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Chapter 3 Review: Solved problems

1. (a) What are the possibilities for the order of an element of Z13×? Explain your answer.

Solution: The group Z13× has order 12, and by Corollary 3.2.11 the order of any element must be a divisor of 12, so the possible orders are
1, 2, 3, 4, 6, and 12.

(b) Show that Z13× is a cyclic group.

Solution: The first element to try is [2], and computing its powers gives us
22 = 4,
23 = 8,
24 = 16 3,
25 = 2 · 24 2 · 3 6, and
26 = 2 · 25 2 · 6 12,
so the order of [2] is greater than 6. By part (a) it must be 12, and thus [2] is a generator for Z13×. We could also write this as
Z13× = < [2]13 >.

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2. Find all subgroups of Z11×, and give the lattice diagram which shows the inclusions between them.

Solution: We first check for cyclic subgroups. For the powers of [2] we get
[2]2=[4],
[2]3=[8],
[2]4=[5],
[2]5=[10],
[2]6=[9],
[2]7=[7],
[2]8=[3],
[2]9=[6],
[2]10=[1].
This shows that Z11× is cyclic, and so by Proposition 3.5.3 the subgroups are as follows:
< [2] > = Z11×;
< [2]2 > = { [1], [2]2, [2]4, [2]6, [2]8 } = { [1],[4],[5],[9],[3] };
< [2]5 > = { [1], [2]5 } = { [1],[10] } and
{ [1] }.
The lattice diagram forms a diamond.

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3. Let G be the subgroup of GL3 (R) consisting of all matrices of the form such that a and b are in R.
Show that G is a subgroup of GL3 (R).

Solution: We have

,

so the closure property holds. The identity matrix belongs to the set, and

,

so the set is closed under taking inverses. (Click here for more details about finding the inverse.)

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4. Show that the group G in Problem 3 is isomorphic to the direct product R × R.

Solution: Define : G -> R × R by

= (a, b).

This is one-to-one and onto because it has an inverse function
: R × R -> G defined by

((a,b)) =

Finally, preserves the respective operations since


    =

    = (a+c, b+d) = (a,b) + (c,d)

    = +

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5. List the cosets of the cyclic subgroup < 9 > in Z20×. Is Z20× / < 9 > cyclic?

Solution: Z20× = { ± 1, ± 3, ± 7, ± 9 }.
Here are the cosets of < 9 > :

< 9 > = { 1,9 }
(-1) < 9 > = { -1, -9 }
3 < 9 > = { 3, 7 }
(-3) < 9 > = { -3, -7}

Since x2 is in < 9 >, for each element x of Z20×, the factor group is not cyclic since it has no element of order 4.

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6. Let G be the subgroup of GL2 (R) consisting of all matrices of the form , and let N be the subset of all matrices of the form .

(a) Show that N is a subgroup of G, and that N is normal in G.

Solution: The set N is nonempty since it contains the identity matrix, and Corollary 3.2.3 shows it to be a subgroup since

.

N is normal in G since

.

(Click here for more details about finding these inverses.)

(b) Show that G / N is isomorphic to the multiplicative group R×.

Solution: Define : G -> R× by = m .
Then we have

.

Since m can be any nonzero real number, maps G onto R×, and
= 1
if and only if m = 1, so N = ker (). The fundamental homomorphism theorem implies that G/N R×.

Comment: The proof of part (b) also proves part (a), since once you have determined the kernel, you know that it is always a normal subgroup. Thus parts (a) and (b) can be proved at the same time, using the argument given for part (b).

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7. Assume that the dihedral group D4 is given as
{ e,a,a2,a3,b,ab,a2b,a3b },
where a4 = e, b2 = e, and b a = a3 b.
Let N be the subgroup < a2 > = { e, a2 }.

(a) Show by a direct computation that N is a normal subgroup of D4.

Solution: We have ai a2 a-i = a2 and

(ai b) a2 (ai b)-1
= ai a-2 b ai b
= ai a-2 a-i b2
= a-2 = a2,

for all i, which implies that N is normal.

(b) Is the factor group D4 / N a cyclic group?

Solution: The cosets of N are

N = { e, a2 },
Na = { a, a3 },
Nb = { b, a2 b }, and
Nab = { ab, a3 b }.

Since b and ab have order 2, and a2 in N, we see that each element in the factor group has order 2, so G/N is not cyclic since it has no element of order 4.

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8. Let G = D8, and let N = { e,a2,a4,a6 }.

(a) List all left cosets and all right cosets of N, and verify that N is a normal subgroup of G.

Solution: The right cosets of N are

N = { e,a2,a4,a6 },
Na = { a, a3, a5, a7 },
Nb = { b, a2b, a4b, a6b },
Nab = { ab, a3b, a5b, a7b }.

The left cosets of N are more trouble to compute, but we get

N = { e,a2,a4,a6 },
aN = { a, a3, a5, a7 },
bN = { b, a6b, a4b, a2b },
abN = { ab, a7b, a5b, a3b }.

The fact that the left and right cosets of N coincide shows that N is normal.

(b) Show that G/N has order 4, but is not cyclic.

Solution: It is clear that there are 4 cosets. We have

(Na)(Na)=Na2=N,
(Nb)(Nb)=Ne=N, and
(Nab)(Nab)=Ne=N,

so each coset has order 2.


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Inverting an n × n matrix

In most linear algebra courses, one general method for deciding whether or not an n × n matrix A has an inverse is to put A and the n × n identity In into a n × 2n matrix and row-reduce. If the rows of the matrix are independent, so that we end up with the identity matrix in the first n columns, then A is invertible, and the right hand columns give the inverse. The following notation is often used.

If the rows of the matrix are not independent, so that we end up with at least one row of zeros when we finish the row-reduction, then the matrix A is not invertible.

Inverting a 2 × 2 matrix

For a 2 × 2 matrix A = , there is an easier way to find the inverse (provided it exists). The first step is to write down the adjoint of A, given by adj(A) = . We have the following matrix equations.

= = (ad-bc)

  and  

= = (ad-bc) .

If ad - bc is nonzero, then we can divide through by ad - bc to get the inverse.

Be careful to first check that the determinant ad - bc is nonzero.

Inverting an upper triangular 2 × 2 matrix

An upper triantular 2 × 2 matrix A = has the following inverse, found by using the general formula for the inverse of a 2 × 2 matrix.

, provided a and c are nonzero.

As a particular examples, we have

, for any a, and

, provided m is nonzero.

Inverting a block upper triangular matrix

Suppose that an n × n matrix can be put into a block form

, where A is a k × k matrix, C is an (n-k) × (n-k) matrix, and B is a k × (n-k) matrix.

If A and C are invertible matrices, then

  and  

,

showing that .

For example, we can write the matrix in the block form , where
A = , B = , and C = 1.
Then the inverse of A is , and so

- A-1 B C-1 = - · 1 = ,

and therefore , as in Problem 3.

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