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Can anyone please help me define clearly using the definition for a group homomorphism that for the given two homomorphisms it holds for all $g_1,g_2\in G$?

We know that the definition gives that there exists two groups, $G$ and $H$ and a function $\psi_1:G\to H$. This is then called a homomorphism if the equality $\psi_1(g_1+g_2)=\psi_1(g_1)+\psi_1(g_2)$ holds for all $g_1,g_2\in G$.

So as a start,we take the first function $\psi_1:\mathbb{Z}_{5}\to\mathbb{Z}_{5}$. If we let $x,y\in\mathbb{Z}_{5}$, then $\psi_1(x+y)=\psi_1(x)+\psi_1(y)$?

Please brief me out on how to do this problem.

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You want to characterize all group homomorphisms from Z5 to itself –  Amr Feb 11 '13 at 0:34
    
Ok yes thanks for the feedback –  Faye Feb 11 '13 at 1:18

3 Answers 3

up vote 3 down vote accepted

The group $\mathbb{Z}_5$ is an additive cyclic group: $(\mathbb{Z}_5, +)$, so the homomorphism property for $\psi_i: \mathbb{Z}_5 \to \mathbb{Z}_5$ is given by

$$\psi_i(x + y) = \psi_i(x) + \psi_i(y)$$

For all homomorphisms, $\psi_i(0) = 0$ (a homomorphism maps identity to identity).

Then choose $\psi_i(1) = x \in \mathbb{Z}_5, x \neq 0$, the the homomorphism property guarantees that this assignment determines the assignment of all elements in $\mathbb{Z}_5$

For each element $x \in \mathbb{Z}_5$, $x \neq 0$, there exists a unique homomorphism determined by $\psi_i(1) = x$.

Hence, there are four such unique homomorphisms corresponding to $\psi_i(1) = i$, $i \in \{ 1, 2, 3, 4\}$

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So this applies to $\psi_2$ as well right? –  Faye Feb 11 '13 at 1:10
    
By the way thanks for the feedback. –  Faye Feb 11 '13 at 1:10
    
Yes...you're welcome... –  amWhy Feb 11 '13 at 1:15
1  
It looks like, from your question, that you need only find two explicit homomorphisms from $\mathbb{Z}_5 \to \mathbb{Z}_5$. There are, in fact, four distinct homomorphisms. Let me know if you have any more trouble in defining two such homomorphisms. –  amWhy Feb 11 '13 at 2:15
    
No im wrong it is asking for the four distinct homomorphisms. –  Faye Feb 11 '13 at 2:45

The operation in ${\bf Z}_5$ is addition, so what you want is $$\psi(x+y)=\psi(x)+\psi(y)$$

You should

  1. prove that $\psi(0)=0$,

  2. prove that if you know $\psi(1)$ then you can work out $\psi(m)$ for all $m$,

  3. see what happens when you take different values for $\psi(1)$.

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Thanks for your feedback –  Faye Feb 11 '13 at 1:09

The group $\mathbb{Z}_5$ is written additively, so the homomorphism property spells as $\psi_1(x + y) = \psi_1(x) + \psi_1(y)$ in this case.

There are several ways to construct such a function. To find them all, try the following:

1) Which possible choices are there for $\psi_1(1)$?

2) Assure yourself, that by the homomorphism property all values $\psi_1(x)$ are determined once that a choice for $\psi_1(1)$ was made.

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Possible choice for $\psi_1$($1$) = 1 itself? –  Faye Feb 11 '13 at 0:41
    
Yes, this works. But there are others. Which value can $\psi_1(1)$ not take? –  azimut Feb 11 '13 at 0:43
    
It cannot take 0 –  Faye Feb 11 '13 at 1:08
    
I'm sorry, in fact $\psi_1(1)$ may take all the values in $\mathbb{Z}_5$, including $0$. (I was thinking that you had asked for *iso*morphisms, where $\psi_1(1) = 0$ would be forbidden). Now fix some choice for $\psi_1(1)$ and figure out what the other values $\psi(x)$ are then. Do this for all the 5 possible choices of $\psi_1(1)$. In the end, you have found all $5$ homomorphisms $\mathbb{Z}_5\to\mathbb{Z}_5$ –  azimut Feb 11 '13 at 1:16

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