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
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
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
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
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 .
(i) In this case, . The isomorphism is give by
(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
In the other two cases , similar Ø and work.