Download Exam 1 – 02/29/12 SOLUTIONS

Exam 1 – 02/29/12
SOLUTIONS
Math 403
Instructor: T. Haines
NOTE: [k] means the problem is worth k points. Total: 100 points.
1. Consider the group (Z/24Z, +)
(a) [5] What is the order of this group?
24. The elements are the equivalence classes [0], [1], . . . , [23].
(b) [5] Does this group have any subgroups of order 7? Why or why not?
No. If it did, then Lagrange’s theorem would imply that 7 divides 24, contradiction.
(c) [5] Find all elements of order 6 (list them).
Note that Z/24Z is a cyclic group, and [1] is a generator. Recall from class that the
24
element [i], which can be written as [1]i , is of order (i,24)
. This order is 6 iff (i, 24) = 4. The
numbers i with this property are 4 and 20. Thus the two elements of order 6 are [4], [20].
(d) [5] Show that the map [x] 7→ [5x] from Z/24Z to itself is well-defined.
We need to show that [x] = [y] ⇒ [5x] = [5y]. But [x] = [y] iff 24|(x − y), which implies
24|5(x − y) and 24|(5x − 5y), and this implies [5x] = [5y].
(e) [5] Show the map [x] 7→ [5x] from Z/24Z to itself is surjective.
Any injective map from a finite set to itself is automatically surjective (Pigeonhole
principle). Thus, it is enough to check the map is injective, i.e that [5x] = [5y] ⇒ [x] = [y].
Suppose [5x] = [5y]. Then 24|5(x − y). Since 5 and 24 are coprime, this implies 24|(x − y),
and thus [x] = [y].
2. Consider the group U24 = ((Z/24Z)× , ·).
(a) [5] Find the order of U24 by listing all of its elements explicitly.
The group consists of the elements [i] ∈ Z/24Z such that (i, 24) = 1. Thus the group is
of order 8, consisting of the elements [1], [5], [7], [11], [13], [17], [19], [23].
(b) [10] Determine the exponent of U24 .
The exponent of a group G is the smallest positive number n such that g n = e for
all g ∈ G. For each of the 8 elements [i] ∈ U24 , one checks by direct computation that
i2 ≡ 1 mod 24. This shows the exponent is 2.
3. Consider the dihedral group D5 of order 10.
(a) [7] List all the elements of order 5.
Say D5 = ha, b | |a| = 2, |b| = 5, aba−1 = b−1 i. Then the elements of order 5 are the
powers of b which are not e, that is the 4 elements
b, b2 , b3 , b4 .
(b) [8] List all the elements of order 2. These are a, ab, ab2 , ab3 , ab4 . There are five of them.
(c) [5] Knowing nothing else, can you conclude from (b) that D5 is not cyclic? Explain.
Yes. Recall from class that in a cyclic group of order n, for each divisor d of n, there are
exactly φ(d) elements of order d. Thus, there is exactly one element of order 2 in a cyclic
group of order 10. From (b) we see that D5 has more than one element of order 2, hence it
cannot be cyclic.
4. [15] Consider the set of matrices H = {A ∈ M2 (Q) | det(A) = 3n for some n ∈ Z}.
Prove or disprove that H is a subgroup of GL2 (Q). You are allowed to use the standard
facts about determinants we mentioned in class.
It is a subgroup. Since 3n 6= 0 for all n, we see A ∈ H ⇒ det(A) 6= 0, and so H is a
subset of GL2 (Q).
Now det(I2 ) = 1 = 30 , so I2 ∈ H.
If A, B ∈ H, we have det(A) = 3n and det(B) = 3m for some integers m, n. Then
AB −1 ∈ H since det(AB −1 ) = det(A)det(B −1 ) = det(A)(det(B))−1 = 3n · 3−m = 3n−m .
This shows that H is a subgroup.
5. [15] If G is a finite group, and H ⊂ G is a non-empty subset which is closed under
multiplication in G, prove that H is a subgroup of G.
It is necessary to prove that e ∈ H (where e ∈ G is the identity in G) and that if h ∈ H,
then h−1 ∈ H, where h−1 is the inverse of h taken in G.
Take any h ∈ H (this exists since H 6= ∅). Then since G is finite, there exists i > j
with hi = hj . Then using cancellation in the group G, we see hi−j = e. Since H is closed,
hi−j ∈ H. Thus e ∈ H.
Now for any given h, the formula hi−j−1 h = hhi−j−1 = e shows that h−1 = hi−j−1 . The
latter is a non-negative power of h, so by closure of H (if i − j − 1 > 0) or by e ∈ H( if
i − j − 1 = 0), we see h−1 ∈ H.
6. (a) [5] Give, with explanation, an example of a nonabelian group of order 24.
The dihedral group D12 has order 24 and is nonabelian since ab = b−1 a and b 6= b−1 .
(b) [5] Give, with explanation, an example of an abelian group which is NOT cyclic.
The group Z/2Z × Z/2Z is abelian or order 4, and has exponent 2, hence has no element
of order 4 and is not cyclic.
Another example is U24 above: it is abelian, and of exponent 2, so has no element of
order 8 (= |U24 |), hence is not cyclic.