These questions are pretty standard, this is the first time I'm trying one on my own, I would like to check my progress:
Using the First Isomorphism Theorem and Lagrange’s Theorem, find all the homomorphisms from $ S_3 \to C_4 $.
My Answer: there are 2, argumentation:
In order to determine all homomorphisms, it is wise to determine the possible sizes of the kernel and the image of such homomorphisms. We know that $|S_3|=3!=6$. Furthermore, for any group homomorphism $f: G \to G'$, corollary 5, enables us to write:
$$ |G|=|\text{im}(f)|\times| \ker(f)|$$
$$ |S_3|=6=|\text{im}(f)|\times| \ker(f)|$$
The product of kernel and image can only be 6, this is a good restriction. We will now use the first isomorphism theorem, which tells us:
$$\ker(f) \triangleleft S_3 $$
$$\text{im}(f) \leq C_4 $$
Hence the kernel and image are always subgroups, respectively of our orginal group and of the group that is mapped to. We can therefore use Lagrange's theorem, which states that the order of a subgroup must always divide the order of the group. Since $|S_3|=6$ and $|C_4|=4$, this tells us that the possible candidates are:
$$ \ker(f)=1, 2, 3, 6 $$
$$ \text{im}(f)=1,2 , 4 $$
We already knew that the product of these two cardinalities/orders had to be $6$. The only possible products that lead to this are for $3 \cdot 2 =6$ and $6 \cdot 1 =6$, so $(|\ker(f)|, |\text{im}(f)|) \in \{(3,2),(6,1)\}$.
The trivial homomorphism that sends every single element of $S_3$ to the identity element in $C_4$ is always a group homomorphism. So, whenever $|\text{Im}(f)|=1$ and $\ker(f)=6$ we have that:
for $e\in C_4$ and $\forall x \in S_3$ the trivial homomorphism $f: S_3 \to C_4$ is defined by:
$$ f(x)=e$$
The other option is that $|\text{im}(f)|=2$ (and kernel of order 3). Since the image is a subgroup by the isomorphism theorem, and by corollary 3, the order of any element in the group needs to divide the order of the group. Thus we can only have elements of order $2$ or $1$ (identity) in the image. We represent $C_4 = \{e, c, c^2, c^3\}$, with generator $c$. We realise that the only element of order $2$ is $c^2$:
$$\text{im}(f) = \langle c^2 \rangle = \{e, c^2\} $$
We also know that the kernel of this homomorphism needs to be of order $3$. Again, the kernel is a subgroup, so the order of the elemens need to divide the order of the group. Elements in the kernel must have orders $1$ or $3$. We know the identity element must always be in the kernel. We first list all elements of $S_3$ and determine the orders so we can decide what can be in the kernel:
$$ S_3 = \{e, ( 1 \ 2 \ 3), ( 1 \ 3 \ 2), ( 1 \ 2), ( 2 \ 3), ( 1 \ 3) \} $$
We notice that the identity is of order 1, and the two "cycles" $( 1 \ 2 \ 3), ( 1 \ 3 \ 2)$ are of order 3, they fit the description. We then define the homomorphism $f: S_3 \to C_4$ that maps:
$$e, ( 1 \ 2 \ 3), ( 1 \ 3 \ 2) \to e'\in C_4 $$
$$ ( 1 \ 2), ( 2 \ 3), ( 1 \ 3) \to c^2 \in C_4$$
Conclusion: There are 2 homomorphismsm the one above and the trivial homomorphism.
Best Answer
Towards the end your proof is a bit roundabout, but it is entirely valid, except perhaps for one point:
At the very end you define a map $f:\ S_3\ \longrightarrow\ C_4$, but you do not verify that it is in fact a homomorphism. At this point, you have only shown that if there exists a homomorphism $f$ with $|\ker(f)|=3$ and $|\operatorname{im}(f)|=2$, then this must be it.
As for the proof being roundabout towards the end; once you have determined that $$(|\ker(f)|, |\text{im}(f)|) \in \{(3,2),(6,1)\},$$ you could finish the argument more clearly by distinguishing two cases;
Then you can conclude that there are precisely two homomorphisms from $S_3$ to $C_4$.