MATH 251
Homework 1
Solutions
1. Let G be a nonempty set and ∗ : G × G → G a function such that
(a) ∀a, b, c ∈ G : (a ∗ b) ∗ c = a ∗ (b ∗ c)
(b) ∃e ∈ G : ∀a ∈ G : a ∗ e = a = e ∗ a
(c) ∀a ∈ G : ∃b ∈ G : a ∗ b = e
Prove the following:
• The element e from (b) is unique. So prove: if e0 ∈ G such that a ∗ e0 = a = e0 ∗ a for all a ∈ G then e0 = e.
Proof : Let e0 ∈ G such that a ∗ e0 = a = e0 ∗ a for all a ∈ G. Using this property with ‘a = e’ we get
e ∗ e0 = e. Using property (b) with ‘a = e0 , we get e0 = e ∗ e0 . Hence e0 = e.
2
• If a, b ∈ G with a ∗ b = e then b ∗ a = e.
Proof : Let a, b ∈ G with a ∗ b = e. By property (c), we have that b ∗ c = e for some c ∈ G. Using
properties (a) and (b), we get
c = e ∗ c = (a ∗ b) ∗ c = a ∗ (b ∗ c) = a ∗ e = a
2
Hence e = b ∗ c = b ∗ a.
• Given a ∈ G, the element b from (c) is unique. So prove: if a, b, c ∈ G with a ∗ b = e = a ∗ c then b = c.
Proof : Let a, b, c ∈ G with a ∗ b = e = a ∗ c. We just showed that b ∗ a = e. Using this and properties (a)
and (b), we get
b = e ∗ b = (b ∗ a) ∗ b = b ∗ (a ∗ b) = b ∗ (a ∗ c) = (b ∗ a) ∗ c = e ∗ c = c
2
2. For a real number x, we denote by [x] the largest integer smaller than or equal to x.
Let G = {x ∈ R : 0 ≤ x < 1}. For x, y ∈ G, we define
x ∗ y = x + y − [x + y]
Prove that ∗ is a binary operation (so x ∗ y ∈ G for all x, y ∈ G) and that (G, ∗) is an abelian group. Clearly
explain which properties you are using to prove this.
Proof : First, we show the following claim:
If t ∈ R and z ∈ Z then [t + z] = [t] + z
Indeed, let t ∈ R and z ∈ Z. Note that [t] ≤ t < [t] + 1 by definition of [t]. So [t] + z ≤ t + z < [t] + z + 1. Since
[t] + z is an integer, it follows from the definition of [t + z] that [t + z] = [t] + z, which proves the claim.
• Binary
If t ∈ R then [t] ≤ t < [t] + 1 and so 0 ≤ t − [t] < 1. Let x, y ∈ G. Then 0 ≤ x + y − [x + y] < 1. Hence
x ∗ y = x + y − [x + y] ∈ G.
• Abelian
Let x, y ∈ G. Since addition of real numbers is abelian, we get:
x ∗ y = x + y − [x + y] = y + x − [y + x] = y ∗ x
1
• Identity
Clearly, 0 ∈ G. Let x ∈ G. Then 0 ≤ x < 1 and so [x] = 0. Since 0 is the identity for the addition of real
numbers, we have
x ∗ 0 = x + 0 − [x + 0] = x − [x] = x − 0 = x
Hence 0 is the identity in G.
• Inverses
Let x ∈ G. If x = 0 then 0 ∗ 0 = 0 since 0 is the identity. So we may assume that 0 < x < 1. Put y = 1 − x.
Then 0 < y < 1 and so y ∈ G. Note that
x ∗ y = x + y − [x + y] = 1 − [1] = 0
So y is the inverse of x.
• Associativity
Let x, y, z ∈ G. Using the fact that addition of real numbers is abelian and associative and the claim, we
get
(x ∗ y) ∗ z
= (x + y − [x + y]) ∗ z
= x + y − [x + y] + z − [x + y − [x + y] + z]
= x + y + z − [x + y] − ([x + y + z] − [x + y])
= x + y + z − [x + y + z]
(+ is associative in R)
(+ is abelian in R, claim)
Once easily checks that x ∗ (y ∗ z) = x + y + z − [x + y + z]. Hence * is associative.
3. Let G be a group. Prove that g −1
−1
2
= g for all g ∈ G.
= e = g −1 ∗ g by definition of the inverse of g. Hence g is an inverse for g −1 .
−1
Since inverses are unique, we get g −1
= g.
2
Proof : Let g ∈ G. Then g ∗ g
−1
4. Let G be a group such that g −1 = g for all g ∈ G. Prove that G is abelian.
Proof : Pick a, b ∈ G. Then (ab)−1 = ab. But (ab)−1 = b−1 a−1 in any group (indeed, b−1 a−1 ab = e). Hence
we have
ab = (ab)−1 = b−1 a−1 = ba
2
So G is abelian.
5. Prove the Subgroup Test :
Let G be a group and ∅ =
6 H ⊆ G. Then H ≤ G if and only if ab−1 ∈ H for all a, b ∈ H.
Proof : Suppose first that H ≤ G. Pick a, b ∈ H. Since H is a group, we have that b−1 ∈ H (H contains
inverses) and ab−1 ∈ H (H is closed under multiplication).
Suppose next that ab−1 ∈ H for all a, b ∈ H. Since H 6= ∅, we can pick x ∈ H. Putting a = b = x, we get that
e = xx−1 ∈ H. Pick h ∈ H. Putting a = e and b = h, we see that h−1 = eh−1 ∈ H. Finally, pick g, h ∈ H.
Putting a = g and b = h−1 , we get that gh = g(h−1 )−1 ∈ H. Since the multiplication of elements in H is clearly
associative (it is associative in G), we get that H ≤ G.
2
2
6. Find all the non-trivial subgroups of Q8 and A4 (a subgroup H of a group G is non-trivial if {e} =
6 H 6= G).
Provide a table of multiplication for the non-cyclic subgroups.
Solution : We first consider Q8 . Since |Q8 | = 8, the order of any proper subgroup of Q8 must be an element of
{2, 4}. Note that Q8 has only one element of order two (namely −1). So the only proper subgroups of Q8 must
be cyclic (a group of order four that is not cyclic has three elements of order two). They are
{1, −1}
{1, −1, i, −i}
{1, −1, j, −j}
{1, −1, k, −k}
= h−1i
= hii = h−ii
= hji = h−ji
= hki = h−ki
Next, we consider A4 . Since |A4 | = 12, the order of any proper subgroup of A4 must be an element of {2, 3, 4, 6}.
We start with the cyclic subgroups of A4 :
{e, (12)(34)}
{e, (13)(24)}
{e, (14)(23)}
{e, (123), (132)}
{e, (124), (142)}
{e, (134), (143)}
{e, (234), (243)}
= h(12)(34)i
= h(13)(24)i
= h(14)(23)i
= h(123)i = h(132)i
= h(124)i = h(142)i
= h(134)i = h(143)i
= h(234)i = h(243)i
Since A4 does not have any elements of order four and only three elements of order two, we get that the only
possible subgroup of A4 of order four is {e, (12)(34), (13)(24), (14)(23)}. It turns out that this is a subgroup
(we have to check that this subset of A4 is closed under multiplication; this follows from the following table of
multiplication ) :
·
e
(12)(34)
(13)(24)
(14)(23)
e
e
(12)(34)
(13)(24)
(14)(23)
(12)(34)
(12)(34)
e
(14)(23)
(13)(24)
(13)(24)
(13)(24)
(14)(23)
e
(12)(34)
(14)(23)
(14)(23)
(13)(24)
(12)(34)
e
The only thing we still have to show is that A4 does not have any subgroups of order six. Suppose that
H ≤ A4 and |H| = 6. Then H contains exactly one subgroup of order three (if H1 and H2 are two different
subgroups of H of order three, then H1 H2 ⊆ H but |H1 H2 | = 9, a contradiction), say {e, (abc), (acb)}. But
(abc)(ab)(cd) = (bdc), (abc)(ac)(bd) = (adb) and (abc)(ad)(bc) = (acd). So H does not contain any elements of
order two, a contradiction.
2
7. Let G be a group, I an index set and Hi ≤ G for all i ∈ I. Prove that ∩i∈I Hi ≤ G.
Proof : Note that ∩i∈I Hi 6= ∅ since e ∈ Hi for all i ∈ I. Pick a, b ∈ ∩i∈I Hi . Then a, b ∈ Hi for all i ∈ I. Since
Hi ≤ G, it follows from the Subgroup Test that ab−1 ∈ Hi for all i ∈ I. So ab−1 ∈ ∩i∈I Hi . By the Subgroup
Test, ∩i∈I Hi ≤ G.
2
8. Rewrite the following elements of D6 in the form sri with 0 ≤ i ≤ 5 :
(a) sr−2 sr5
(b) r−3 sr4 sr−2
3
Solution : Recall that sri = r−i s for all i ∈ Z, s2 = e and r6 = e.
(a) sr−2 sr5 = ssr2 r5 = s2 r7 = r
2
(b) r−3 sr4 sr−2 = r−3 r−4 ssr−2 = r−7 s2 r−2 = r−7 r−2 = r−9 = r3
: a, b, c ∈ R; a 6= 0, c 6= 0 . Prove that H ≤ GL(2, R).
1 0
a b
x y
Solution : Note that H 6= ∅ since
∈ H. Pick
,
∈ H. Then we have that
0 1
0 c
0 z
9. Put H =
a
0
b
c

a b
0 c
x
0
y
z
−1
=
a b
0 c
1

 x

0
 
a
y
−
 
xz  =  x
1  
0
z

bx − ay

xz
∈H

c
z
2
So by the Subgroup Test, H ≤ G.
10. Put H = {e, (124), (142)}.
(a) Write down all the left cosets of H in A4 .
(b) Find a left transversal to H in A4 .
Solution : (a) We easily get that all the left cosets of H in A4 are
{e, (124), (142)} , {(123), (14)(23), (234)} , {(134), (13)(24), (132)}
(b) One example of a left transversal to H in A4 is {e, (123), (134), (243)}.
4
and {(243), (12)(34), (143)}
2