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Transcript
```Math 417
Exam 2 (Solutions)
Prof. I.Kapovich April 13, 2009
Problem 1.[20 points]
For each of the following statements indicate whether it is true or false. You DO
(1) If G is a group and H / G is a normal subgroup then for every g ∈ G and
every h ∈ H we have gh = hg.
(2) For every n ≥ 2 there exists an onto homomorphism f : Dn → Z2 .
(3) For every n ≥ 2 and every σ ∈ Sn we have σ (10n)! = .
(4) The subset GL(2, R) ⊆ M2 (R) is a subring of M2 (R).
(5) If G is a group and H / G is a normal subgroup then for every g ∈ G the
order of the element g ∈ G is equal to the order of the element gH in the
factor group G/H.
(1) False. The correct conclusion is that gH = Hg for every g ∈ G. However, if
H is a normal subgroup of G, this does not necessarily imply that H is contained
in the center of G. For example, this is not the case for AN / An (for n ≥ 3) and
for SL(n, R) / GL(n, R) (for n ≥ 2).
(2) True. The subgroup H = hai ≤ Dn has order n and index [Dn : An ] =
= 2n
n = 2 in Dn . Therefore H is normal in Dn and the factor group Dn /H
is a cyclic group of order 2, which is therefore isomorphic to Z2 . The quotient
homomorphism Dn → Dn /H is onto, and when composed with an isomorphism
between Dn /H and Z2 , it yields an onto homomorphism Dn /H → Z2 .
|Dn |
|H|
(3) True. Since |Sn | = n!, we have σ n! = for every σ ∈ Sn . However,
(10n)! = 10n · (10n − 1) · · · · · (n + 1) ·n · (n − 1) · · · · · 2 · 1 = m · n!
|
{z
}
m
is divisible by n!. Therefore every σ ∈ Sn we have σ (10n)! = (σ n! )m = .
0 0
(4) False. The zero matrix
does not belong to GL(2, R) and hence
0 0
GL(2, R) is not a subgroup of the additive group of M2 (R).
(5) False. For example, if G is an arbitrary group and H = G then H / G and
G/H is a trivial group, so that every element of G/H has order 1. However, if
|G| > 1, then not every element of G has order 1.
Problem 2.[20 points]
(1) Determine whether the groups Z2 × Z8 and Z16 are isomorphic. Justify
(2) Let R∗ = {x ∈ R : x 6= 0} and R+ = {x ∈ R : x > 0}. Determine whether
the groups (R∗ , ·) and (R+ , ·) are isomorphic. Justify your answer.
Solution.
(1) The groups Z2 × Z8 and Z16 are not isomorphic. Indeed, the group Z16 has
an element of order 16 (namely 1) but Z2 × Z8 has no elements of order 16. To see
the latter, recall that for every a ∈ Z2 , b ∈ Z8 and g = (a, b) ∈ Z2 × Z8 we have
|g| = lcm(|a|, |b|). The possible values of |a| are 1 and 2 and the possible values
1
2
of |b| are 1, 2, 4, 8. Hence the maximal possible value of lcm(|a|, |b|) is 8, so that
Z2 × Z8 indeed has no elements of order 16.
(2) The groups (R∗ , ·) and (R+ , ·) are not isomorphic. The group (R+ , ·) has no
elements if order 2 since if x > 0 then x2 = 1 if and only if x = 1. However, (R∗ , ·)
has an element of order 2, namely the element −1 ∈ R∗ .
Problem 3.[20 points]
(1) Prove that if G is a group and H / G, K / G are normal subgroups in G then
H ∩ K is again a normal subgroup of G.
Show the details of your work.
(2) Give an example of a group G and a normal subgroup H /G such that |H| > 1
and that the group G/H is non-abelian. Justify that your example has the required
properties.
Solution.
(1) Recall first that by the Normality Test (Theorem 3 in Ch 2.8) a subgroup
Y ≤ G is normal in G if and only if for every g ∈ G we have g −1 Y g ⊆ Y , that is,
for every g ∈ G and every y ∈ Y we have g −1 yg ∈ Y .
There was a theorem prooved in class (see Theorem 3 in Ch 2.3) that the intersection of two subgroups is always a subgroup, so that H ∩ K ≤ G.Thus we only
need to verify that the subgroup H ∩ K is normal in G. Let g ∈ G and a ∈ H ∩ K
be arbitrary. We need to show that g −1 ag ∈ H ∩ K.
Since H / G and a ∈ H, we have g −1 ag ∈ H. Since K / G and a ∈ K, we have
−1
g ag ∈ K. Hence a ∈ H ∩ K, and therefore H ∩ K / G by the Normality Test, as
required.
(2) Let K be any nonabelian group (e.g. K = S3 ) and let G = K × K. Consider
H = {(k, 1) ∈ K × K : k ∈ K}. It was verified in one of the homework problems
that H / G and that G/H ∼
= K. Since K was chosen to be nonabelian, this example
has all the required properties.
We recall the argument why G/H ∼
= K. Consider the map f : G → K defined
by f (k, k 0 ) = k for any k, k 0 ∈ K. Then it is easy to check that f is an onto
homomorphism and that Ker(f ) = H. Hence H / G and by the Isomorphism
Theorem we have G/H ∼
= K.
Problem 4.[20 points]
Let R 6= {0} be a ring such that r2 = r for every r ∈ R. Prove that R is
commutative and that char(R) = 2.
Solution.
Note: This was one of the homework problems.
For every r ∈ R we have
r + 1 = (r + 1)2 = r2 + 2r + 1 = r + 2r + 1 = 3r + 1
=⇒
2r = 0.
Thus for every r ∈ R we have 2r = 0. Since R 6= 0, this implies that char(R) = 2.
Note that since char(R) = 2, we have 2r = 0, that is r = −r for every r ∈ R.
Let a, b ∈ R be arbitrary. Then
3
a + b = (a + b)2 = a2 + ab + ba + b2 = a + ab + ba + b =⇒
ab + ba = 0.
Hence ba = −ab for every a, b ∈ R. Since r = −r for every r ∈ R, we have
−ab = ab and hence ba = ab for every a, b ∈ R. Thus R is commutative, as
required.
Problem 5.[20
points]
a b
Let G = {
: a, b, c ∈ R, a 6= 0, c 6= 0}, so that G is a subgroup of GL(2, R)
0 c
and thus is a group
withrespect to matrix multiplication (you do not have to verify
1 b
this). Let K = {
: b ∈ R}. Prove that K is a normal subgroup of G and
0 1
∗
∗
∼
that G/K = R × R , where R∗ = R − {0} is the multiplicative group of nonzero
real numbers.
Solution.
a b
Consider the function f : G → R × R defined as f
= (a, c).
0 c
We claim that f is an onto homomorphism and that Ker(f ) = K.
First, check that f is a homomorphism:
aa1 ab1 + bc1
a b a1 b1
= (aa1 , cc1 ) =
=f
f
0
cc1
0 c 0 c1
a b
a1 b1
(a, c)(a1 , c1 ) = f
f
0 c
0 c1
∗
∗
Thus indeed f is a homomorphism.
x 0
The map f is onto since for any x, y ∈ R we have f
= (x, y).
0 y
The kernel of f is:
∗
a b
a b
∈G:f
= (1, 1) =
Ker(f ) =
0 c
0 c
1 b
a b
∈ G : (a, c) = (1, 1) =
: b ∈ R = K.
0 c
0 1
Thus indeed K = Ker(f ) and hence K / G is a normal subgroup.
We have verified that f : G → R∗ ×R∗ is an onto homomorphism with Ker(f ) =
K. Therefore by the Isomorphism Theorem G/K ∼
= R∗ × R∗ , as required.
```