Solution

Math 4281
Final Exam
Page 1 of 9
Name:
Problem
Points
1
10
2
10
3
8
4
6
5
10
6
10
7
10
8
10
Total
74
Score
Instructions
• You may use a scientific calculator and one sheet (two-sided) of handwritten notes. Other notes,
books, electronic devices, and graphing calculators are not allowed.
• If you need more space to solve a problem, use the back of the page, and indicate that you have
done so.
• Read each question carefully, and work the problems in an order that will maximize your score.
• Good luck!
Math 4281
Final Exam
Page 2 of 9
1. Let Q be the following group of order 8: Q = {±1, ±i, ±j, ±k}, where i2 = j 2 = k 2 = −1, ij = k,
and −1 behaves as you’d expect, i.e. (−1)2 = 1 and (−1)x = −x for any x; also −1 commutes with
every group element. This is called the quaternion group.
(a) (2 pt). What are i−1 , j −1 , and k −1 ?
Solution. Multiplying the equation i2 = −1 on both sides by i−1 gives i = (−1)(i−1 ). Now multiply
by −1 on both sides to get −i = i−1 . By the same reasoning, j −1 = −j and k −1 = −k.
(b) (4 pt). Compute ji and kj (hint: what happens when you invert both sides of ij = k?)
Solution. Inverting both sides of ij = k gives (ij)−1 = j −1 i−1 = k −1 . From (a), we know that
j −1 i−1 = (−j)(−i) = (−1)(−1)ji = ji, while k −1 = −k. Thus, ji = −k.
As for kj, start with ij = k and multiply through on the right by j: ij 2 = kj, or kj = i(−1) =
−i.
(c) (4 pt). The subset {1, −1} is a normal subgroup of Q, which you don’t have to prove. Is Q/{1, −1}
cyclic? Justify your answer.
Solution. No, it is not cyclic. The four cosets of H = {1, −1} are
H = {1, −1}
iH = {i, −i}
jH = {j, −j}
kH = {k, −k}.
We have (iH)2 = (jH)2 = (kH)2 = (−1)H = H, since −1 ∈ H. Thus no element of Q/H can have
order 4, meaning Q/H is not cyclic.
Math 4281
Final Exam
Page 3 of 9
2.
(a) (5 pt). Find all solutions x ∈ Z100 to the equation 97x ≡ 4 (mod 100).
Solution. Once we find the multiplicative inverse of 97 mod 100, we will be essentially done, since
then x ≡ 97−1 · 4 (mod 100). This inverse is guaranteed to exist because gcd(97, 100) = 1, and we
find it with the Euclidean algorithm applied to 97 and 100:
100 = 1 · 97 + 3
97 = 32 · 3 + 1
3=3·1
Recall that the goal here is to write 1 in the form 100a + 97b for integers a, b. We do this by going
back up the list of equations produced by the Euclidean algorithm as follows:
1 = 1 · 97 − 32 · 3
= 1 · 97 − 32 · (100 − 1 · 97)
= 33 · 97 − 32 · 100.
But if 33 · 97 − 32 · 100 = 1, then 33 · 97 ≡ 1 (mod 100). In other words, 97−1 ≡ 33 (mod 100).
Now
x ≡ 97−1 · 4 ≡ 33 · 4 ≡ 132 ≡ 32 (mod 100).
(And indeed, 97 · 34 = 3104 ≡ 4 (mod 100).)
(b) (5 pt). Let g(x) denote the coset g(x) + (x3 + 4x2 + x + 3) in the ring Z7 [x]/(x3 + 4x2 + x + 3).
Express
x2 + 4 · x2 + 3
in the form ax2 + bx + c, where a, b, c ∈ Z7 .
Solution. We have x2 + 4 · x2 + 3 = x4 + 5. To write this in the form ax2 + bx + c, we need to
reduce modulo x3 + 4x2 + x + 3—in other words, divide x4 + 5 by x3 + 4x2 + x + 3. Doing this gives
a quotient of x + 3 and a remainder of x2 + x + 3, so x4 + 5 = x2 + x + 3.
Math 4281
Final Exam
Page 4 of 9
3. Let G be the group {id, (1 2), (3 4), (1 2)(3 4)}, a subgroup of S4 . Let X be the set of binary strings
of length 4, so X = {0000, 0001, 0010, . . . , 1111} has 16 elements. The group G acts on X by permuting
entries; for example, (1 2) swaps the first two entries in a string:
id ·0110 = 0110,
(1 2) · 0110 = 1010,
(3 4) · 0110 = 0101,
(1 2)(3 4) · 0110 = 1001.
(a) (4 pt). Write down all the orbits of the action of G on X.
Solution. Any two orbits are disjoint or equal, so we can do this as follows. Choose an element
x ∈ X, and write down its orbit (i.e. apply every group element to x). Then choose an element in
X that wasn’t in the first orbit, and write down its orbit. Continue like this until you’ve exhausted
X. Here are the orbits:
{0000}
{0001, 0010}
{0011}
{0100, 1000}
{0101, 1001, 0110, 1010}
{0111, 1011}
{1100}
{1101, 1110}
{1111}
(b) (4 pt). Use Burnside’s theorem to verify that you have the right number of orbits in part (a).
Solution.
• The identity element of G fixes all 16 strings in X.
• (1 2) fixes any string in X starting with 00 or 11, and there are 23 = 8 of these.
• Likewise, (3 4) fixes 8 strings.
• (1 2)(3 4) fixes the 4 strings 0000, 0011, 1100, and 1111.
Burnside’s theorem then says that there are 41 (16 + 8 + 8 + 4) = 9 orbits, as we saw in (a).
Math 4281
Final Exam
Page 5 of 9
4. (6 pt). Prove that a group homomorphism is injective if and only if its kernel is {e}.
Proof. Let φ : G → H be a group homomorphism. Suppose φ is injective. If x ∈ ker φ, then we have
φ(x) = e = φ(e). Since φ is injective, this means x = e. Since x was arbitrary, this means ker φ = {e}.
Conversely, suppose ker φ = {e}, and that φ(x) = φ(y). Then
e = φ(x)φ(y)−1 = φ(x)φ(y −1 ) = φ(xy −1 ),
so xy −1 ∈ ker φ = {e}. Thus, xy −1 = e, so x = y. This shows that φ is injective.
Math 4281
Final Exam
Page 6 of 9
5. We can view symmetries of a regular polygon as permutations of the vertices. For instance, clockwise
rotation of a square by 90◦
1
2
4
1
4
3
3
2
replaces 1 by 4, 2 by 1, 3 by 2, and 4 by 3, so we could view this symmetry as the permutation 4123
(in one-line notation) or (1 4 3 2) (in cycle notation).
(a) (6 pt). For each symmetry in D4 , write down the corresponding permutation as described above,
in one-line notation and in cycle notation.
Solution. I’ll assume that our square is labeled like the square on the left above. Then:
• The identity symmetry is 1234 = ( ), the identity permutation.
• Rotation by 90◦ clockwise is 4123 = (1 4 3 2), as in the example.
• Rotation by 180◦ is 3412 = (1 3)(2 4).
• Rotation by 270◦ is 2341 = (1 2 3 4).
• Reflection across the usual vertical line is 2143 = (1 2)(3 4).
• Reflection across the usual horizontal line is 4321 = (1 4)(2 3).
• Reflection across the line through 1 and 3 is (2 4).
• Reflection across the line through 2 and 4 is (1 3).
(b) (4 pt). Compute the sign of each permutation from part (a).
Solution. In the order listed above, the signs are 1, −1, 1, −1, 1, 1, −1, −1. Here we use the facts
that sgn(στ ) = sgn(σ) sgn(τ ) and that the sign of a k-cycle is 1 if k is odd, −1 if k is even. For
example, (1 3) has sign −1, and (1 3)(2 4) has sign (−1)(−1) = 1.
Math 4281
Final Exam
Page 7 of 9
6.
(a) (6 pt). Decide which of the following subsets S of a ring R is a subring, an ideal, both, or neither.
You don’t have to justify your choices. Abbreviations: I = “ideal but not a subring”, SR = “subring
but not an ideal”, B = “both a subring and an ideal”, N = “neither a subring nor an ideal”.
(i) R = Z, S = all multiples of 3:
I
(ii) R = Z, S = all prime numbers:
(iii) R = R, S = Q:
I
SR
SR
I
B
SR
B
B
N
N
N
(iv) R = all functions R → R, S is the set of 2π-periodic functions, i.e. those f ∈ R such that
f (x + 2π) = f (x) for all x:
I SR B N
(v) R = all functions R → R, S is the set of functions f ∈ R such that f (x) = 0 for all x ∈ [0, 2π]:
I SR B N
Solution. There’s a bit of a trick question here: every ideal is a subring, so “I” is never an answer.
(i) B; in any ring, the set of multiples of a fixed element forms an ideal (keyword: “principal
ideal”)
(ii) N; the set of primes isn’t closed under addition or multiplication
(iii) SR; Q is certainly a ring using
operations of R, so it’s a subring, but it isn’t an ideal.
√ the usual√
/ Q. (Any ideal containing 1 must be the whole
For instance, 1 ∈ Q and 2 ∈ R, but 2 ∈
ring!)
(iv) SR; The sum, difference, or product of two 2π-periodic functions is still 2π-periodic, and 0 is
2π-periodic, so this is a subring. It’s not an ideal, because if f is 2π-periodic and g is some
strange function, there’s no reason for f (x)g(x) to be periodic.
(v) B; If f is zero on [0, 2π] and g is any function, then f (x)g(x) is still zero on [0, 2π].
(b) (4 pt). Decide whether each of the following functions is a group homomorphism. Again, you don’t
have to justify your choices.
(i) φ : Z × Z → Z, φ((x, y)) = x + y:
yes
(ii) φ : Z × Z → Z, φ((x, y)) = xy:
(iii) φ : S5 → Z, φ(σ) = σ(1):
(iv) φ : (Z[x], +) → Z, φ(f (x)) = f (1):
yes
yes
no
no
no
yes
no
Solution.
• Yes: φ((x, y) + (a, b)) = φ((x + a, y + b)) = (x + a) + (y + b) is equal to φ((x, y)) + φ((a, b)) =
(x + y) + (a + b).
• No: for instance, φ((1, 1) + (1, 1)) = φ((2, 2)) = 4 is not equal to φ((1, 1)) + φ((1, 1)) = 2.
• Not even close: this doesn’t even map the identity permutation 12345 to the identity 0 ∈ Z
(it maps it to 1), so it can’t be a homomorphism.
• Yes: φ(f (x) + g(x)) = φ((f + g)(x)) = (f + g)(1) is equal to φ(f (x)) + φ(g(x)) = f (1) + g(1).
Math 4281
Final Exam
Page 8 of 9
7.
(a) (4 pt). Is Z3 [x]/(x2 − x + 1) a field? What about Z3 [x]/(x2 + x − 1)? Justify your answers.
Solution. Asking whether Z3 [x]/(f (x)) is a field is equivalent to asking whether f (x) is irreducible.
Since f (x) in both cases has degree ≤ 3, this is equivalent to asking whether f (x) has any roots in
Z3 . Setting x = 0, 1, 2 will show that x2 − x + 1 has x = 2 as a root, and that x2 + x − 1 doesn’t
have any roots.
(b) (2 pt). For part (b) and part (c), let R be one of the rings from part (a), your choice (the answers
to (b) and (c) are the same either way, and neither choice makes the questions harder). What is
the characteristic of R?
Solution. Regardless of whether R is a field, it still has a multiplicative identity 1 = 1+(your choice of f ),
and the characteristic of R is the smallest positive integer n such that n · 1 = 0. But 3 = 0 in Z3 [x],
so we still have 3 = 0 in R. Thus the characteristic of R is 3. (Well, at most 3. What we have to
rule out is that 1 = 0 or 2 = 0; equivalently, 1 ∈ (f ) or 2 ∈ (f ). But f has degree 2, so (f ) doesn’t
contain any constants.)
(c) (4 pt). Since (R, +) is an abelian group of order 9, it must be isomorphic to either Z9 or Z3 × Z3 .
Which one, and why?
Solution. Because R has characteristic 3 by (b), we have 3r = 0 for all r ∈ R. This means (R, +)
doesn’t contain an element of order 9, so it can’t be cyclic, i.e. can’t be isomorphic to Z9 .
Math 4281
Final Exam
Page 9 of 9
8. Let G and H be two groups, and let π : G × H → H be the function defined by π((g, h)) = h.
(a) (4 pt). Show that π is a homomorphism.
Proof. Take g1 , g2 ∈ G and h1 , h2 ∈ H. Then
π((g1 , h1 )(g2 , h2 )) = π((g1 g2 , h1 h2 )) = h1 h2 = π((g1 , h1 ))π((g2 , h2 )).
(b) (4 pt). Show that ker(π) is the subset G × {eH } = {(g, eH ) : g ∈ G} of G × H. Show that
im(π) = H.
Proof. We have (g, h) ∈ ker(π) if and only if eH = π((g, h)) = h , if and only if (g, h) ∈ G × {eH }.
For the image, suppose h ∈ H. Then h ∈ im(π) because, for instance, π((eG , h)) = h.
(c) (2 pt). Conclude from (b) that (G × H)/(G × {eH }) ' H.
Proof. By the first isomorphism theorem, (G × H)/ ker(π) ' im(π). Putting in the results of (b)
gives (G × H)/(G × {eH }) ' H.