English | Español

Try our Free Online Math Solver!

Online Math Solver

 

 

 

 

 

 

 

 
 
 
 
 
 
 
 
 

 

 

 
 
 
 
 
 
 
 
 

Please use this form if you would like
to have this math solver on your website,
free of charge.


Solutions to Algebra I Problem Set 5

1. We need to show that if G is a finite abelian group , and if f : G → K is a surjective
homomorphism to an abelian group K , then G has a subgroup which is isomorphic
to K. (If we let K = G/H we get the desired result).

Since G is a finite abelian group , there are integers each a power of a
prime
number such that



By putting the summand which corresponds to the same prime number, we can
assume that



for a prime number p and integers . Then we can write K as



such that . To prove the statement, it is enough to show that
for every . Assume on the contrary that this is not true, and assume
that l is the largest index for which . For , let be the element
whose summands are all zero expect a 1 at the i-th place (0,... , 1, ... , 0), and for
, let be the element whose summands are all zero except a 1 at the
i-th place.

For , let be such that . Then let . Then
are in the subgroup of G generated by , and so the same is true
for their images : They are in the subgroup generated by l-1 elements
. This is a contradiction since none of the are
zero .

2. Assume that H is a finite subgroup of Q/Z. First note that if where
gcd(a, b) = 1, then : since gcd(a, b) = 1, there are integers x and y such
that ax + by = 1, so . Now let Let now

Then H is the cyclic group generated by : if , then so is .
Let now a = gcd(d, ). Then there are integers x and y such that . So
. But , and since , by our
choice of , lcm (, d) should be equal to , so d is a divisor of , and . So
H is cyclic and its order is .
Therefore, the only subgroup of Q/Z is the subgroup generated by .

3. Assume R is commutative and let I be the set of nilpotent elements. To show that
R is an ideal, we need to show that

• For a, b ∈ I, a + b ∈ I: If an = 0 and bm = 0, then (a + b)nm = 0.
• For a ∈ I, -a ∈ I: If an = 0, then (-a)n = -an = 0.
• For a ∈ I and r∈ R: ra ∈ I: If an = 0, then (ra)=rnan = 0.

If R is not commutative, then I is not necessarily an ideal: Let R the ring of 2
by 2 matrices with real entries . Then and are nilpotent, but
which is not nilpotent as and .

4.

(i) In this case, . The isomorphism is give by
, and


(ii) The isomorphism is given by


(1 3 4), and .

5. (b) Assume G is not cyclic. It is enough to consider the case where H is of the form
or where m and n are not relatively prime. In the third case,
define Ø: G → G by , and Ø(0, 1) = (0, 1). Define : G →G
by . Then and Ø extend to homomorphisms of G.
And we have and . Hence End(G) is not
commutative.

In the other two cases , similar Ø and work.

Prev Next