Sample Midterm
Math 402, Summer 2004
1) (7=5+2) Let Rx denote the group of non-zero real numbers under multiplication. Let P denote the subset of positive real numbers.
a) Prove that P is a subgroup of Rx.
Proof: We must check that P is
closed under the group operation and inverses, and that it contains the identity.
Alternatively, we can use the homework problem from the first homework and just
check that if x, y are in P, then xy-1 is
also in P. If y is a positive real, then y-1=1/y is also a positive
real. If x is a positive real, then the product xy-1 must be in P. Therefore P is a subgroup.
b) Describe the cosets of P in Rx, as sets.
The set P of positive reals itself is the trivial coset.
If x is any real in Rx but not in
P, it is a negative real, so xP =-P= the set of
negative reals. These are the only cosets since -P contains all the elements in Rx but not in P.
2) (7=4+3) Let H= {1, -1, i, -i, j, -j, k, -k} be the quartenion group.
a) List all its proper subgroups. Show that there are no others.
There is one subgroup of order
2: {1, -1}=<-1>
All other elements of H have order
4, so they generate subgroups of order 4. There are 3 such subgroups: <i>={1,-1, i, -i}, <j>={1, -1, j, -j}, and <k>={1, -1, k, -k}.
Any other subgroup would have order
larger than 4, so it would be equal to the whole group (since it must divide
|H|=8).
b) Prove that the subgroup {1, -1, i, -i} is normal.
Let S denote this subgroup.
1 and -1 commute with all the
elements in H. Since i is in H and H is closed as a
subgroup, i S i-1=S. So it suffices to
check the normality condition for j and k:
jSj-1=jS(-j)=-jSj= {-j1 j, -j(-1) j, -ji j , -j(-i)j}={1, -1, k j=i, kj=-i}=S.
kSk=kS(-k)=-kSk={-kk, -k(-1)k, -kik, -k(-i)k}={1, -1, -(-j)k=jk=i, kik=-jk=-i}=S.
Hence xSx-1=S for all
x in H, so S is normal.
3) (6) Let G and H be two groups and f: GàH be a group isomomorphism. Show that for any element x in G, the order of x equals the order of f(x).
Suppose the order of x is n. Then
n is the smallest integer for which xn=1. By
the homomorphism property, f(xn)=fn(x)=1.
Therefore the order of f(x) divides n. But if the order of f(x) were to be some
number m smaller than n, then fm(x)=f(xm)= 1, so xm
would be in the kernel but not the
identity. Hence f would not be injective.
This shows that the order of the
image of x mut be precisely n.
4) (5) Prove that the direct product of two infinite cyclic groups is not infinite cyclic.
Done in class in a fair amount of detail.
5) (5) Prove that in any group the orders of ab and ba are equal for any elements a and b.
Suppose the order of (ab) is n. Then multiplying n times: (ab)(ab)…(ab)=1.
Use associativity to get a(ba)(ba)…(ba)b=1,
with n-1 copies of ba in the middle. Multiply by a-1
on the left and by a on the right to get (ba)n=1. Hence
the order of (ba) divides the order of (ab). But the
argument is symmetric, so the order of ab must also divide
the order of ba. Thus they must be equal.
6) (10, 2 each) All or nothing, no proof:
a) Give an example of a group G with at least 4 elements that has no proper subgroups (i.e. no subgroups other than {1} and itself).
Any
cyclic group of prime order would make a good example.
b) Let f :G-> G’ be a surjective group homomorphism. Suppose |G|=153 and |G’|=51. What is |Ker f|?
|Ker f|=3, by the first isomorphism thm
and the counting formula.
c) What is the automorphism group of Z/5Z?
Cyclic group
of order 4
d) Give an example of an element of order 6 in the symmetric group S5
(123)(45)
e) Give an example of a subgroup H and a group G such that H is not normal in G.
In Dn={1, x, x2,
…, xn-1, y, xy, x2y, …, xn-1y},
the subgroup {1,y}