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# 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.

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