Download MA 294 Midterm 1

MA 294 Midterm 1
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• The exam consists of 4 questions on 4 pages, plus this cover sheet. If you discover that you
are missing any pages, please notify the instructor right away.
Question
Points
1
35
2
35
3
15
4
15
Total:
100
Score
MA 294 Midterm 1
Page 1 of 4
1. For any positive integer n, define the set P to be the set of all quadratic polynomials with real
coefficients; that is,
P = {ax2 + bx + c | a, b, c ∈ R}.
(a) (10 points) Prove that P is a group under the operation of addition.
Solution: We have closure under addition because
(ax2 + bx + c) + (dx2 + ex + f ) = (a + d)x2 + (b + e)x + (c + f ) ∈ P.
The identity is 0 (i.e., 0x2 + 0x + 0), the inverse of ax2 + bx + c is −ax2 − bx − c, and
the associativity of addition of polynomials comes from the associativity of addition of
real numbers.
(b) (5 points) Is P under addition an abelian group? Prove your assertion.
Solution: Yes, because (ax2 + bx + c) + (dx2 + ex + f ) = (dx2 + ex + f ) + (ax2 + bx + c)
by the commutativity of multiplication of the real numbers.
MA 294 Midterm 1
Page 2 of 4
(c) (10 points) Is P a group under the operation of multiplication? Prove your assertion.
Solution: No, because (ax2 + bx + c)(dx2 + ex + f ) is a polynomial of degree 4, and
so is not an element of P .
(d) (10 points) Define Q to be the set of all linear polynomials with real coefficients; that is,
Q = {bx + c | b, c ∈ R}. Is Q a subgroup of P under addition? Prove your assertion.
Solution: Yes. Q is certainly a subset of P (it’s the subset where a = 0). S1 is verified
because (bx + c) + (ex + f ) = (b + e)x + (c + f ) ∈ Q, and S2 is verified because the
inverse of bx + c is −bx − c ∈ Q.
MA 294 Midterm 1
Page 3 of 4
2. Let G = {1, 3, 5, 7}, and define a multiplication operator that works just like multiplication in
Z8 ; for example, we’ll say that 5 · 7 = 3 in G since 5 × 7 = 35 and 35 ≡ 3 (mod 8).
(a) (15 points) Write out the multiplication table for G and use it to prove that G is a group.
1 3 5 7
1
3
5
7
Solution: The multiplication table is:
1
3
5
7
1
1
3
5
7
3
3
1
7
5
5
5
7
1
3
7
7
5
3
1
We have closure since each entry in the multiplication table is also an element of G.
The identity is 1, and x2 = 1 for each x ∈ G, making each element its own inverse.
Associativity of multiplication comes from the associativity of multiplication in Z8 .
(b) (8 points) Find the order of each element of G.
Solution: From the multiplication table, we can see that 3, 5, and 7 each have order
2 (since their squares are equal to 1) and of course 1 has order 1.
(c) (12 points) In class we showed that every group of order 4 is isomorphic to C4 or C2 × C2 .
Which of these is G isomorphic to? Prove your assertion.
Solution: G is isomorphic to C2 × C2 under the isomorphism φ : G → C2 × C2 given
by:
φ(1) = (1, 1),
φ(3) = (x, x),
φ(5) = (1, x),
φ(7) = (x, 1)
You could check that this is indeed an isomorphism (by showing it is bijective and a
homomorphism). Alternatively, you can observe that G has no element of order 4, but
that C4 does have an element of order 4 (its generating element), so it is not possible
for G to be isomorphic to C4 , and G must be isomorphic to C2 × C2 .
MA 294 Midterm 1
Page 4 of 4
3. (15 points) Prove that the group A × B is isomorphic to the group B × A. [Remember that
the group operator on A × B is defined by (a1 , b1 ) · (a2 , b2 ) = (a1 a2 , b1 b2 ).]
Solution: We will show that φ : A × B → B × A given by φ((a, b)) = (b, a) is an isomorphism.
It is one-to-one: if φ((a, b)) = φ((a0 , b0 )), then (b, a) = (b0 , a0 ), and we would have a = a0 and
b = b0 so (a, b) = (a0 , b0 ).
It is onto: given any (b, a) ∈ B × A, simply reverse the coordinates and φ maps (a, b) to
(b, a).
It is a homomorphism, since
φ((a, b))φ((a0 , b0 )) = (b, a)(b0 , a0 ) = (bb0 , aa0 ) = φ((aa0 , bb0 )) = φ((a, b)(a0 , b0 )).
Therefore, since φ is a homomorphism, one-to-one, and onto, it must be an isomorphism,
and A × B is isomorphic to B × A.
4. (15 points) Is it possible for there to exist a group G that has fewer than 24 elements but has
subgroups of orders 6 and 8? Justify your answer.
Solution: By Lagrange’s Theorem, we must have that 6 | |G| and 8 | |G|, so |G| must be
at least as large as the least common multiple of 6 and 8, which is 24.