HKU/MATH3301 (Algebra I)/Test1 - Suggested Solution/2Oct15/ 1

1. (4 points) Let G = {a, b, c, d} be a set of 4 elements and its multiplication table be

∗ a b c da a b c db b c d ac c d c bd d a b c

Does G under the operation ∗ form a group? Explain your answer.

Ans. No, G is not a group. From the table, a2 = a and c2 = c. Suppose G is a group.

Then multiplying the inverses of a and c respectively, we get a = e andc = e where

e ∈ {a, b, c, d} denotes the identity of G. This implies a = c, contradiction arises.

2. (10 points) [In Parts (b) and (c) below, e denotes the identity of the group.]

(a) Compute the orders of all elements in Z2 × Z3.

(b) Let G be a group and a ∈ G have order k. Show that if ah = e, then k divides h.

(c) Let G be a group of order 2015. Show that no element x ∈ G\{e} satisfies x201 = e.

Ans.

(a) ord(0, 0) = 1, ord(0, 1) = ord(0, 2) = 3, ord(1, 0) = 2, ord(1, 1) = ord(1, 2) = 6.

(b) See Tutorial 3, Qn 7. Note that the tutorial solution is very (too!) brief.

One should emphasize the minimality of the order. Below is an example.

By division algorithm, h = qk + r where 0 ≤ r < k. Now, ah = e implies

(∗): (ak)qar = e.

As ord(a) = k, ak = e and there is no integer 0 < r < k such that ar = e.

The equation (∗) implies ar = e with 0 ≤ r < k and thus r = 0. Hence k|h.

Classmates’ mistakes :

• ord(0, 0) = 0.

• Z2 × Z3∼= Z6 is a cyclic group (this is correct!)

All elements in Z6 are of order 6 (wrong)

• x201 = e implies ord(x) = 201

• ∵ 201 = 3 · 67, ∴ ord(x) = 3 or 67

HKU/MATH3301 (Algebra I)/Test1 - Suggested Solution/2Oct15/ 2

3. (10 points)

For each of the following either give an example or say that no such example exists.

NO explanation is required.

(a) A group G that ab 6= ba for some a, b ∈ G.

(b) A non-cyclic finite abelian group.

(c) Two non-isomorphic finite groups of the same order.

(d) A subgroup H of a group G with their orders satisfying 1 < |H| < |G| = 17.

(e) An infinite nonabelian subgroup H of a group G with the index [G : H] = 2.

Ans.

(a) GL(2,R)

(b) Z2 × Z2

(c) Z4 and Z2 × Z2

(d) No such example.

Here is an explanation (although it is not required in the question):

By Lagrange theory, the order of a subgroup must divide the order of the group.

Now |G| = 17 is a prime. Only 1 and 17 divide 17. Hence the possible order of a

subgroup in G is 1 or 17.

(e) Example 1. G = {A ∈ GL(2,R) : detA = 1 or − 1} and H = SL(2,R).

Example 2. G = SL(2,R)× Z2 and H = SL(2,R)× {0}

Classmates’ mistakes :

• Group of nonzero matrices under multiplication. (It is not a group)

• Mn(R) with matrix multiplication. (It is not a group)

• G = the group of 2× 2 invertible matrices and H = the subgroup of matrices in G

whose determinant = 1. (But [G : H] 6= 2)

• Perm(X) is non-abelian. (How about the case X = {1} or X = {1, 2}?!)

• Z×4 is non-cyclic. (Note |Z×4 | = 2. Could we have a non-cyclic group of order 2?)

HKU/MATH3301 (Algebra I)/Test1 - Suggested Solution/2Oct15/ 3

4. (10 points) Prove or disprove with justification the following statements.

(a) If xy = x−1y−1 for all x, y in the group G, then G is abelian.

(b) There is a non-abelian subgroup of the product group G1 ×G2 where |G1| = 4 and

|G2| = 11.

Ans.

(a) True. (It is a variant of Assignment 1 Qn 5!)

Take y = e, then we have x = x−1 for all x, i.e. x2 = e for all x ∈ G. Now argue as

in Assignment 1, Qn 5 to conclude G is abelian.

(b) False.

As |G1| = 4, we know G1 is abelian (indeed, G1∼= Z4 or Z2 × Z2).

As |G2| = 11 which is a prime, G2 is a cyclic group (see Lecture Notes).

Thus G2 is abelian.

Hence G1 × G2 is abelian (see Tutorial 2, Qn 5). Thus any subgroup of G1 × G2,

which is abelian, must be abelian.

Classmates’ mistakes :

• (xy)−1 = x−1y−1

• If H ⊂ G1 × G2 is a subgroup, then H = H1 × H2 where H1, H2 are subgroups of

G1, G2 respectively. (Wrong. Figure out a counterexample please!)

• xz2 = z2x for all x, z ∈ G ⇒ xy = yx by taking y = z2 for y ∈ G. (Can we always

express an element x in a group G into the form x = z2 where z ∈ G?)

• ord(x) = ord(y) = 2 ⇒ x = y

• Some classmates commented that for part (b) it suffices to give a counterexample.

This is not sufficient! Think about the following question:

There is a non-abelian subgroup of the product group G1×G2 where |G1| = 6

and |G2| = 11.

What will be the answer?

The answer is yes, and a justification is: Consider G1 = S3 and G2 = Z11 and take

H = G1 × {0}. That means the question is understood as: There exist groups G1

and G2 of order 4 and 11 such that G1×G2 has a non-abelian subgroup. To disprove

it, you have to assure no such a pair of groups, not just to find one pair.

HKU/MATH3301 (Algebra I)/Test1 - Suggested Solution/2Oct15/ 4

5. (6 points) Let H be a subgroup of the group G and K a subgroup of H.

(a) Show that [G : K] = [G : H][H : K] if G is finite. (b) Does Part (a) hold if G is an

infinite group while both [G : H] and [H : K] are finite? Justify your answer.

Ans.

(a) By Lagrange theorem, [G : K] = |G||K| = |G|

|H||H||K| = [G : H][H : K].

(b) Yes, the formula still holds when [G : H] and [H : K] are finite (even though G is

infinite).

Proof. As [G : H] and [H : K] are finite, we may write

G/H = {a1H, · · · , amH} and H/K = {b1K, · · · , bnK}.

Then G =m⋃i=1

aiH and H =n⋃

j=1

bjK. Want to show:

G/K = {aibjK : i = 1, · · · ,m, j = 1, · · · , n}.

Observe that G =m⋃i=1

n⋃j=1

aibjK.

Thus it suffices to show aibjK 6= apbqK whenever (i, j) 6= (p, q).

Suppose aibjK = apbqK.

This implies aibjk ∈ apbqK for some k ∈ K and hence ai ∈ apbqKb−1j .

As bq, bj ∈ H and K ⊂ H, this implies ai ∈ apH. Thus i = p and also bjk ∈ bqK.

This implies bj ∈ bqK, so j = q.

In summary, the set {aibjK : i = 1, · · · ,m, j = 1, · · · , n} of mn elements is the set

of all left K-cosets in G. Thus [G : K] = mn = [G : H][H : K].

End