Download Homework 7 - Material from Chapters 6-7

Math 400
SOLUTIONS
Homework 7 - Material from Chapters 6-7
1. Show that U (8) is not isomorphic to U (10).
Solution: There are many ways to do this, but the easiest is probably to note that
U (10) is cyclic, but U (8) is not.
2. Show that S4 is not isomorphic to D12 .
Solution: Again, there are many solutions. Note that D12 contains R30 , a rotation
by 30 degrees. This is an element of order 12. However, it is quick to see (by looking
at cycle structures) that there is no element of order 12 in S4 , so the groups are not
isomorphic (even though they’re both nonabelian and have order 24).
1 n 3. Consider the group of matrices
n ∈ Z under matrix multiplication. (I’m
0 1 telling you it’s a group – you don’t need to prove it.) What familiar group is this
isomorphic to? Explain why you chose your answer, but you don’t need a full proof of
the isomorphism.
1 n
1 m
1 n+m
Solution: Note that
=
, so multiplying such matrices
0 1
0 1
0
1
is basically the same as adding their upper right entries. Thus this group under matrix
multiplication should be isomorphic to Z (under addition.)
4. Suppose ϕ is an isomorphism from some group G to Z, a ∈ G, and ϕ(a) = 4. What is
ϕ(a3 )?
Solution: Using the operation-preserving property of ϕ (or the theorem giving many
properties of isomorphisms), and remembering that the operation of Z is addition, we
have ϕ(a3 ) = ϕ(a) + ϕ(a) + ϕ(a) = 4 + 4 + 4 = 12.
5. Show that Z under addition is NOT isomorphic to Q under addition.
Solution: There are lots of ways to do this; the easiest to note is that Z is cyclic (since
h1i = Z ) but Q is not (as proved on a previous homework).
6. Suppose ϕ is an isomorphism from D4 to itself such that ϕ(R90 ) = R270 and ϕ(V ) = V .
Determine ϕ(D) and ϕ(H).
Solution: (I’m using the Cayley table for D4 on page 31 in the book; if you use one
with a different convention (rotations CW vs. CCW, for example) things might be
slightly different in your computations, but the answers should be the same.)
Note D = R90 V , so ϕ(D) = ϕ(R90 )ϕV = R270 V = D0 . Now, note that H = R90 D, so
ϕ(H) = ϕ(R90 )ϕ(D) = R270 D0 = H.
7. Inner automorphisms:
(a) Let G be a group and x ∈ G a fixed element. Define a function ϕx : G → G by
ϕx (a) = xax−1 for all a ∈ G. Prove that ϕx is an isomorphism. (It is called the
inner automorphism of G induced by x.)
Math 400
SOLUTIONS
Solution: First, note that if ϕx (a) = ϕx (b), then xax−1 = xbx−1 so that cancellation gives a = b. Thus ϕx is one-to-one.
Now, given any b ∈ G, we have x−1 bx is also in G, and ϕx (x−1 bx) = x(x−1 bx)x−1 =
b. Since b was arbitrary, ϕx is onto.
Finally, let a, b ∈ G. Then ϕx (ab) = xabx−1 = xaebx−1 = xax−1 xbx−1 =
ϕx (a)ϕx (b), so ϕx is O-P. Thus ϕx : G → G is an isomorphism.
(b) Under what condition(s) is φx the trivial isomorphism (that is, the identity function)?
Solution: Well, ϕx is the trivial isomorphism if ϕx (a) = a for all a ∈ G. This is
equivalent to xax−1 = a, i.e. xa = ax, for all a ∈ G. This statement is equivalent
to x ∈ Z(G) (the center of G).
Thus, ϕx is the trivial isomorphism of G if and only if x ∈ Z(G).
8. Let H be the subgroup of A4 given by H = {ε, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. (You
don’t need to prove that H is a subgroup.) Find all the left cosets of H in A4 .
Solution: We must find all sets σH, where σ ∈ A4 . (The Cayley table on page 107
may be helpful, although you can do it by hand.) We have
εH = H
(1 2 3)H = {(1 2 3), (1 3 4), (2 4 3), (1 4 2)}
(1 3 2)H = {(1 3 2), (2 3 4), (1 2 4), (1 4 3)}
Since we’ve now covered all the elements of A4 , these are all the cosets of H in A4 .
9. Let H be the same as in the previous problem. How many left cosets of H are there
in S4 ? (Do this without finding the cosets.)
Solution: Since |S4 | = 24 and |H| = 4, the number of left cosets is |S4 : H| = 24/4 =
6.
10. Let H be all the multiples of 3, which is a subgroup of Z. Find all the left cosets of H
in Z.
Solution: Remember that the operation here is addition, so instead of “aH” we’ll
have a + H, for elements a ∈ Z. We have:
0 + H = H = Multiples of 3
1 + H = 1 + Multiples of 3 = integers which are equal to 1 (mod 3)
2 + H = 2 + multiples of 3 = integers which are equal to 2 (mod 3)
Since every integer is in one of these sets (because of the division algorithm, though
you don’t need to note that ), these are all the cosets of H in Z.
11. Suppose a has order 12. Find all the left cosets of ha4 i in hai.
Solution: The cosets are
eH = H = {e, a4 , a1 2}
Math 400
SOLUTIONS
aH = {a, a5 , a9 }
a2 H = {a2 , a6 , a1 0}
a3 H = {a3 , a7 , a1 1}.
12. Suppose G is a group of order 145, and a, b are nonidentity elements of G with different
orders. Prove that the only subgroup of G that contains both a and b is G itself.
Solution: Note that the divisors of |G| are 1, 5, 29, 145. Thus these are the only
possible orders of subgroups of G (by Lagrange’s theorem).
Since the order of an element is equal to the order of the cyclic subgroup it generates,
this means a and b each have order equal to 5, 29, or 145.
If either |a| or |b| = 145, then any subgroup containing that element also must contain
all its powers (due to closure), so must have order 145 and thus be all of G.
Therefore the only case left to consider is that where a, b have orders 5 and 29 (and
without loss of generality, we may assume |a| = 5.) Now let H ≤ G be a subgroup
containing a and b. The same argument as above tells us that |a| and |b| be divisors
of |H|, which is itself a divisor of 145. The only possibility for |H| such that |H| is
divisible by both 5 and 29 is |H| = 145. So H contains all the elements in G, so H = G
as required.
13. Let G be a group of order 84. What are the possible orders for the subgroups of G?
Solution: By Lagrange’s theorem, the possible orders are the positive divisors of 84:
1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84.
14. Suppose K is a proper subgroup of H and H is a proper subgroup of G. If |K| = 42
and |G| = 420, what are the possible orders of H?
Solution: Using Lagrange’s theorem, we have 42 is a divisor of |H| and |H| is a
divisor of 420. The integers satisfying this are 42, 84, 210, 420. The “proper” subgroup
requirement eliminates 42 and 420, so that |H| = 84 or 210.
15. Suppose H, K are subgroups of a group G with relatively prime orders. Prove that
H ∩ K = {e}.
Solution: We’ve seen on a previous homework that H ∩ K is a subgroup of G. Thus
it’s also a subgroup of H and a subgroup of K. Thus Lagrange’s theorem tells us that
|H ∩ K| is a divisor of both |H| and |K| (in other words, a common divisor.) Since
these orders are relatively prime, |H ∩ K| = 1 and thus H ∩ K is the trivial group.
16. Suppose H, K are subgroups of a finite group G, with H ⊆ K ⊆ G. Prove that
|G : H| = |G : K| |K : H|.
Solution: Since G is finite, we have |G : H| =
other two indexes in the problem. Thus we have
|G : K| |K : H| =
|G|
,
|H|
and similar expressions for the
|G| |K|
|G|
=
= |G : H|
|K| |H|
|H|
Math 400
SOLUTIONS
17. Suppose a group contains elements of orders 1 through 9. What is the smallest possible
order of the group?
Solution: Since the order of an element is the order of its cyclic subgroup, and orders
of subgroups must be divisors of the order of the group, such a group’s order must
be divisible by each of the integers 1 through 9. The smallest such positive integer is
9 · 8 · 7 · 5 = 2520, so this is the smallest possible order for the group.