MATH 3005 Homework Han-Bom Moon
Homework 11 Solution
Chapter 10 - Group Homomorphisms.
due: Dec. 10.
1. Let G and H be two groups.
(a) Show that the map π : G ⊕ H → G defined by π(g, h) = g is a homomor-
phism. This is called the projection of G ⊕ H to G. Find the kernel and the
image of π.
For two elements (g1 , h1 ), (g2 , h2 ) ∈ G ⊕ H,
π((g1 , h1 )(g2 , h2 )) = π(g1 g2 , h1 h2 ) = g1 g2 = π(g1 , h1 )π(g2 , h2 ).
So π is a homomorphism.
ker π = {(g, h) ∈ G ⊕ H | g = π(g, h) = e} = {(e, h) | h ∈ H}.
Im φ = G, because for every g ∈ G, π(g, e) = g.
Note that ker φ ≈ H (why?).
(b) Show that the map ι : G → G ⊕ H given by ι(g) = (g, e) is a homomorphism.
Find the kernel and the image of ι.
For g1 , g2 ∈ G,
ι(g1 g2 ) = (g1 g2 , e) = (g1 , e)(g2 , e) = ι(g1 )ι(g2 ).
Thus ι is a homomorphism.
ker ι = {g ∈ G | (g, e) = ι(g) = (e, e)} = {e}.
Also
Im ι = {(g, e) | g ∈ G}.
Note that Im ι ≈ G (can you prove it?).
2. Construct a nontrivial (which means that the image is not {e}) homomorphism
from G to G. In each case, show that it is a homomorphism, and compute the
kernel and the image.
There are many possible examples. The example below is just one of them.
1
MATH 3005 Homework Han-Bom Moon
(a) G = Z35 , G = Z5 .
Define φ : Z35 → Z5 as φ(x mod 35) = (x mod 5). First of all, this map
is well-defined because if x1 mod 35 = x2 mod 35, then 35|x1 − x2 and so
5|x1 − x2 and x1 mod 5 = x2 mod 5. This is a homomorphism, because
φ(x+y mod 35) = x+y mod 5 = x mod 5+y mod 5 = φ(x mod 35)+φ(y mod 35).
ker φ = {x ∈ Z35 | x mod 5 = 0} = {x ∈ Z35 | 5|x} = h5i
Im φ = Z5 , because for any x mod 5 ∈ Z5 , φ(x mod 35) = x mod 5.
(b) G = Z5 , G = Z35 .
Define φ : Z5 → Z35 as φ(x mod 5) = 7x mod 35. If x1 mod 5 = x2 mod 5,
5|x1 − x2 and thus 35|7x1 − 7x2 . So 7x1 mod 35 = 7x2 mod 35. Therefore φ
is a well-defined map.
φ(x+y mod 5) = 7(x+y) mod 35 = 7x mod 35+7x mod 35 = φ(x mod 5)+φ(y mod 5).
So φ is a homomorphism.
ker φ = {x ∈ Z5 | 7x mod 35 = φ(x mod 5) = 0 mod 35} = {0}
Im φ = {7x mod 35} = h7i.
(c) G = (R, +), G = GL(2, R).
" #
cos x − sin x
Sol 1. Define φ : R → GL(2, R) as φ(x) = .
sin x cos x
" #
cos(x + y) − sin(x + y)
φ(x + y) =
sin(x + y) cos(x + y)
" #
cos x cos y − sin x sin y − sin x cos y − cos x sin y
=
sin x cos y + cos x sin y cos x cos y − sin x sin y
" # " #
cos x − sin x cos y − sin y
= · = φ(x)φ(y)
sin x cos x sin y cos y
So φ is a homomorphism.
" # " #
cos x − sin x 1 0
ker φ = {x ∈ R | = φ(x) = }
sin x cos x 0 1
= {x ∈ R | cos x = 1} = {2πk | k ∈ Z}
" #
cos x − sin x
Im φ = { | x ∈ R}.
sin x cos x
If you are interested in, show that Im φ = SL(2, R) ∩ O(2).
2
MATH 3005 Homework Han-Bom Moon
It seems that the computation
" is technical,
# but if you recall the geometric
cos x − sin x
meaning of the matrix (which is a counterclockwise rota-
sin x cos x
tion by the angle x), this is very natural.
Here are two more examples. I will skip the computation.
" #
1 x
Sol 2. Define φ : R → GL(2, R) as φ(x) = .
0 1
" #
1 x
ker φ = {0}, Im φ = { }.
0 1
" #
ex 0
Sol 3. Define φ : R → GL(2, R) as φ(x) = .
0 ex
ker φ = {0}, Im φ = {cI | c > 0}.
3. Let G and G be two finite groups and let φ : G → G be a surjective homomor-
phism. Show that if G has an element of order n, then so does G.
Let a ∈ G has order n. Because φ is onto, there is x ∈ G such that φ(x) = a.
Then n = |a| = |φ(x)| | |x|. Therefore |x| = kn for some k ∈ N. Take xk . Then
|xk | = |x|/k = n.
4. By using the first isomorphism theorem, show that Z ⊕ Z/h(1, 1)i ≈ Z.
Let φ : Z ⊕ Z → Z be a map defined by φ(a, b) = a − b. Then for (a1 , b1 ), (a2 , b2 ) ∈
Z ⊕ Z,
φ((a1 , b1 ) + (a2 , b2 )) = φ(a1 + a2 , b1 + b2 ) = a1 + a2 − b1 − b2
= a1 − b1 + a2 − b2 = φ(a1 , b1 ) + φ(a2 , b2 )
so φ is a homomorphism.
For any a ∈ Z, φ(a, 0) = a − 0 = a. So φ is onto.
Also
ker φ = {(a, b) ∈ Z ⊕ Z | a − b = φ(a, b) = 0} = {(a, b) ∈ Z ⊕ Z | a = b}
= {(a, a) ∈ Z ⊕ Z} = h(1, 1)i.
By the first isomorphism theorem, Z ⊕ Z/h(1, 1)i ≈ Z.
5. Let m, n be two relatively prime positive integers. Define a map φ : Z → Zm ⊕ Zn
as φ(a) = (a mod m, a mod n).
3
MATH 3005 Homework Han-Bom Moon
(a) Show that φ is a group homomorphism.
For every a, b ∈ Z,
φ(a+b) = (a+b mod m, a+b mod n) = (a mod m, a mod n)+(b mod m, b mod n)
= φ(a) + φ(b).
Thus φ is a homomorphism.
(b) By using the first isomorphism theorem, show that Zmn ≈ Zm ⊕ Zn .
If a ∈ ker φ, then (a mod m, a mod n) = φ(a) = (0, 0). So a = 0 mod m
and a = 0 mod n. So m|a and n|a. Because gcd(m, n) = 1, this implies
that mn|a. Therefore a ∈ hmni. Conversely, if a ∈ hmni, then a = kmn for
some k ∈ Z. Then φ(a) = φ(kmn) = (kmn mod m, kmn mod n) = (0, 0).
Therefore ker φ = hmni.
Now we show that φ is onto. Because gcd(m, n) = 1, there are x, y ∈ Z
such that xm + yn = 1. Take (a, b) ∈ Zm ⊕ Zn . Then axm + ayn = a and
bxm + byn = b. Let c := bxm + ayn. Then
φ(c) = (c mod m, c mod n) = (bxm + ayn mod m, bxm + ayn mod n)
= (ayn mod m, bxm mod n) = (a mod m, b mod n).
So φ is onto and Im φ = Zm ⊕ Zn .
By the first isomorphism theorem, Zm ⊕ Zn ≈ Z/hmni ≈ Zmn .
You’ve proved a famous theorem in number theory, so called ‘Chinese re-
mainder theorem’. Find wikipedia for the statement of the theorem.
(c) By using (b), show the following result: Show that there is an integer x such
that x ≡ 3 mod 5 and x ≡ 12 mod 17.
We are looking for x ∈ Z such that φ(x) = (3 mod 5, 12 mod 17). By (b), φ
is onto, therefore there must be such x. If you follow the construction in (b),
then you can check that x = 63 is such a number.
6. Prove that there is no nontrivial homomorphism from A5 to Z7 ⊕ Z7 .
For a homomorphism φ : A5 → Z7 ⊕ Z7 , |Im φ| | |A5 | = 60 and |Im φ| | |Z7 ⊕
Z7 | = 49. So |Im φ| is a positive common divisor of 60 and 49. Because 60 and
49 are relatively prime, |Im φ| = 1. Therefore Im φ = {e} and φ is a trivial
homomorphism.
