Model Answers for Test 1 with grading scheme
#1(a): let G be a set
with a binary operation *
G is a group if
1. there exists an element e
such that e*x=x*e=x
for every x in G
2. for every a in G
there exists an element denoted a-1
such that a*a-1=e and a-1*a=e
3. for every a,b,c in G
a*(b*c) = (a*b)*c
(1p)
(1p)
#1(b): Let H be a subset of a group G
H is a subgroup if
1. e є H
2. for every a and b in H, a*b є H
3. for every a in H, a-1 є H
(1p)
(1p)
(2p)
(1p)
(1p)
(1p)
(2p)
(1p)
(1p)
(1p)
(2p)
(2p)
#1(c): GCD Theorem: Suppose a and b are non-zero integers.
(1p)
Then there exist integers m and n
(2p)
Such that gcd(a,b) = ma+nb
(2p)
Moreover, the gcd(a,b) is the smallest positive linear combination of a and b
(2p)
-------------------------------------------------------------------------------------------#2: Given a and b, let S={am+bn | m,n є Z}.
(3p)
let d be the smallest positive element of S.
(2p)
Claim 1: d divides a:
(2p)
By Division Algorithm we can write a=qd+r with 0≤ r < d
If r>0, then r=a-qd=a-q(am+bn)=a(1-qm)-bqn, which means that r є S, positive but
smaller than d. So that can’t be and r=0, and d divides a.
(3p)
Claim 2: similarly, d divides b
Claim 3: if d’ is another common divisor of a and b, then d’ divides d:
(2p)
Write a=d’s and b=d’t, and d=am +bn.
Then d=d’sm+d’tn=d’(sm+tn), so d’ divides d.
Therefore d is the greatest of all common divisors. QED
(3p)
#3: this was done well by all
#4.
(20p)
(a) it turns out that a9=1 and a10=2
(10p)
(b) it turns out that sum of digits=2 mod 11. This means that the error is a digit
that is either 2 too big or 9 too small.
the weighted sum of all digits =6 mod 11. This shows that the error is in the third
position, because the error of 2 is multiplied by the weight 3 of the third position to yield
the 6.
So the correct number should be 7324501877
(10p)
a b
#5. if the inverse matrix is A-1= 
 , then it satisfies the equation:
c d 
0 2 a b  1 0
=
 , which can be written out as


2 1  c d  0 1
 2c=1 mod 3, hence c=2
 2a+c=0 mod 3, hence a=2
 2d=0 mod 3, hence d=0
 2b+d=1 mod 3, hence b=2
 2 2
So the inverse matrix is A-1= 
.
2 0
The checking is done multiplying A with A-1, and getting the identity matrix.
(8p)
(2p)
(B) each entry of any matrix in GL(2,Z3) are either 0, 1 or 2. Hence the maximum
number of matrices that can exist in this group is 34=81. However, not every matrix is
invertible. In fact, every matrix with ad-bc=0 is not invertible. One can check that there
are 33 different ways how ad-bc can be 0. So GL(2,Z) has exactly 48 members. Any
estimate near that number would have been sufficient.
(5p)
#6(a) False. Many examples possible. (Do specify the group and the elements.)
One possibility is G=D4, with a=H and b=V, because a2=b2=e, but a≠b.
(8p)
(b)True. Proof: Suppose a and b are relatively prime.
Then, by the GCD Thm, there is a linear combination am+bn=1.
Now write b=qa+r, so that r=b mod a.
Then am+bn=1 implies am+(qa+r)n=1.
This means that a(m+qn) +r n=1. This shows that there is linear combination of a and r
that equals 1. By the GCD Thm this means that a and r are relatively prime.
(12p)
(c) True. Proof: need to prove three things
1. e є H. Checks because e*a=a*e
2. if x є H, then x-1 є H. Checks because if x*a=a*x, multiply both sides by x-1 on
the right and on the left, then we get a*x-1=x-1*a
3. if x,y є H, then x*y є H. This also checks, because
(x*y)*a=x*(y*a)=x*(a*y)=(x*a)*y=(a*x)*y=a*(x*y)
(12p)
(d) False. Consider for example Z={0,1,2,3,4,5}.
Let H={0,3} and K={0,2,4} be two subgroups. Then their union
H  K ={0,2,3,4} is not a subgroup.
(8p)
Notes:
 in statements of definitions or theorems, pay close attention to the `quantifiers’
 completeness, thoroughness and precision are key in pure math.
 getting `the gist’ is not enough, because it leaves too much ambiguity, and
ambiguity is public enemy #1. You’ll never write a good proof if you only know
the `gist’. . .
 Tests are meant to teach, not just assess. So if you don’t understand why I
subtracted a point, please ask. If you don’t ask, it can lead to grievances, as well
as missed opportunities for learning.