# 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 a^{n} = 0 and b^{m} = 0, then (a + b)^{nm} = 0.

•
For a ∈ I, -a ∈ I: If a^{n} = 0, then (-a)^{n} = -a^{n} = 0.

•
For a ∈ I and r∈ R: ra ∈ I: If a^{n} = 0, then (ra)=r^{n}a^{n} = 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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