-Hmomorphism @:Z_2->Z_4 &Z_2->Z_6

[sutupidmath]

Urgent!-Hmomorphism @:Z_2-->Z_4 &Z_2-->Z_6

Homework Statement

THis is probbably very easy, just i am kinda bogged down:

(a) Show that the mapping $$\theta:Z_2-->Z_6$$ with $$\theta(\bar 0)=\bar 0, \theta(\bar 1)=\bar 3$$

Is an injective ring homomorphism?
(b) Show that there is no injective ring homomorphism $$\theta:Z_2-->Z_4$$
Proof:

(a) Well, i said since $$\bar 0 =/=\bar 3$$ in Z_6 then such theta is injective.
Now to establish homomorphism, i proceded

$$\theta(\bar 0 \bar 1)=\theta(\bar 0)=\bar 0$$
$$\theta(\bar 0)\theta(\bar 1)=\bar 0$$ so they are equal

$$\theta(\bar 0+\bar 1)=\thta(\bar 1)=\bar 3$$
$$\thetea(\bar 0)+\theta(\bar 1)=\bar 3$$

So, i concluded that we have a homomorphism

Well, this didn't cause me any problems, as far as my understanding goes. However, i am having trouble on the second part:

(b) I started like this:

In order for $$\theta:Z_2-->Z_4$$ to be a homomorphism, if $$\theta(\bar a)=\bar b, \bar a \in Z_2, \bar b \in Z_4$$ then we should have the following $$o(\bar b)|o(\bar a)$$ ( i think there is a theorem that says this)

So, looking at the orders of the elements in both rings, we notice that the only such possibility is:

$$\theta(\bar 0)=\bar 0, \theta(\bar 1)=\bar 2$$

Now, i am failing to show that this is not an injective homomorphism.

Here is what i am doing:

$$\theta(\bar 0 \bar 1)=\theta(\bar 0)=\bar 0$$

$$\theta(\bar 0)\theta(\bar 1)=\bar 0$$

$$\theta(\bar 0 +\bar 1)=\theta(\bar 1)=\bar 2$$

$$\theta(\bar 0)+\theta(\bar 1)=\bar 2$$

So, by this reasoning it looks to me that such a mapping is a homomorphism.

However, now everything that i am trying is also 'showing' that such a theta is also injective.

SO, how to show that theta is not injective/?



Many thanks in return!

 

[Office_Shredder]

$$\theta(1) = 2$$ gives us

$$\theta(1^2) = \theta(1)^2 = 2^2 = 4 = 0$$

Whoops

 

[sutupidmath]

$$\theta(1) = 2$$ gives us

$$\theta(1^2) = \theta(1)^2 = 2^2 = 4 = 0$$

Whoops

Well since $$\theta:\bar 1->\bar 0, and \bar 1-> \bar 2$$ this, from my point of view, would mean that theta is not a mapping at all right, since it is violating the definition of a mapping, right? To show that a mapping is not injective wouldn't we need something like this:

$$If, x_1=/=x_2=/>\theta(x_1)=/=\theta(x_2)$$

Or, if $$\theta(a)=\theta(b)=/>a=b$$ ??

$$=/>$$ stands for "Does not follow"

So, how would that tell us that theta is not injective? Pardone my ignorance!