# Algebra Final

• May 15th 2008, 07:59 PM
matt90
Algebra Final
I need some help with these review problems I have for an upcoming final.

How many homomorphisms are there of Z onto Z?
How many homomorphisms are there of Z into Z?
How many homomorphisms are there of Z into Z_2?

How would you go about approaching this type of problem?

Are the following groups isomorphic?
Z_2 X Z_12 and Z_4 X Z_6

I think these are because the lcm(2,12)=lcm(4,6). Would this be a right assumption?

Z_8 X Z_10 X Z_24 and Z_4 X Z_10 X Z_40
I think these are because the lcm(8,10,24)=lcm(4,12,40)
• May 15th 2008, 09:15 PM
Jhevon
Quote:

Originally Posted by matt90
I need some help with these review problems I have for an upcoming final.

How many homomorphisms are there of Z onto Z?

There are only 2.

define $\displaystyle \theta_1(x) = x$ for $\displaystyle x \in \mathbb{Z}$
and $\displaystyle \theta_2 (x) = -x$ for $\displaystyle x \in \mathbb{Z}$

check that these are homomorphisms

also, they will be the only ones. since any homomorphism has to be of the form $\displaystyle \theta (x) = nx$ for some integer n. this only works with n = +/- 1, otherwise the map would not be onto.

Quote:

How many homomorphisms are there of Z into Z?
from directly above, we can choose n to be any integer here, so there are an infinite amount.

Quote:

How many homomorphisms are there of Z into Z_2?
there are 2.

$\displaystyle \theta (x) = [0]_2$ for all $\displaystyle x \in \mathbb{Z}$

or define a mapping based on $\displaystyle \theta (1) = [1]_2$, in which case $\displaystyle \theta (x) = \left \{ \begin{array}{lr} \text{[0]} & \mbox{ if } x \mbox{ is even} \\ & \\ \text{[1]} & \mbox{ if } x \mbox{ is odd} \end{array} \right.$

Quote:

Are the following groups isomorphic?
Z_2 X Z_12 and Z_4 X Z_6

I think these are because the lcm(2,12)=lcm(4,6). Would this be a right assumption?
Z_2 x Z_12 = Z_2 x Z_3 x Z_4

and Z_4 x Z_6 = Z_4 x Z_2 x Z_3

these are the same thing, just rearranged. thus they are isomorphic

Quote:

Z_8 X Z_10 X Z_24 and Z_4 X Z_10 X Z_40
I think these are because the lcm(8,10,24)=lcm(4,12,40)
the orders are different. thus they are not isomorphic
• May 15th 2008, 09:26 PM
Isomorphism
Quote:

Originally Posted by Jhevon
There are only 2.

define $\displaystyle \theta_1(x) = x$ for $\displaystyle x \in \mathbb{Z}$
and $\displaystyle \theta_2 (x) = -x$ for $\displaystyle x \in \mathbb{Z}$

check that these are homomorphisms

also, they will be the only ones. since any homomorphism has to be of the form $\displaystyle \theta (x) = nx$ for some integer n. this only works with n = +/- 1, otherwise the map would not be onto.

from directly above, we can choose n to be any integer here, so there are an infinite amount.

there are 2.

What is the justification for this form $\displaystyle \theta (x) = nx$ being the only one?
• May 15th 2008, 09:32 PM
matt90
yeah i don't understand that part
• May 15th 2008, 09:34 PM
matt90
oh i made a mistake on the last part it should be Z_12

Z_8 X Z_10 X Z_24 and Z_4 X Z_12 X Z_40
I think these are because the lcm(8,10,24)=lcm(4,12,40)
• May 15th 2008, 09:35 PM
Jhevon
Quote:

Originally Posted by Isomorphism
What is the justification for this form $\displaystyle \theta (x) = nx$ being the only one?

1 is a generator of Z, if i can figure out where it goes, i can figure out where anything goes.

so i start with some homomorphism $\displaystyle f$, for which $\displaystyle f(1) = n$ .........$\displaystyle n$ is just some integer.

then, $\displaystyle f(x) = f( \underbrace{1 + 1 + 1 ... + 1}_{\text{x times}}) = f(1) + f(1) + \cdots f(1) = nx$

is something wrong with that reasoning?
• May 15th 2008, 10:27 PM
Isomorphism
Quote:

Originally Posted by Jhevon
$\displaystyle f(x) = f( \underbrace{1 + 1 + 1 ... + 1}_{\text{x times}}) = f(1) + f(1) + \cdots f(1) = nx$

is something wrong with that reasoning?

Nothing wrong, of course :D

I could not figure out the reason, so I asked.

Thanks :)
• May 16th 2008, 12:39 PM
ThePerfectHacker
Quote:

Originally Posted by Isomorphism
Nothing wrong, of course :D

I could not figure out the reason, so I asked.

Thanks :)

In general if a group is generated by $\displaystyle \left< a_1, a_2, ... , a_k\right>$. Then any homomorphism can be determined by $\displaystyle \phi(a_1), ... ,\phi(a_k)$.