MAT301 Solutions to Practice Problems for Term Test 2
(1) Prove that C∗/R∗ has infinite order.
Solution 1
Note that since C∗ is abelian the subgroup R∗CC∗ is normal and
hence the factor group C∗/R∗ is well defined.
Next note that cosets given z1, z2 ∈ C∗ the cosets z1R and z2R
are equal iff z1z2 ∈ R. Therefore to show that C∗/R∗ is infinite it’s
enough to produce zn 6= 0, n > 0 such that znzm /∈ R for any n 6= m.
Let zn = n+ i, n = 1, 2, . . . .. Then for n 6= m we have
zn
zm
=
n+ i
m+ i
=
(n+ i)(m− i)
(m+ i)(m− i) =
mn+ 1 + i(m− n)
m2 + 1
/∈ R
Therefore there are infinitely many distinct left cosets of R∗ in C∗
and hence C∗/R∗ is infinite.
Solution 2
Let z = cosα+ i sinα where α =
√
2 · 2pi.
Then zn = cos(n
√
2 · 2pi) + i sin(n√2 · 2pi).
Note that since
√
2 is irrational we can never have that n
√
2 ·2pi =
pik for any integer k, n with n 6= 0.
Since sinx = 0 only for x = pik it follows that sin(n
√
2 · 2pi) 6= 0
for any n 6= 0 and therefore zn 6∈ R for any n 6= 0 and hence zR∗ has
infinite order in C∗/R∗.
(2) Let G = (R∗, ·).
Find all subgroups of G of index 2.
Solution
Let H ⊂ G be a subgroup of index 2. Since G is abelian H C G
is normal. Therefore G/H ∼= Z2 and every element gH in G/H
satisfies (gH)2 = eH which is equivalent to g2 ∈ H. Therefore for
any x ∈ R∗ we have that x2 ∈ H. Since for any x > 0 we have that
x = (
√
x)2 it follows that all positive real numbers are in H.
We claim that H = (0,∞). Indeed, if H contains a negative
number x0 then any other negative number x can be written as
x = x0 · t where t = xx0 > 0. since both x0, t ∈ H it follows that
x ∈ H also. but this means that H = G which contradicts the
assumption that H has index 2. Therefore H contains no negative
numbers and hence H = (0,∞).
Lastly, observe that (0,∞) is indeed a subgroup of G of index 2.
1
2Answer: The only subgroup of G of index 2 is H = (0,∞).
(3) Let n ≥ 3 and H C Sn be a normal subgroup such that (12) ∈ H.
Prove that H = Sn.
Hint: Use that every element of Sn can be written as a product
of transpositions.
Solution
Let σ ∈ Sn be arbitrary. Recall that by the formula for conjugation
on Sn proved in class we have that σ(12)σ
−1 = (σ(i), σ(j)).
Since H C Sn we have that σ(12)σ−1 = (σ(i), σ(j)) ∈ H. Since σ
can be chose to be arbitrary for any transposition (ij) ∈ Sn we can
find a σ such that σ(1) = i, σ(2) = j. Therefore (ij) ∈ H for any
i 6= j, 1 ≤ i 6= j ≤ n. Since every element of Sn can be written as
a product of transposition and H is closed under multiplication it
follows that H = Sn.
(4) Let G = (Z,+). Let H = 〈5〉 and K = 〈7〉.
(a) Prove that G = HK
Solution
We need to show that any element of Z¯ can be written as h+ k
where h = 5m, k = 7n for some integer m,n. In other words we
need to show that any integer can be written as 5m+ 7n. This
is true since gcd(5, 7) = 1. Explicitly, 1 = 5 · 3− 7 · 2. Therefore
any l ∈ Z can be written as l = 5 · (3l)− 7 · (2l).
(b) Is G the internal direct product of H and K?
Solution
No, because H ∩K 6= {0}. In particular 35 = 7 · 5 ∈ H ∩K.
Answer: No.
(5) Let G be a non abelian group of order 125 such that |Z(G)| > 1.
Prove that G/Z(G) ∼= Z5 ⊕ Z5.
Solution
We have that |G| = 125 = 53. By Lagrange’s theorem |Z(G)| divides
|G| = 125 so it can only be equal to 1, 5, 25 or 125. Since we are given
that |Z(G)| > 1 we know that |Z(G)| 6= 1. Since G is not abelian
we know that |Z(G)| 6= 125. So either |Z(G)| = 5 or |Z(G)| = 25.
Suppose |Z(G)| = 25. Then |G/Z(G)| = 125/25 = 5 which is
prime. Hence G/Z(G) is cyclic. But then by Theorem 9.3 (this the-
orem was also proved in class) G is abelian. This is a contradiction
and therefore |Z(G)| = 5.
3Hence |G/Z(G)| = 125/5 = 25 = 52. By classification of groups
of order p2 for prime p (Theorem 9.7 which was also proved in calss)
it follows that G/Z(G) ∼= Z5 ⊕ Z5 or G/Z(G) ∼= Z25. Suppose
G/Z(G) ∼= Z25. Since Z25 is cyclic, again using Theorem 9.3 we
conclude that G is abelian. This is a contradiction and therefore
G/Z(G) ∼= Z5 ⊕ Z5.
(6) Let H,K ⊂ G be subgroups. Suppose H C G,K C G. Prove that
HK is a normal subgroup of G.
Solution
First let us show that HK is a subgroup of G.
Let us verify the two-step subgroup test. Let h1k1, h2 ∈ HK
where h1, h2 ∈ H, k1, k2 ∈ K. Then (h1k1)(h2k2) = h1(k1h2)k2 =
h1(h
′
2k1)k2 = (h1h
′
2)(k1k2) ∈ HK where we used that k1H = Hk1
since H CG. This shows that HK is closed under the operation.
Next let hk ∈ HK where h ∈ H, k ∈ K. Then (hk)−1 = k−1h−1 =
h′k−1 ∈ HK where h′ ∈ H because k−1H = Hk−1.
This shows that HK ⊂ G is a subgroup.
Let us verify that it is normal.
Let g ∈ G, h ∈ H, k ∈ K. Then g(hk)g−1 = ghg−1gkg−1 = h′k′
where h′ = ghg−1 ∈ H since H C G and k′ = gkg−1 ∈ K since
K C G. Hence g(hk)g−1 ∈ HK and therefore g(HK)g−1 ∈ HK.
Since g ∈ G was arbitrary, by the normal subgroup test this implies
that HK CG.
(7) Let m,n > 1 be relatively prime. Let ϕ : Zm ⊕ Zn → Zm ⊕ Zn be
an automorphism.
Prove that there are automorphisms ϕ1 ∈ Aut(Zm) and ϕ2 ∈
Aut(Zn) such that ϕ(a¯, b¯) = (ϕ1(a¯), ϕ2(b¯)).
Hint: Use that automorphisms preserve orders of elements.
Solution
Let x = (1¯, 0¯) ∈ Zm ⊕ Zn and b = (0¯, 1¯) ∈ Zm ⊕ Zn. Let ϕ(x) =
(k¯, l¯). Since ϕ is an automorphism we have that m = |x| = |ϕ(x)|
By the general formula for orders of elements in external direct
products we have that |(k¯, l¯)| = lcm(|k¯|, |l¯). Since |k¯| divides m and
|k¯| divides n and gcd(m,n = 1 we have that gcd(|k¯|, |l¯)) = 1 and
hence |(k¯, l¯)| = lcm(|k¯|, |l¯|) = |k¯| · |l¯|.
Thus we must have m = |ϕ(x)| = |(k¯, l¯)| = lcm(|k¯|, |l¯|) = |k¯| ·
|l¯|. But |l¯| is a divisor of n and hence it is relatively prime to m.
Therefore the only way this equality can hold if |l¯| = 1 which can
only be if l¯ = 0¯.
4Therefore ϕ(1¯, 0¯) = (k¯, 0¯). Moreover we must have that gcd(k,m) =
1 since otherwise |ϕ(x)| = mgcd(k,m) < m.
By a similar argument ϕ((0¯, 1¯)) = (0¯, s¯) where gcd(s, n) = 1.
Since ϕ preserves operation it follows that ϕ(a¯, b¯) = (k¯ · a¯, l¯ · b¯). Set
ϕ1 : Zm → Zm be multiplication by k¯ and ϕ2 : Zn → Z2 be multi-
plication by s¯. The maps ϕ1, ϕ2 are automorphisms by a theorem
from class since gcd(m, k) = gcd(n, s) = 1.
Thus we have proves the desired claim that ϕ(a¯, b¯) = (ϕ1(a¯), ϕ2(b¯)).

(8) Let G be a group of order 30. Suppose Z(G) = {e}.
Prove that G can not be represented as the internal direct product
of two nontrivial subgroups.
Hint: First show that if G = H×K and both |H| > 1 and |K| > 1
then |H| or |K| is prime.
Solution
Suppose G = H × K where |H| > 1, |K| > 1. Then 30 = |G| =
|H| · |K|. The only ways to factor 30 as a product of two integers
bigger than 1 are 30 = 2 · 15 = 3 · 10 = 5 · 6. Note that in each
of the cases one of the factors is prime. Therefore |H| or |K| is
prime. WLOG |H| = p is prime where p = 2, 3 or 5. Since groups
of prime order are cyclic it follows that H is abelian. But then
any element h ∈ H commutes with any element of H and it also
commutes with any element of K because G = H ×K by a theorem
about internal direct products from class. Therefore H ⊂ Z(G).
Indeed if g ∈ G then g = h′k′ where h′ ∈ H, k′ ∈ K and hence
gh = (h′k′)h = h′k′h = h′hk′ = hh′k′ = hg.
This is a contradiction with the assumption that Z(G) = {e}.
Therefore G can not be represented as the internal direct product of
two nontrivial subgroups.
(9) Suppose G1 ⊕G2 ⊕ . . .⊕Gn is cyclic.
Prove that each Gi is cyclic.
Solution 1
Let G = G1 ⊕ G2 ⊕ . . . ⊕ Gn. Then each Gi is isomorphic top
a subgroup of G via ϕi : Gi → G the canonical map onto the i’th
factor;
ϕi(x) = (e1, . . . , ei−1, x, ei+1, . . . , en). here ek ∈ Gk is the unit.
Since a subgroup of a cyclic group is cyclic it follows that each Gi is
cyclic.
Solution 2
5Let a = (a1, . . . , an) be a generator for G then for any g =
(g1, . . . , gn) ∈ G there is a k ∈ Z such that ak = g which means
that gi = a
k
i for all i = 1, . . . n. Since gi ∈ Gi can be arbitrary it
follows that any element of Gi is a power of ai and hence Gi = 〈ai〉
is cyclic.
(10) Let ϕ : Z30 → Z6 ⊕ Z5 be an isomorphism such that ϕ(3¯) = (3¯, 2¯).
Find all the possibilities for ϕ(1¯). Justify your answer.
Solution
Let ϕ(1¯) = (k¯, l¯). Then ϕ(3¯) = (3k¯, 3l¯) = (3¯, 2¯). Hence 3k = 3
mod 6, 3k − 3 = 6n, k − 1 = 2n, k = 1 mod 2. Also 3l = 2 mod 5.
Multiplying by 2 we get 6l = 4 mod 5, l = 4 mod 5.
This means that ϕ(1¯) must be one of the following (1¯, 4¯), (3¯, 4¯), (5¯, 4¯).
By the general formula |(g1, g2)| = lcm(|g1|, |g2|). Applying it we
get |(1¯, 4¯)| = lcm(6, 5) = 30 = |Z6 ⊕ Z5|. Therefore |〈(1¯, 4¯)〉| = 30 =
|Z6 ⊕ Z5| and hence 〈(1¯, 4¯)〉 = Z6 ⊕ Z5 .
Which means that (1¯, 4¯) is a generator of the cyclic group Z6⊕Z5.
It was proved in lectures that given two cyclic groups of the same
order G1 = 〈a〉, G2 = 〈b〉 with |a| = |b| = n the map G1 → G2 given
by ϕ(ak) = bk is an isomorphism. Note that it sends a to b.
Therefore there exists an isomorphism ϕ : Z30 → Z6 ⊕ Z5 such
that ϕ(1¯) = (1¯, 4¯).
Similarly, we compute |(1¯, 5¯)| = lcm(6, 5) = 30. By the same
argument as above there is an isomorphism ϕ : Z30 → Z6 ⊕ Z5 such
that ϕ(1¯) = (1¯, 5¯).
Lastly, we compute |(3¯, 4¯)| = lcm(2, 5) = 10 6= 30. Since isomor-
phisms preserve orders of elements there is no isomorphism ϕ : Z30 →
Z6 ⊕ Z5 such that ϕ(1¯) = (3¯, 4¯).
Therefore the only possibilities for ϕ(1¯) are (1¯, 4¯) and (5¯, 4¯)
Answer: The possibilities for ϕ(1¯) are (1¯, 4¯) and (5¯, 4¯).
(11) Determine the order of (Z⊕ Z)/〈(2, 2)〉. Is this group cyclic?
Solution
Let G = Z⊕Z, H = 〈(2, 2)〉. Note that H = {(2l, 2l) | l ∈ Z}. Let
a = (1, 0) ∈ G. We claim that |aH| =∞ in G/H. Indeed |aH| = k
if k is the smallest positive number such that ak ∈ H. However, for
k > 0 we have that ak = (k, 0) 6= (2l, 2l) for any k > 0, l ∈ Z. Hence
ak /∈ H for any k > 0 and therefore |G/H| =∞.
We claim that G/H is not cyclic. Indeed if it’s cyclic then it must
be infinite cyclic since |G/H| =∞. But every infinite cyclic group is
isomorphic to Z. In particular it has no nontrivial elements of finite
order since Z doesn’t. On the other hand let b = (1, 1) Then b /∈ H
6but 2b ∈ H which means that |bH| = 2 in G/Z. Therefore G/H Z
and hence G/H is not cyclic.
Answer: |(Z⊕ Z)/〈(2, 2)〉| =∞. (Z⊕ Z)/〈(2, 2)〉 is not cyclic.
(12) Prove that (1 3 5) belongs to [A5, A5].
Solution
Let a, b, c, d, e be distinct numbers. Let σ = (a b c), τ = (c d e). Let
us compute [σ, τ ] We have σ−1 = (a c b) and τ−1 = (c e d).
Hence [σ, τ ] = στσ−1τ−1 = (a b c)(c d e)(a c b)(c e d) = (a d c). Set-
ting a = 1, d = 3, c = 5, b = 2, e = 4 we get that [(1 2 5), (5 3 4)] =
(1 3 5).
It was shown in class that a k-cycle with odd k is even. Therefore
both (1 2 5) and (5 3 4) are in A5 and hence (1 3 5) = [(1 2 5), (5 3 4)]
is in [A5, A5].
(13) 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)|. .
(14) 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.
7Next, 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.
Since |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.
(15) 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.
(16) 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.