MA 294 Midterm 2
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• You have 1 hour to complete this exam.
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• If you run out of space, feel free to use the back of any page for extra work, but make sure to
clearly indicate on the front of that page that you will be writing on the back.
• The exam consists of 4 questions on 5 pages, plus this cover sheet. If you discover that you
are missing any pages, please notify the instructor right away.
Question
Points
1
24
2
12
3
32
4
32
Total:
100
Score
MA 294 Midterm 2
Page 1 of 5
1. (a) (8 points) Let G = {1, (12), (34), (12)(34)} and let X = {1, 2, 3, 4}. Compute the orbit
and stabilizer for each element of X.
Solution: G1 = G2 = {1, 2}, G3 = G4 = {3, 4}, G1 = G2 = {1, (34)}, and G3 =
G4 = {1, (12)}.
(b) (16 points) Now let G be a group of permutations that acts on a finite set X (not necessarily the examples from part a!), and suppose that there are exactly two orbits of the
same size. Prove that for every x ∈ X there is exactly one nonidentity permutation that
fixes x.
Solution: The wording of this question on the midterm was confusing and omitted a
key assumption: that |G| = |X|. (As a result, everyone was given full credit on this
problem.) If there are exactly two orbits of the same size, then |Gx| = |X| /2 = |G| /2
for each x ∈ X. Therefore, using the orbit-stabilizer equation, we get that |G| =
|Gx | · |Gx| = |Gx | · |G| /2, so |Gx | = 2. Since 1 is always in the stabilizer, there must
be exactly one nonidentity permutation in Gx no matter what x is.
MA 294 Midterm 2
Page 2 of 5
2. (12 points) Recall that Sn is the group of all permutations on {1, 2, . . . , n}. Show that if
σ, τ ∈ Sn , then σ −1 τ −1 στ is always an even permutation.
Solution: Suppose that σ decomposes into k transpositions and τ decomposes into l transpositions. Then σ −1 τ −1 στ decomposes into k + l + k + l = 2(k + l) transpositions, so it is
an even permutation. Several people said that σ −1 τ −1 στ = τ −1 σ −1 στ = 1, but this is not
true in general because Sn is a non-abelian group. (Only disjoint cycles commute!)
MA 294 Midterm 2
Page 3 of 5
3. (32 points) How many “really different” ways are there to color the sides of a square using the
colors red, green, and blue? Here, “really different” means that one configuration cannot be
rotated and/or flipped into another. For example, here are three ways to color the sides of the
square:
A. Each of the four sides is colored red.
B. The top side is colored red, the bottom side is colored green, and the left and right sides
are both colored blue.
C. The top side is colored green, the bottom side is colored red, and the left and right sides
are both colored blue.
Note that B and C are not “really different” since C is just B rotated by 180◦ , but that A is
“really different” from both B and C.
Solution: The permutation group here is equivalent to D8 .
Permutation type
Identity
Rotation
Rotation 2x
Rotation 3x
Flip (horizontal axis)
Flip (vertical axis)
Flip (SW-NE diagonal)
Flip (NW-SE diagonal)
Configurations Notes
fixed
81
Each of the 34 = 81 possible configurations
3
The 3 configurations where each side is the
same color
9
The top and bottom must be the same
color, and the left and right must be the
same color
3
The 3 configurations where each side is the
same color
27
The left and right can be anything, but top
and bottom must be the same
27
The top and bottom can be anything, but
left and right must be the same
9
The left and bottom must be the same, and
the top and right must be the same
9
The top and left must be the same, and
the bottom and right must be the same
Since |G| = |D8 | = 8, the total number of really different colorings is
81 + 3 + 9 + 3 + 27 + 27 + 9 + 9
= 21.
8
MA 294 Midterm 2
Page 4 of 5
4. Let R be a ring. Recall that an element u ∈ R is a unit if it has a multiplicative inverse.
(a) (8 points) Prove that (−1)2 = 1.
Solution: First note that 0 = (1 + (−1))(−1) = 1(−1) + (−1)2 . By ring property B,
1(−1) = −(1 · 1) = −1. Therefore, (−1)2 is the additive inverse of −1, or 1.
(b) (8 points) Prove that if u is a unit in R then so is −u.
Solution: Since u is a unit, there must be a u−1 ∈ R such that uu−1 = 1. We’ll show
that −u−1 is the multiplicative inverse of u. We have that (−u)(−u−1 ) = uu−1 = 1 by
ring property C; the reverse product comes from the same argument. Therefore −u is
a unit with multiplicative inverse −u−1 .
MA 294 Midterm 2
Page 5 of 5
(c) (8 points) Prove that 0 is not a unit.
Solution: By ring property A, 0a = 0 for all a ∈ R, so there is no a ∈ R that would
make 0a = 1. Therefore, there is no multiplicative inverse for 0.
(d) (8 points) Suppose that a, b ∈ R and that the multiplicative inverse of 1 − ab is x. Show
that the multiplicative inverse of 1 − ba is 1 + bxa.
Solution: Since the multiplicative inverse of 1 − ab is x, we know that (1 − ab)x =
x(1−ab) = 1. Now all we have to show is that (1−ba)(1+bxa) = (1+bxa)(1−ba) = 1.
We can compute
(1 − ba)(1 + bxa) = 1 − ba − bxa − babxa
= 1 − ba + b(x − abx)a
= 1 − ba + bx(1 − ab)a
= 1 − ba + ba
= 1.
(1)
(2)
(3)
(4)
(5)
We use the distributive property to get (1), we factor out a b on the left and an a on
the right to get (2) (remember that multiplication is not necessarily commutative in R
so we have to be careful about multiplication on the left vs. right), we factor out an
x to get (3), and then in (4) we make use of the fact that x(1 − ab) = 1 since x and
1 − ab are multiplicative inverses. The reverse direction is essentially the same since
addition is commutative:
(1 + bxa)(1 − ba) = 1 − ba − bxa − babxa
= 1 − ba + b(x − abx)a
= 1 − ba + bx(1 − ab)a
= 1 − ba + ba
= 1.