Download Math 400 Spring 2016 – Test 3 (Take

Math 400 Spring 2016 – Test 3 (Take-Home Solutions)
1. (8) Suppose G is a group, H is a subgroup of G, and K is a subgroup of H. (Thus we know
K is a subgroup of G as well.) If K C H and H C G is it necessarily true that K C G? If
so, prove it. If not, give a counterexample (and show that it is, in fact, a counterexample.)
Solution: It’s false. The smallest (lowest order) counterexample is G = D4 , H = {R0 , R180 , v, h},
K = {R0 , v}. (Prove each subgroup is normal in the one larger by noting index 2; prove
K 6C G by noting dvd 6∈ K.
2. (8) Let G be the quaternion group (also known as Q8 ) that you worked with on Tests 1 and
2; I’m including the Cayley table below. Let B be the subgroup hbi of G. Show that B C G
and that G/B ≈ Z2 . (Note: In fact, Q8 is the smallest example of a non-Abelian group with
the property that all of its subgroups are normal. You don’t need to prove this.)
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Solution: Note hbi has order 4, and therefore |Q8 : hbi| = 2, so B = hbi is normal in Q8 .
One way to prove the isomorphism is using the first iso theorem, but it’s quicker to do the
following: |Q8 /B| = 2 so this factor group has order 2. Thus every subgroup has order 1
(trivial) or 2 (whole thing), so by a problem on the first test the factor group is cyclic. Every
cyclic group of order n is isomorphic to Zn , so this factor group is isomorphic to zz2 .
r 0 3. (12) Let G be the group GL2 (R). Let H =
r ∈ R∗ .
0 1 (a) Prove that H ≤ G.
r 0
0 1
s 0
0 1
Solution: Let A =
,B =
be arbitrary elements of H. Then AB =
rs 0
1/r 0
−1
∈ H since rs is a nonzero real number. Also, A =
∈ H.
0 1
0 1
Therefore H is closed under the group operation (matrix multiplication) and inverses,
so H ≤ G.
(b) (corrected) Prove that H 6B G.
r 0
a b
Solution: Let A =
∈ H and X =
∈ G. Then the bottom-left entry
0 1
c d
cdr − cd
, and for any r 6= 1 this value is not 0, so XAX −1 6∈ H. Therefore
of XAX −1 is
ad − bc
H 6C G.
4. (8) Suppose φ : G → H is a group homomorphism, |G| = 120, and |H| = 30. Find all possible
values for |Ker φ|, and for each of these possible values of |Ker φ|, find the associated value
of |φ(G)|.
Solution: From the first isomorphism theorem we have G/(ker φ) ≈ φ(G), so that |G| =
| ker φ||φ(G)|. We also know that φ(G) ≤ H, so that |φ(G)| divides |H|. Since |G| = 120
and |H| = 30, this leads to the following list of possibilities:
| ker φ| |φ(G)|
120
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2
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40
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10
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8
4
30
5. (12) Show that it is impossible to have a one-to-one homomorphism from a group of order
40 to a group of order 60. Then generalize – what is a requirement on group orders for a
one-to-one homomorphism to be possible? Prove your generalization.
Solution: The generalization is: If φ : G → H is a one-to-one homomorphism of finite
groups, then |G| divides |H|.
Proof: If φ is one-to-one, we have |G| = |φ(G)| (from set theory, or use 1st iso thm). But
we know for any homomorphism of finite groups that the order of the image, |φ(G)|, divides
the order of the codomain, |H|. This proves the statment.
6. (6) Find all isomorphism classes of Abelian groups of order 1500. Remember to justify your
answer!
Solution: We have 1500 = 22 · 3 · 53 . By the Fundamental Theorem of Finite Abelian
Groups, any Abelian group of order 1500 is isomorphic to a direct product of Zn , where n
is a prime power, and the product of the n’s is 1500. Thus the isomorphism classes are:
Z4 ⊕ Z3 ⊕ Z125
Z4 ⊕ Z3 ⊕ Z5 ⊕ Z25
Z4 ⊕ Z3 ⊕ Z5 ⊕ Z5 ⊕ Z5
Z2 ⊕ Z2 ⊕ Z3 ⊕ Z125
Z2 ⊕ Z2 ⊕ Z3 ⊕ Z5 ⊕ Z25
Z2 ⊕ Z2 ⊕ Z3 ⊕ Z5 ⊕ Z5 ⊕ Z5
7. (6) Suppose G is an Abelian group of order 32. In computing its elements, you come across
an element of order 16 and two different elements of order 2. Explain why no further work
is necessary to determine the isomorphism class of G, and determine the isomorphism class.
Solution: By the Fund. Thm. of Fin. Ab. Gps, G is isomorphic to one of: Z32 , Z16 ⊕
Z2 , Z8 ⊕ Z4 , Z8 ⊕ Z2 ⊕ Z2 , Z4 ⊕ Z4 ⊕ Z2 , Z4 ⊕ Z2 ⊕ Z2 ⊕ Z2 , and (Z2 )(5 times). Since G has an
element of order 16, the only possibilities are Z32 and Z16 ⊕ Z2 . However, the cyclic group
Z32 only has one element of order 2, namely 16. Therefore G ≈ Z16 ⊕ Z2 .