NAME:___SOLUTIONS_________
MIDTERM
Math 402, Summer 2004
Please READ each
question CAREFULLY and ask for clarification if necessary. Write proofs in clear, full
sentences. Any claim you make must be justified, except in problem 5. If you
use a result from the text, mentioned in class, or proven in the homework,
clearly state what result you are using.
There are 6 questions
worth 50 points. Good luck!
1.(8pts) Let
G be an abelian group. Prove that the subset H={g Є
G | g2=1} is a subgroup.
We need to verify three conditions (associativity is inherited):
1) Closure under multiplication: Given any two elements x and y in H, their product satisfies (xy)2=x2y2=1G hence xy is also in H.
The first equality holds because G is abelian so xy=yx, while the latter equality holds because x and y are in H so x2=y2=1.
2) Closure under inverses: If x is in H then x2=1, so the inverse element of x2 is also 1. That is, 1= (x2)-1=(x-1)2. Hence x-1 is also in H.
3) Identity: Since 1 Є G satisfies 12=1, 1 Є H
This proves that H is a subgroup of G
REMARK: The number 2
here was purely arbitrary. Problem generalizes to:
In any abelian group, and for any positive
integer n,
the set of points
whose order divides n forms a subgroup
2. (7pts.) Prove that if f: G→H is a group
homomorphism, it must map the identity 1G of G to the identity 1H
of H. (hint: 1G=1G1G)
This is proved in the text.
Let y = f(1G) in H and notice that using the properties of the identity and the fact that f is a homomorphism: y=f(1G) =f(1G1G)=f(1G)f(1G)=yy. Multiplying by y-1 on both sides, y=1H.
3.(10=5+5)
a) Find all the
proper subgroups of G=Z/12Z. For each subgroup, list its elements and
generators. Indicate how you know that there are no other subgroups.
The elements of Z/12Z are {0,1,2,3,4,5,6,7,8,9,10,11} & the operation is addition modulo 12.
The elements {1, 5, 7, 11} have order 12 and generate the whole group. Any subgroup that contains them must be the whole group.
6 has order 2. It generates <6>={0,6} a subgroup of order 2.
4 has order 3. It generates the subgroup <4>={0, 4, 8} =<8> of order 3.
3 has order 4. It generates the subgroup <3>={0, 3, 6, 9}=<9> of order 4 .
2 has order 6. It generates the subgroup <2>={0, 2, 4, 6, 8, 10}=<10> of order 6.
Since we have accounted for all possible elements of G, we have listed all the cyclic subgroups of G. There are no other because all the subgroups of a cyclic group must be cyclic.
(alternatively: any other subgroup must contain at least 2 relatively prime numbers. These would generate the whole group so the subgroup would not be proper).
b) Prove that Z/12Z
is isomorphic to a direct product of two (non-trivial) groups.
Z/12Z ≈ Z/3Z x Z/4Z
Method I:
The element (1,1) in Z/3Z x Z/4Z has order equal to the least common multiple of the orders of 1 in Z/3Z and 1 in Z/4Z, i.e lcm(3,4)=12. This was part of the worksheet and it’s also in your text.
Therefore, (1,1) must generate a subgroup of order 12 in Z/3ZxZ/4Z. Since the order of Z/3ZxZ/4Z is 3x4=12, (1,1) must generate the entire group, i.e. Z/3ZxZ/4Z is cyclic of order 12. Hence it is isomorphic to Z/12Z.
Method II:
Theorem in book (and done in class) showed that Z/nZxZ/mZ ≈ Z/mnZ if and only if m and n are relatively prime numbers. Since 3 and 4 are relatively prime, Z/3Z x Z/4Z ≈ Z/12Z .
4. (10pts., 4+3+3)
a) Suppose that G is a group, and x and y
are two elements of G satisfying xy=yx-1.
Suppose that y has order
2. Prove that (yx)2=1.
(yx)2=y(xy)x=y(yx-1)x= (yy)(x-1x)=(1)(1)=1. I used the relation between x and y in the second equality and the fact that y has order 2 in the fourth.
b) The dihedral group Dn
is generated by two elements x and y, where x has order n>2, y has order 2
and xy=yx-1. Let H be the cyclic subgroup
generated by x. Show that all elements in the left coset
yH have order 2.
The left coset yH={y, yx, yx2, …, yxn-1}. We have shown above that (yx)2=1, and we know that y has order 2. For 1<i<n, a similar argument can be used to get:
(yxi)2 =(yxi) (yxi)=yxi-1(xy)xi=yxi-1(yx-1)xi=…=yyx-ixi=1.
Hence every element in the coset satisfies (yxi)2=1.
Remark: I did not take points off this
time, but one should really explain why the elements have order 2 and not 1:
(yxi)2=1 means that yxi has order 2 or 1. But the only element of order 1 is the identity. The identity is not in the coset yH (because it is in the group H and, since y is not in H, cosets H and yH are disjoint.)
c) List all elements of D4 in
terms of x and y. List which ones have order 2.
D4={1, x, x2, x3, y, yx, yx2, yx3}=H (disjoint union) yH.
H is cyclic of order 4 so it has only one element of order 2, namely x2.
Hence the elements of order 2 in the dihedral group D4 are: x2, y, yx, yx2, yx3
5. (10pts, 2 each)
Credit/No credit. Answer the following questions with no justification necessary:
a) Let G be a group and H, K two subgroups. What condition ensures that |HK|=|H| |K|?
H ∩K ={1}. (from the solution to homework pbl 2.8.9, for example)
b) Compute the conjugate element: (123) (12) (123)-1 in S3.
=(123)(12)(132)=(123)(13)=(23)
c) How many group homomorphisms can you have from Z/5Z to Z/3Z?
just one, the trivial one that maps everything to zero.
(homework problem 2.7.1)
d) Write a 2 x 2 matrix corresponding to a rotation by 90o about the origin of the plane.
What is its order?
| cos 90 -sin90|, i.e. |0 -1| It has order 4
| sin 90 cos 90| |1 0|
(There is also the clockwise rotation matrix
|0 1|
|-1 0|)
e) How many non-isomorphic abelian groups of order 27 are there?
3 (namely Z/27Z, Z/3Z x Z/9Z and Z/3Z x Z/3Z x Z/3Z)
(Classification Theorem for Fin. Gen. Abelian groups: 9=3x3x3=3x9, neither relatively prime)
6.(5pts) Let N be a normal subgroup of index 5
in a group G. Prove that g5 Є N, for all elements g of G.
Since N is normal of index 5, the quotient set G/N is a group of order 5.
Let g be an element of G. Then the order of the coset gN must divide the order of the group G/N, so the coset satisfies (gN)5=N=1G/N in the quotient group G/N.
Hence g5N=N for all g in G. But this is equivalent to g5 Є N.
REMARK: There was nothing special
about 5 here. Pick your favorite index and generalize.