HW2 Solutions
Section 16
13.) Let G be the additive group of real numbers. Let the action of θ ∈ G in the real
plane R2 be given by rotating the plane counterclockwise about the origin through θ
radians. Let P be a point other than the origin in the plane.
(a.) Show that R2 is a G-set.
Note that elements of our set are points other than the origin in the plane R2 (the
origin is fixed). To show that R2 is a G-set, we need to show that the action
defined above satisfy the two criteria for a group action.
• The additive identity is 0 ∈ R = G. Rotating by 0 radians means that the
plane stays fixed. In particular every point in the plane is fixed and so
0 ∗ P = P.
• Now let θ1 , θ2 ∈ G. We then have to show that (θ1 + θ2 ) ∗ P = θ1 ∗ (θ2 ∗ P ).
Note that acting on P by θ1 + θ2 is exactly the same as first rotating by θ2
and then by θ1 . And so the equation holds.
(b.) Describe geometrically the orbit containing P .
Let d be the distance from the point P to the origin. Then the orbit of P is the
circle of radius d centered at the origin.
(c.) Find the group GP .
The group GP is the subgroup of G = R that fixes the point P . Note that
rotating by integer multiples of 2π rotates the plane back to its original position.
Therefore GP = h2πi ≤ R.
Extra Problem: Let G act on itself by conjugation.
(a.) Show that every element in the same conjugacy class has the same order.
Recall that the order of an element is the smallest positive integer n ∈ Z+ such
that an = e where e ∈ G is the identity element. Let a ∈ G such that |a| = n for
some positive integer n ∈ Z+ . Also let b ∈ G be in the conjugacy class of a. That
means that there exists g ∈ G such that b = gag −1 . We then consider
bn = (gag −1 )n = (gag −1 )(gag −1 ) · · · (gag −1 )
|
{z
}
−1
n of these
−1
= ga(g g)a(g g) · · · (g −1 g)ag −1
= gan g −1 = geg −1 = e.
But we’re not done because this just shows that the order of b is ≤ n. Assume
that there exists m ∈ Z+ with m < n such that bm = e. Since b = gag −1 we can
solve for a and get a = g −1 bg. Applying the exact same argument as above, we
would get that am = e. But this contradicts the assumption that the order of a is
n (remember, n is the smallest such integer such that an = e). So we indeed have
|b| = n.
(b.) Determine the conjugacy classes of the symmetric group on 3 letters.
I’m going to use cycle notation here
S3 = {(1), (1 2), (2 3), (1 3), (1 2 3), (1 3 2)}.
You can use the book to figure out to what ρ’s and µ’s these elements correspond.
By part (a) we can cut down our work a little bit since we now know that only
elements of the same order might be in the same conjugacy class. So at the very
least, we know that the identity is in its own conjugacy class.
There are only two elements of order 3, namely (1 2 3) and (1 3 2). So we think
these guys might be in the same conjugacy class. So we just have to find an
element σ ∈ S3 such that σ(1 2 3)σ −1 = (1 3 2). Note that all the cycles of length
2 are self-inverse.
(1 2)(1 3 2)(1 2)−1 = (1 2)(1 3 2)(1 2) = (1 2)(2 3) = (1 2 3)
So (1 2 3) and (1 3 2) are indeed in the same conjugacy class.
Now we wonder if all the order 2 elements are in the same conjugacy class. So we
just guess and check.
(2 3)(1 2)(2 3) = (2 3)(1 2 3) = (1 3)
And so we know that (1 3) is in the same conjugacy class as (1 2). Similarly we
have that
(1 3)(1 2)(1 3) = (1 3)(1 3 2) = (2 3)
and so (2 3) is also in the same conjugacy class as (1 2) as well.
Therefore there are three conjugacy classes
{e}, {(1 2), (2 3), (1 2)}, {(1 2 3), (1 3 2)}.
Section 18
20.) Consider the matrix ring M2 (Z2 .
(a.) Find the order of the ring, that is, the number of elements in it.
Note that there are only two elements in Z2 = {0, 1}. In a 2 × 2 matrix there are
4 entries and for each entry there are two possiblities (either 0 or 1). And so
|M2 (Z2 )| = 24 = 16.
(b.) List all units in the ring.
Hopefully you recall from matrix algebra that a matrix has an inverse if and only
if it has nonzero determinant. But you only know this when the entries of your
matrix are real numbers. It turns out to be true in this case since Z2 is a field but
since we haven’t shown that, we can’t use that fact. So we will use brute force.
We first
note
that the identity element in this ring is indeed the identity matrix
1 0
I=
which is its own inverse and therefore a unit.
0 1
Suppose that the matrix has a row of 0’s,
0 0
c d
0
0
=
a b
e f
ac + be ad ∗ bf
which can never be the identity matrix no matter what a, b, c, d, e, f are. We can
make similar arguments for if the second row or either column consists entirely of
0’s. Therefore the only things left to check are the elements are
0 1
0 1
1 0
1 1
1 1
,
,
,
, and
.
1 0
1 1
1 1
0 1
1 0
So let’s take one of these guys and multiply it by an arbitrary matrix and set it
equal to the identity matrix I. Suppose we have that
1 0
0 1
a b
c
d
=
=
.
0 1
1 1
c d
a+c b+d
For this equation to be true we must have that
c = 1and d= 0 giving
us that
0 1
1 1
a = 1 and b = 1. Thefore we know that both
and
are units
1 1
1 0
since they are inverses of each other. We can make similar arguments for the
other 3 matrices in question and will find that they are all self-inverse.
38.) Show that a2 − b2 = (a + b)(a − b) for all a and b in a ring R if and only if R is
commutative.
Remember that a ring R being commutative means that the multiplication is
commutative. Let a, b ∈ R be arbitrary and let’s take the right-hand side of the
equation in question and expand it out. By the distributive laws we have
(a + b)(a − b) = a(a − b) + b(a − b) = a2 − ab + ba − b2 .
Furthermore we have that
a2 − ab + ba − b2 = a2 − b2
if and only if −ab + ba = 0 which is true if and only if ba = ab. Note that this is exactly
what it means for R to be commutative since our choice of a and b were arbitrary.
Since all of the statements above were “if and only if” statements, we are done.
44.) An element a of a ring R is idempotent if a2 = a.
(a.) Show that the set of all idempotent elements of a commutative ring is closed
under multiplication.
Let a, b ∈ R be idempotents, that is a2 = a, b2 = b. Then to show that the set of
all idempotents is closed under multiplication, we just need to show that ab is also
idempotent, that is (ab)2 = ab. Note that (ab)2 = (ab)(ab) = a(ba)b. Since R is
commutative, we have that ba = ab and so
(ab)2 = (ab)(ab) = a(ba)b = a(ab)b = a2 b2 = ab
since a and b are idempotents by assumption.
(b.) Find all idempotents in the ring Z6 × Z12 .
Note that an element (a, b) ∈ Z6 × Z12 is idempotent if and only if (a, b)2 = (a, b). Note
that (a, b)2 = (a, b)(a, b) = (a2 , b2 ) and so (a, b) is idempotent if and only if a2 = a in Z6
and b2 = b in Z12 . We proceed by just checking the squares of all elements in these
rings.
In Z6 , we have:
02 = 0
12 = 1
22 = 4
32 = 3
42 = 4
.
52 = 1
In Z12 , we have:
02 = 0
12 = 1
22 = 4
32 = 9
42 = 4
52 = 1
62 = 0 92 = 9
72 = 1 102 = 4 .
82 = 4 112 = 1
Therefore, the idempotents of the ring Z6 × Z12 are the sixteen ordered pairs in the set
{0, 1, 3, 4} × {0, 1, 4, 9}.
48.) Show that a subset S of a ring R gives a subring of R if and only if the following
hold:
i.) 0 ∈ S;
ii.) (a − b) ∈ S for all a, b ∈ S;
iii.) ab ∈ S for all a, b ∈ S.
Note that this is an if and only if statement so we have to show the implication both
ways. Let us first assume that S is a subring of R. Therefore (i) and (iii) automatically
hold. If b ∈ S then we must have −b ∈ S since S is a subgroup under addition.
Therefore (a − b) = a + (−b) ∈ S for all a, b ∈ S, verifying criteria (ii).
Now let us assume that (i), (ii), and (iii) hold. We will then show that S is a subring.
Note that we have that the associativity of addition and multiplication are inherited
from R. We now show that S is a subgroup under addition. Since 0 ∈ S, then for all
a ∈ S we have that 0 − a = −a ∈ S, that is S is closed under additive inverses.
Therefore we also have that for all a, b ∈ S, a + b = a − (−b) ∈ S again by criterion (ii).
And so we indeed have that S is a subgroup under addition. All that’s left to show
then is that S is closed under mulitplication, but that is exacly criterion (iii).