Homework 1 Solutions

1 . Suppose that G is a group such that x2 = e for all x ∈ G. Show that G is an abelian group.

We recall definition of abelian group .

Definition. A group G is said to be an abelian group if xy = yx for all x, y ∈ G .

Proof. for any x, y ∈ G. since (xy) was in G, by hypothesis (xy)2 = e. we have (xy)(xy) = e .
Then x(xy)(xy) = xe = x. Since xx = e , yxy = x. finally we have yx = (yx)e = (yx)(yy) =
(yxy)y = xy. Thus xy = yx .

2. Let G be a group. For any a ∈ G , define a0 = e. For any positive integer n, define an = an−1 a,
For any positive integer m , define a−m = (am )−1 .
(a) For any n ∈ Z , show (an )−1 = a−n = (a−1 )n .
(b) For any n, k ∈ Z , an ak = an+k .
(c) For any n, k ∈ Z , (an )k = ank .

For proof of 2 we need the following lemmas :

Lemma. Let G be a group . For any a ∈ G and any m ∈ Z , the followings are true .

aam = am+1 = am a
a−1 am = am−1 = am a−1

Proof (a). For left (=) . For any n ∈ Z .

if n = 0 , then (an )−1 = (a0 )−1 = e−1 = e = a−0 = a−n .
if n > 0 , then by definition listed in above ,we have a−n = (an )−1 .
if n < 0 , then by definition (an )−1 = [(a−n )−1 ]−1 = a−n .
For right (=) . for any m ∈ Z .
if m = 0 then a−m = a−0 = a0 = e = (a−1 )0 = (a−1 )m .
if m > 0 . we use induction on m .
For m = 1 : a−m = a−1 = (a−1 )1 = (a−1 )m .
Suppose m holds . (a−1 )m+1 |{z}
= (a−1 )m a−1 =
|{z} a−m a−1 |{z}
= a−m−1 = a−(m+1) .
by def by induction hypothesis by lemma
−1 m −1 −m −1 m −1 −m
if m < 0 then (a ) |{z}
= ((a ) ) =
|{z} (a ) =
|{z} a .
by def since −m>0 by left equality

1
Proof (b). for any n, k ∈ Z , we consider the following cases .
(n = 0 or k = 0), (n > 0 and k > 0), (n < 0 and k < 0), (n < 0 and k > 0), (n > 0 and k < 0)
case 1 : n = 0 or k = 0 , This is trivial. you should verify this by yourself .
case 2 : n > 0 and k > 0 . we induction on k .
for k = 1 , then an ak = an a1 = an a |{z}
= an+1 = an+k
by def

Suppose k holds .an ak+1 = an ak a = (an ak )a =

|{z} an+k a |{z}
= an+k+1 = an+(k+1) .
by induction hypothesis by def

= (a−n )−1 (a−k )−1 = (a−k a−n )−1 |{z}

case 3 : n < 0 and k < 0 . an ak |{z} = (a−n−k )−1 |{z}
= an+k .
by def by case2 by a

case 4 : n < 0 and k > 0 . we induction on k .

For k = 1 . an ak = an a1 = an a |{z}
= an+1 = an+k .
by lemma

Suppose k holds . an ak+1 = an ak a =

|{z} an+k a |{z}
= an+(k+1) .
by induction hypothesis by lemma

case 5 : n > 0 and k < 0 . we induction on n .

For n = 1 . an ak = a1 ak = aak |{z}
= a1+k = an+k .
by lemma

Suppose n holds . an+1 ak |{z}

= a1 an ak = aan ak =
|{z} aan+k |{z}
= a(n+1)+k .
by case 2 by induction hypothesis by lemma

Proof (c). For any n, k ∈ Z .

case 1 : k = 0 . Then (an )k = (an )0 = e = a0 = ank .
case 2 : k > 0 . we induction on k
For k = 1 . (an )k = (an )1 = an = ank .
Suppose k holds . (an )k+1 = (an )k an =
|{z} ank an |{z}
= ank+n = an(k+1) .
by induction hypothesis by b

= ((an )−k )−1 |{z}

case 3 : k < 0 . Then (an )k |{z} = (a−nk )−1 |{z}
= ank .
by def by case 2 by a

2
3 . Let R be set of all real numbers and let n ∈ N . Let

GL(n, R) = {A ∈ Mn (R) : A is invertible}

Prove GL(n, R) is group under matrix multiplication . Let

SL(n, R) = {A ∈ GL(n, R) : det(A) = 1}

.
Show SL(n, R) is subgroup of GL(n, R) and for n > 1 , SL(n, R) is not an abelian group .
we review some elementary properties about linear algebra .

Definition. Let n ∈ N and A ∈ Mn (R) .

A is said to be an invertible matrix iff there exist a matrix B ∈ Mn (R) such that AB = In = BA .
where In denotes identity matrix .

Propsition. Let n ∈ N and A, B, C ∈ Mn (R) .

1 : det(AB) = det(A)det(B) .
2 : (AB)C = A(BC) (associative law)

Proof. For proving GL(n, R) is a group . we need to verify the followings .

1 : closed under matrix multiplication .
2 : The matrix multiplication is associative .
3 : identity element exists .
4 : Every element in GL(n, R) has inverse .
For convenience we write G = GL(n, R) .
For 2 : this is true from above propsition .
For 1 : for any A, B in G . Then AC = CA = In and BD = DB = In for some C, D ∈ Mn (R).Take
E = DC , then E ∈ Mn (R) .we have

(AB)E = A(BE) = A(B(DC)) = A((BD)C) = A(In C) = AC = In

E(AB) = (EA)B = ((DC)A)B = (D(CA))B = (DIn )B = DB = In

By definition of G , AB ∈ G . Therefore G is closed under matrix multiplication .

For 3 : In ∈ G and apparently , In is multiplicative identity for G .
For 4 : for any F ∈ G . Then F K = KF = In for some K ∈ Mn (R) . By definition , K is also an
invertible matrix . Thus K ∈ G and F K = KF = In .
By 1 ,2 ,3 ,4 we conclude that G = GL(n, R) is a group under matrix multiplication .

3
Proof (Proving subgroup SL(n, R)). For proving SL(n, R) is a subgroup of GL(n, R) . we need to
show the followings .
1 . SL(n, R) ⊆ GL(n, R) .
2 . SL(n, R) ̸= ∅
3 . for any A, B ∈ SL(n, R) ,AB −1 ∈ SL(n, R)
For convenience , we write S = SL(n, R) .
For 1 , this is true by its definition .
For 2 , we can easily see that In in S .
For 3 , for any A, B ∈ S . Note : we have BB −1 = In . by properties listed in above , we have
1 = det(In ) = det(BB −1 ) = det(B)det(B −1 ) . since B ∈ S , 1 = det(B −1 ) . so we have the following
.

det(AB −1 ) = det(A)det(B −1 ) = 1 × 1 = 1

by definition of the set S , AB −1 ∈ S . Therefore , we have S = SL(n, R) is a subgroup of

GL(n, R) .

For proving SL(n, R) is not an abelian group for all n > 1 . we need to define two matries .
Let n > 1 and n ∈ N . we definte matries E12 , E21 in Mn (R) by the following definition .
For any i, j ∈ {1, 2, 3, . . . , n}
( (
1 if i = 1 and j = 2 1 if i = 2 and j = 1
E12 (i, j) = and E21 (i, j) =
0 if for otherwise 0 if for otherwise
Some important properties about E12 and E21 is
1 : E12 E12 = 0n
2 : E21 E21 = 0n
3 : E12 E21 = E11
4 : E21 E12 = E22
Where 0n is n by n matrix with all entries are zero . E11 , E22 are matries defined similarly as matries
defined above . Now we shall prove that SL(n, R) is not an abelian group for n > 1 .

Proof. For any n ∈ N with n > 1 . Write A = In + E12 and B = In + E21

Since A was an upper-triangular matrix with determinant 1 , A is invertibe matrix and with de-
terminant 1 . Then A ∈ SL(n, R) . Similarly , B ∈ SL(n, R) . Now we consider the followings
.

AB = (In + E12 )(In + E21 ) = In + E12 + E21 + E12 E21 = In + E12 + E21 + E11
BA = (In + E21 )(In + E12 ) = In + E12 + E21 + E21 E12 = In + E12 + E21 + E22

But (AB)(1, 1) = 2 and (BA)(1, 1) = 1 . Thus AB ̸= BA . There SL(n, R) is not an abelian group
.

4
 4 . Let R be the set of real numbers and let n be an integer with n ≥ 1.
Let O(n, R) = {A ∈ GL(n, R)|AAt = In } where At is the transpose of A.
Show that O(n, R) is a subgroup of GL(n, R). Note that O(n, R) is called the orthogonal group of
degree n over R.
We now need two lemmas listed in your linear algebra book .

Lemma. Let n ∈ N and A, B ∈ Mn (R)

if AB = In then A is invertible and B is the invertible matrix of A .

Lemma. Let n ∈ N and A, B ∈ Mn (R)

(AB)t = B t At

Proof (4). For proving O(n, R) is a subgroup of GL(n, R) , we need to show the followings .
1 : O(n, R) ⊆ GL(n, R)
2 : O(n, R) ̸= ∅
3 : for any A, B ∈ O(n, R) , AB −1 ∈ O(n, R) .
For convenience , write O = O(n, R)
For 1 : This is true from its definition .
For 2 : Note : In ∈ GL(n, R) and In Int = In . Thus In ∈ O
For 3 : for any A, B ∈ O Then AAt = In and BB t = In .From lemma listed in above , we know B t
is the invertible matrix of B . Hence B t = B −1 .we have the following

(AB −1 )(AB −1 )t = (AB −1 )((B −1 )t At ) = (AB −1 )((B t )t At ) = (AB −1 )(BAt ) = AAt = In

By definition of O , AB −1 ∈ O . By 1, 2, 3 we can conclude that O = O(n, R) is a subgroup of

GL(n, R) .

5
5 . Prove that the set of all 3 × 3 matrices with real entries of the form
 
1 a b
0 1 c
0 0 1

is a group under the usual matrix multiplication.

Proof (5). Let T be set of all matrix with form listed in above .
For proving T is a group , we need to show the followings .
1 : T is closed under usual matrix multiplication .
2 : The usual matrix multiplication is associative .
3 : T contains a multiplicative identity .
4 : For any element in T has inverse .
any A, B ∈ T . we write them by
For 1 : For
   
1 a b 1 d e
A= 0 1
 c and B = 0 1 f  where a, b, c, d, e, f ∈ R
0 0 1 0 0 1
 
1 d + a e + af + b
Now , we multiply them directly . AB = 0 1 f + c  so AB ∈ T
0 0 1
For 2 : we know matrix multiplication is associate .
For 3 : clearly I3 ∈ T and it is multiplicative identity for T .
For 4 : for any C ∈ T . Write C by
 
1 g h
C = 0 1 i  , Where g, h, i ∈ R .
0 0 1
 
1 −g −h + gi
Take D = 0 1 −i  then D ∈ T
0 0 1
clearly CD = DC = In .
By 1, 2, 3, 4 ,we can conclude that T is group under usual matrix multiplication .

6
6 : Let G be a finite group. Prove that, given a ∈ G, there is a positive integer n, depending on a,
such that an = e .
Before we prove this statement , we need a lemma in set theory .

Lemma. Let A be finite set and N is set of all positive integers .

Then there is no one-to-one function f from N into A .

(we assume that you know definition of finite set .)

Proof (6). For any a ∈ G . Since for any positive integer n , an ∈ G , we can define a function f .
For any n ∈ N, f (n) = an .Since G was a finite set ,By above lemma
we have f is not one-to-one function . By definition of 1-1 function , we have f (i) = f (j) and i ̸= j
for some i, j ∈ N . This shows ai = aj and i − j ̸= 0 . Without loss of generality we assume that
i > j , so ai−j = ai a−j = aj a−j = a0 = e and i − j ∈ N .

For 7 , Since it was very trivial , we skip its proof .

7
8 . Suppose that G is a nonempty set closed under an associative operation such that the following
conditions hold.
(a) Given a, y ∈ G, there is an element x ∈ G such that ax = y.
(b) Given a, w ∈ G, there is an element u ∈ G such that ua = w .
Show that G is a group.

Proof (8). For proving G is a group , we need to show the followings .

1 : G contains an identity element .
2 : For any element in G has inverse .
For 1 : Since G was nonempty , we fixed an element d in G .
By hypothesis a, b , we have that ed = d and du = e for some e, u ∈ G .
We now show e is an identity element for G . For any g ∈ G .
By hypothesis (b) , write g = te for some t ∈ G and by hypothesis (a) , write g = dv for some v ∈ G
.
We now have the followings .

1 :g = te (1)
2 :g = dv (2)

we have :

ge = g(du) = (gd)u |{z}

= ((te)d)u = (t(ed))u |{z}
= (td)u = t(du) |{z}
= te = g
by 1 by ed = d by du = e

eg |{z}
= e(dv) = (ed)v |{z}
= dv = g
by 2 by ed = d

Thus eg = ge = g . Therefore ,e is an identity element for G .

For 2 , for any x ∈ G , by hypothesis , we have yx = e and xz = e for some y, z ∈ G .
Then we have y = ye = y(xz) = (yx)z = ez = z . Thus yx = e = xy and y ∈ G .
By 1, 2 , we can conclude that G is a group .

For 9 , it is not difficult to show so we skip its proof .

8
10 . Let G be a group. Suppose that{Ai |i ∈ Λ} is a family of subgroups of G.
Show that ∩i∈Λ Ai is a subgroup of G .

Proof (10). Write S = ∩i∈Λ Ai . We need to show the followings .

1:S⊆G.
2 : S ̸= ∅ .
3 : for any a, b ∈ S , ab−1 ∈ S .
For 1 : this is trivial .
For 2 : Since for any i ∈ Λ, Ai was a subgroup of G ,e ∈ Ai for all i ∈ Λ .
This means e ∈ S .
For 3 : for any a, b ∈ S , (WTS : ab−1 ∈ S .)
For any i ∈ Λ . Since a, b ∈ S , a, b ∈ Ai . Note : Ai is a subgroup of G .
This shows b−1 ∈ Ai . Since Ai was closed under operation defined on G , ab−1 ∈ Ai . Since i was
arbitrary , ab−1 ∈ S . By 1, 2, 3 , we can conclude that S = ∩i∈Λ Ai is a subgroup of G .

11 . Let k be a positive integer. Let G be an abelian group.

Suppose that H = {a ∈ G|ak = e} . Show that H is a subgroup of G.

Proof (11). For proving H is a subgroup of G . we need to show the followings .

1: H⊆G.
2 : H ̸= ∅ .
3 : for any a, b ∈ H , ab−1 ∈ H .
For 1 : This is true from its definition .
For 2 : it is not difficult to check that e ∈ H .
For 3 : for any a, b ∈ H , (WTS : ab−1 ∈ H)

(ab−1 )k =
|{z} ak b−k = e(bk )−1 = e−1 = e
G is abelian
−1
By definition of H ,ab ∈ H . By 1, 2, 3 ,we can conclude that H is a subgroup of G .

9
12 . Let G be a group and let a ∈ G. Let CG (a) = {x ∈ G|xa = ax} be the centralizer of a in G and
let Z(G) be the center of G. Prove that Z(G) = ∩a∈G CG (a).

Proof (12). Before we prove this statement ,we recal definition of Center of G .

Z(G) = {x ∈ G | xg = gx for all g ∈ G}

For (⊆) . for any x ∈ Z(G) . for any a ∈ G , By above definition , xa = ax . Thus x ∈ CG (a) .
Since a was arbitrary , x ∈ ∩a∈G CG (a) . Thus Z(G) ⊆ ∩a∈G CG (a) .
For (⊇) . for any y ∈ ∩a∈G CG (a). for any g ∈ G . Since y ∈ CG (g) , yg = gy . Because g is
arbitrary , y ∈ Z(G) .
Combine above statements , we can conclude that Z(G) = ∩a∈G CG (a) .

13 . Let A, B be subgroups of a group G. Now we define the set AB = {ab | a ∈ A, b ∈ B}.

(a) Prove that if G is abelian, then AB is a subgroup of G.
(b) If G is non-abelian, is it still true that AB is a subgroup of G ?

Proof (a). Suppose G is an abelian group .

For convenience , write C = AB . We need to show the followings .
1:C⊆G.
2 : C ̸= ∅ .
3 : For any a, b ∈ C ,ab−1 ∈ C .
For 1, 2 trivally ture .
For 3 : for any a, b ∈ C . write a = uv, b = ts for some u, t ∈ A and some v, s ∈ B .
Then ab−1 = (uv)(s−1 t−1 ) = (ut−1 )(vs−1 ) . Since A, B were subgroups of G , ab−1 ∈ C .
|{z}
G is abelian

By 1, 2, 3 , we have C = AB is a subgroup of G .

Proof (b). The answer is no. For a counterexample, we consider symmetry group of {1, 2, 3} ,
namely S3 . First we list its all elements :
           
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3
ζ1 = ζ2 = ζ = ζ4 = ζ5 = ζ6 =
1 2 3 2 1 3 3 3 2 1 1 3 2 2 3 1 3 1 2

Note : ζ2 ζ3 = ζ6 and ζ3 ζ2 = ζ5 . This means S3 is non-abelian group

Let A =< ζ2 > and B =< ζ3 > . Since ord(ζ2 ) = 2 = ord(ζ3 ) , o(A) = 2 = o(B) .
So AB = {ζ1 , ζ2 }{ζ1 , ζ3 } = {ζ1 , ζ2 , ζ3 , ζ2 ζ3 } = {ζ1 , ζ2 , ζ3 , ζ6 } .
Note : ζ3 , ζ2 ∈ AB but ζ5 = ζ3 ζ2 not in AB . So AB is not closed under operation defined in S3 .
Thus AB is not a subgroup of S3 . (you can also use Lagrange’s theorem to verify this fact .)

10