(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, R−120} which is a cyclic subgroup of order 3.
Next, since conjugation is an automorphism it follows that if
x ∈ D12 and g ∈ H then (xgx−1)3 = xg3x−1 = e. Hence
xHx−1 ⊂ 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 RαFlR
−1
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 = k−1h−1. We have that x = k−1 ∈ K and y =
h−1 ∈ H since both H and K are subgroups. as before xyx−1 =
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= 1¯.
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 ϕ : Zmn → Zm ⊕ Zn be given
by
ϕ(k mod mn) = (k mod m, k mod n)
Prove that ϕ is an isomorphism.
Solution
Let us first check that ϕ 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 ϕ is well defined.
Next, ϕ(k¯1 + k¯2) = ((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) = ϕ(k¯1) +ϕ(k¯2). This shows that
ϕ preserves operation.
Let us show that ϕ is 1− 1.
Suppose ϕ(k¯1) = ϕ(k¯2). 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. k¯1 = k¯2 in Zmn.
This proves that ϕ is 1-1.
4Since |Zmn| = mn = |Zm ⊕ Zn| this means that ϕ is an injec-
tive map between two finite sets with the same number of elements.
Hence ϕ must be onto.
This verifies that ϕ 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 = 〈9¯〉,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) = {1¯, 3¯, 7¯, 9¯, 11, 13, 17, 19} ⊂ Z20.
We have that 3¯2 = 9¯ /∈ K, 33 = 27 = 7 mod 20 /∈ K, 34 = 81 = 1
mod 20. Hence |3¯K| = 4.
On the other hand 3¯2 = 9¯ ∈ H, 7¯2 = 49 mod 20 = 9 mod 20 ∈
H, 9¯2 = 1¯ ∈ H, 112 = −92 = 1¯ ∈ H, 132 = −72 = 9¯ ∈ H, 172 =
−32 = 9¯ ∈ H, 192 = −12 = 1¯ ∈ H. This shows that every element of
G/H has order at most 2.
Therefore G/H G/K.
Answer: G/H G/K.