g?1| = |g| it follows that H is closed under inverses.

We have verified that H is a subgroup of G.

Since G is abelian every subgroup of G is normal.

Answer: H CG is a normal subgroup.

(d) Let G be a finite group, a, b ∈ G. Let H = 〈a, b〉.

then which of the following holds

a) |H| = |a| · |b|

b) |H| = lcm(|a|, |b|)

1

2c) |H| is divisible by lcm(|a|, |b|)

d) H need not be divisible by lcm(|a|, |b|).

Solution

Let G = S3, a = (12), b = (13). Then |a| = |b| = 2. But it’s

easy to check that 〈a, b〉 = S3 which has order 6. This means

that a) and b) are false in general.

Next observe that since a ∈ H by Lagrange’s theorem we have

that |a| divides |H|. Similarly |b| divides |H|. Hence lcm(a, b)

divides |H| as well.

Answer: |H| is divisible by lcm(|a|, |b|)

(e) How many normal subgroups of order 2 are there in D10?

Solution

Subgroups of order 2 correspond to elements of order 2.

In D10 there are 11 elements of order 2: 10 reflections and R180.

The subgroup 〈R180〉 = {R0, R180} is equal to the center of D10

and hence is normal.

We claim that none of the subgroups generated by reflections

are normal. Indeed, let Fl be a reflection, let α = 2pi/10. Then

Rα ∈ D10.

But RαFlR1

α = FRα(l) by a formula from class. Note that

Rα(l) 6= l and hence RαFlR?1α = FRα(l) /∈ 〈Fl〉 = {R0, Fl}.

Hence 〈Fl〉 is not normal in D10.

Therefore

Answer: There is exactly 1 normal subgroup of order 2 in D10.

(2) Let G be a group and let H,K CG be normal subgroups.

Prove that HK is a normal subgroup of G.

Solution

Let us first verify that HK satisfies the subgroup test.

Let hh′ ∈ H, k, k′ ∈ K. We claim that (hk)(h′k′) ∈ HK also.

Since H CG we have that kh′k?1 = h′′ ∈ H and therefore kh′ =

h′′k. Hence hkh′k′ = h(kh′)k′ = h(h′′k)k′ = (hh′′)(kk′) ∈ HK

suince hh′′ ∈ H and kk′ ∈ K. This shows that HK is closed under

the noperation.

Next (hk)1 = k1h1. We have that x = k1 ∈ K and y =

h1 ∈ H since both H and K are subgroups. as before xyx1 =

y′ ∈ H since H CG and hence xy = yx′ ∈ HK.

This verifies that HK is closed under inverses. Thus HK is a

subgroup of G.

3It remains to show that this subgroup is normal. Let g ∈ g, h ∈

H, k ∈ K then ghg?1 = h′ ∈ H and gkg?1 = k′ ∈ K. since H,K are

normal. Therefore g(hk)g?1 = ghg?1gkg?1 = h′k′ ∈ H.

This verifies that g(HK)g?1 ? HK.

Therefore HK CG by the normal subgroup test.

(3) Let n > 2.

Prove that |U(n)| is even.

Hint: Use Lagrange’s theorem.

Solution

Note that gcd(n ? 1, n) = 1 since if d divides both n and n ? 1

then it also divides n? (n? 1) = 1.

Therefore n? 1 ∈ U(n).

Next (n?1) = ?1 mod n and hence (n?1)2 = (?1)2 = 1 mod n.

This can also be seen more directly since (n? 1)2 = n2? 2n+ 1 = 1

mod n.

Since n > 2 we have that n? 1 > 1 and hence n? 1 6= 1ˉ.

This means that n? 1 has order 2 in U(n).

By a corollary to Lagrange’s theorem we have that 2 = |n? 1|

divides |U(n)|. .

(4) Let m,n > 1 be relatively prime. Let ? : Zmn → Zm ⊕ Zn be given

by

?(k mod mn) = (k mod m, k mod n)

Prove that ? is an isomorphism.

Solution

Let us first check that ? is well defined and preserves the operation.

if k1 = k2 mod mn then mn divides k1 ? k2. Hence both m,n

divide k1 ? k2 and therefore k1 = k2 mod m and k1 = k2 mod n.

This shows that ? is well defined.

Next, ?(kˉ1 + kˉ2) = ((k1 + k2) mod m, (k1 + k2) mod n)) =

(k1 mod m+k2 mod m, k1 mod n+k2 mod n) = (k1 mod m, k1

mod n) + (k2 mod m, k2 mod n) = ?(kˉ1) +?(kˉ2). This shows that

? preserves operation.

Let us show that ? is 1? 1.

Suppose ?(kˉ1) = ?(kˉ2). This means that k1 = k2 mod m and

k1 = k2 mod m . Therefore both m and n divide k1 ? k2.

Since gcd(m,n) = 1 this implies that mn divides k1 ? k2 as well,

i.e. kˉ1 = kˉ2 in Zmn.

This proves that ? is 1-1.

4Since |Zmn| = mn = |Zm ⊕ Zn| this means that ? is an injec-

tive map between two finite sets with the same number of elements.

Hence ? must be onto.

This verifies that ? is a bijection.

(5) Let G be a group such that G/Z(G) is abelian. Prove that for any

a, b, c ∈ G it holds that [[a, b], c] = e.

Solution

Let H = Z(G). Since G/H is abelian we have that [aH, bH] = H.

Recall that (aH)1 = a?1H and (bH)?1 = b?1H.

ThereforeH = [aH, bH] = (aH)(bH)(aH)?1(bH)?1 = (aH)(bH)(a?1H)(b?1H) =

aba?1b?1H = [a, b]H. Therefore z = [a, b] ∈ Z(G). This means

that z commutes with any element of G and hence [z, c] = e i.e.

[[a, b], c] = e.

(6) Let G = U(20), H = 〈9ˉ〉,K = 〈11〉 be subgroups of G.

Are G/H and G/K isomorphic?

Solution

Recall that in general |aH| is the smallest positive n such that

(aH)n = H which is equivalent to anH = H or an ∈ H.

Let us compute orders of various elements of G/H and G/K.

First, we have that U(20) = {1ˉ, 3ˉ, 7ˉ, 9ˉ, 11, 13, 17, 19} ? Z20.

We have that 3ˉ2 = 9ˉ /∈ K, 33 = 27 = 7 mod 20 /∈ K, 34 = 81 = 1

mod 20. Hence |3ˉK| = 4.

On the other hand 3ˉ2 = 9ˉ ∈ H, 7ˉ2 = 49 mod 20 = 9 mod 20 ∈

H, 9ˉ2 = 1ˉ ∈ H, 112 = ?92 = 1ˉ ∈ H, 132 = ?72 = 9ˉ ∈ H, 172 =32 = 9ˉ ∈ H, 192 = ?12 = 1ˉ ∈ H. This shows that every element of

G/H has order at most 2.

Therefore G/H G/K.

Answer: G/H G/K.

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