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Many of my homework problems are of the form:

Find all morphisms between $\text{Group 1}$ and $\text{Group 2}$.

How does one in general approach these problems? We have looked at the morphism theorem, but it doesn't seem to be of much help here.

This is my likely wrong attempt on one such problem.

Find all morphisms from $\mathbb{Z}$ to $D_4$.

I essentially just do some inspection, and come to a conclusion. I haven't shown that these are all possible morphisms.

By the definition of morphism, we are searching for all functions $f: \mathbb{Z} \to D_4$ satisfying $f(a+b) = f(a)f(b)$.

I remark that the above structure looks like standard exponentiation, since $x^{a+b} = x^a x^b$. After investigating, I find that morphisms of the form $f(a) = d^{ka}$ where $d \in D_4$ and $k \in \mathbb{Z}$ seem to work.

Simple investigation shows that this is indeed a morphism:

$f(a+b) = d^{k(a+b)} = f(a)f(b) = d^{ka}d^{kb} = d^{k(a+b)}$

How do I find the rest or prove that these are the only ones? Thanks in advance.

WeierstrassSauce
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    Hint: Let $f: \mathbb{Z} \rightarrow G$ be any homomorphism of groups. Let $g = f(1) \in G$. Then $f(2) = f(1 + 1) = f(1) \cdot f(1) = g^2$. You can generalize this argument to compute $f(n)$ for all $n \in \mathbb{Z}$. – Michael Joyce Mar 19 '13 at 22:38
  • Ah, thanks! So in effect, by induction all such morphisms can be given by $f(n) = g^n$ for any element in $D_4$, right? That was surprisingly easy!

    Is there a general way to approach these sorts of problems? What about a problem like, find all morphisms from $C_3 \to C_4$?

     – WeierstrassSauce Mar 19 '13 at 23:12
  • I'd like to understand this question better. What does $D_4$ mean? – Doug Spoonwood Mar 20 '13 at 00:28
  • A homomorphism $f: C_3 \rightarrow C_4$, is determined by the image of a generator of $C_3$. Let $a$ denote a generator of $C_3$. Then we must have $f(a)^3 = e$ in $C_4$, which limits the possibilities of what the homomorphism can be. – Michael Joyce Mar 20 '13 at 02:09
  • I came across this and rather than starting a new question I was wondering if we have $f(a)^3=e$ then is the only morphism $f(a^r)=e$? If so, why is that the only one? – Yousuf Soliman Jul 12 '13 at 19:06

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In the case $\mathbb{Z}$, $D_4$, notice that a homomorphism from $\mathbb{Z}$ to $D_4$ is completely determined by the image of $1$ (i.e. the generator of the abelian group $\mathbb{Z}$).

Daniel Robert-Nicoud
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