How many homomorphisms are there from $\mathbb D_5$ to $\mathbb A_4$? Find one of them which is nottrivial.
My try:
To be honest, I don't really understand how to find homomorphisms between groups, so I will present what I was able to deduce:
$$\mathbb D_5 = <\rho,\epsilon: \rho^5=1,\epsilon^2=1, \epsilon\rho\epsilon=\rho^{-1}>=$$ $$=<1,\rho,\rho^2,\rho^3,\rho^4,\epsilon,\epsilon\rho,\epsilon\rho^2,\epsilon\rho^3,\epsilon\rho^4> $$ $$\mathbb A_4=\left\{\text{id},(12)(34),(13)(24),(14)(23)\right\}$$
$\mathbb D_5$ is generated by $\rho$ and $\epsilon$. Let $f:\mathbb D_5 \rightarrow \mathbb A_4$ so we have: $$\text{ord}(\rho)=5 \Rightarrow \text{ord}(f(\rho))=1 \text{ or } \text{ord}(f(\rho))=5$$ $$\text{ord}(\epsilon)=2 \Rightarrow \text{ord}(f(\epsilon))=1 \text{ or } \text{ord}(f(\epsilon))=2$$ Then we have: $$\text{ord}(\text{id})=1$$ $$\text{ord}((12)(34))=\text{ord}((13)(24))=\text{ord}((14)(23))=2$$ That's why we must have: $$f(\rho)=\text{id}$$ and $\epsilon$ can go to any item from $\mathbb A_4$. So exists $4$ homomorphisms from $\mathbb D_5$ to $\mathbb A_4$.
It is correct solution?
If so, how can I determine specific homomorphism now?
First of all, your $A_4$ misses a lot of elements. You miss the $3$-cycles.
Let be a morphism $f:D_5\to A_4$. Let $a=f(\rho)$, $b=f(\epsilon)$. Then $a^5=Id,b^2=Id$ and $bab^{-1}=a^{-1}$ since a morphism preserves products and inverses.
There is no element of order $5$ in $A_4$, so $a=1$. The elements of order $1$ or $2$ is $A_4$ are $Id$ and the product of two transpositions with disjoint supports, so $b$ is one of these $4$ elements. The relation $bab^{-1}=a^{-1}$ is trivially satisfied since $a=Id$.
So there is at most $4$ morphisms.You need to show that , conversely, if $b$ has order $1$ or $2$ in $A_4$, there exists a unique morphism $f:D_5\to A_4$ such that $f(\rho)=1$ and $f(\epsilon)=b$.
You have two possibilities , depending on your knowledge on group presentations.
First proof. $D_5 = <\rho,\epsilon: \rho^5=1,\epsilon^2=1, \epsilon\rho\epsilon=\rho^{-1}>=F/R$, where $F$ is the free group on two letters $\{\rho,\epsilon\}$ and $R$ is the normal subgroup generated by $\rho^5,\epsilon^2, \epsilon\rho\epsilon\rho.$ The general theory says that morphisms that morphisms $D_5=F/R\to A_4$ are exactly "evaluation morphisms", where the images $a$ and $b$ of $\rho $ and $\epsilon$ satisfy the imposed relations $R$, that is $a^5=Id,b^2=Id$ and $bab^{-1}=a^{-1}$. So we are done here.
[More precisely: The factorisation theorem tells you that morphisms $F/R\to A_4$ comes from a unique morphism $F\to A_4$ which contains $R$ in its kernel.
Since $F$ is free, morphisms $F\to R$ are exactly "evaluation morphisms", and are uniquely determined by the images $a$ and $b$ of $\rho $ and $\epsilon$ respectively. Saying that the corresponding morphism contains $R$ in it kernel is equivalent to ask $a^5=Id,b^2=Id$ and $bab^{-1}=a^{-1}$, and we are done.]
Second proof. From scratch. First, you need to convince yourself that an element of $D_5$ can we written in a unique way as $\rho^m \epsilon^n, 0\leq m\leq 4,n=0,1$. I let you do it.
In particular, if $f:D_5\to A_4$ is a morphism, we have $$f((\rho^m\epsilon^n))=f(\rho)^m f(\epsilon)^n=b^n.$$ What remains to do is that the map $f_b: D_5\to A_4$ defined by $$f_b(\rho^m\epsilon^n)=b^n, 0\leq m\leq 4,n=0,1$$ is indeed a morphism.
The key observation now is that for all $r\in\mathbb{Z}$ and $s\in\mathbb{Z}$ we have $f_b(\rho^r \epsilon^s)=b^s$. Notice that there is something to prove since a priori we only know what happens for $0\leq r \leq 4$ and $s=0,1$.
To do so, write $r=4q+m, 0\leq m\leq 4$ and $s=2q'+n, n=0,1$, and use the fact that $b^2=1$.
Now we can prove that $f_b$ is a morphism .We have $(\rho^m\epsilon^n)(\rho^k\epsilon^\ell)=\rho^{m+(-1)^nk}\epsilon^{n+\ell}$, so $$f_b((\rho^m\epsilon^n)(\rho^k\epsilon^\ell))=f_b(\rho^{m+(-1)^nk}\epsilon^{n+\ell})=b^{n+\ell}=b^n b^\ell=f_b(\rho^m\epsilon^n)f_b(\rho^k\epsilon^\ell)).$$
Note that we used the key obervation, since the integers $m+(-1)^nk$ and $n+\ell$ may live outside $\{0,\ldots,4\}$ and $\{0,1\}$ respectively.
All in all , the morphisms you seek are exactly the maps $f_b:D_5\to A_4$, where $b=Id, (1 2)(34), (13)(24), (14)(23)$