Show that $S_4/V$ is isomorphic to $S_3 $, where $V$ is the Klein Four Group. (i) Show that $S_4/V$ is isomorphic to $S_3 $, where $V$ is the Klein Four Group.
I understand isomorphism to be a bijective homomorphism but I'm unsure how one would go about proving this. $S_4/V$ has order 6 and I think that will be of use but that's as far as I can go.
(ii) Let $G = D_7$ be the dihedral group of order 14 and let $H = C_7$ be the cyclic group of order 7. Find all the homomorphisms from $G$ to $H$.
Are there any homomorpisms at all?
 A: Let me add what is possibly the trivial solution to (i).
There is a distinguished copy of $S_3$ in $S_4$, given by the stabilizer of $4$
$$
S_3 = \{1, (123), (132), (12), (13), (23)  \}.
$$
This clearly intersects trivially
$$
V = \{1, (12)(34), (13)(24), (14)(23)  \}.
$$
Since $S_3 \cap V = \{1\}$, we have $S_4 = S_3 V$.
(From a mildly higher point of view, $V$ is a transitive subgroup, so $S_4 = S_3 V$, as $S_3$ is a one-point stabilizer, and $S_{3}\cap V = \{1\}$ as $V$ acts regularly.)
Now one of the isomorphism theorems tells you that
$$
S_4/V = S_3 V / V \cong S_{3}/(S_{3}\cap V) = S_3/\{1\} \cong S_3.
$$
A: Just a “visual” construction of the isomorphism between $\mathfrak S(4)/V_4$ and $\mathfrak S(3)$...
It's quite well known that $4 = 2 +2$. Concretely, it means that if you have four things, you can always group them into two pairs. What's remarkable is that there are exactly three such groupings:
$$ \{1, 2, 3, 4\} = \{1, 2\} \sqcup \{3,4\} = \{1, 3\} \sqcup \{2,4\} = \{1, 4\} \sqcup \{2,3\}.$$
If I call these three groupings $\mathbf{Order}$, $\mathbf{Parity}$ and $\mathbf{5sum}$, every permutation of $\{1,2,3,4\}$ induces a permutation of these three things. For example, the $4$-cycle $(1234)$ induces the transposition $(\mathbf{Order}\ \mathbf{5sum})$. So I get a homomorphism
$$ \mathfrak S(4) \to \mathfrak S(\{\mathbf{Order}, \mathbf{Parity}, \mathbf{5sum}\})\simeq \mathfrak S(3).$$
It's quite easy to show that this morphism is surjective (we already found the  transposition $(\mathbf{Order}\ \mathbf{5sum})$ in its image, it's not hard to find another transposition or a $3$-cycle, and that will be enough to generate the whole of $\mathfrak S(3)$.)
Let's give a proof that the kernel is exactly $V_4$: let $\sigma$ be a nontrivial permutation in it. So for example, there are two elements $a \neq b$ such that $\sigma(a) = b$ (let's call $c$ and $d$ the two remaining elements) In that case, because $\sigma$ preserves the grouping $\{a,b\} \sqcup \{c,d\}$, it must send $b$ back to $a$. And because it preserves $\{a,c\} \sqcup \{b,d\}$ and it exchanges $a$ and $b$, it must exchange $c$ and $d$ as well. So $\sigma = (a\, b)(c\, d) \in V_4$.
Therefore, the factorisation theorem gives you an isomorphism $\mathfrak S(4)/V_4 \to \mathfrak S(3)$.
Of course, this proof is not the shortest (well, it certainly isn't the shortest to write, but if you're allowed to make a lot of pictures or to play with four actual tokens, essentially all the arguments become self-evident). But it really makes the isomorphism concrete. There's an isomorphism between $\mathfrak S(4)/V_4$ and $\mathfrak S(3)$ because $4$ equals $2+2$ in $3$ different ways...
A final remark: if $n \neq 4$, the only proper normal subgroup of $\mathfrak S(n)$ is $\mathfrak A(n)$. That means that the situation I described here is pretty exceptional. If you want $\mathfrak S(n)$ to act on fewer than $n$ objects, the only interesting actions are :


*

*for $n$ arbitrary, $\mathfrak S(n)$ acts on two tokens with the following rule: even permutations do nothing, odd ones swap the two tokens (you have to admit this action is pretty boring).

*one exceptional case: $\mathfrak S(4)$ acts on three tokens, as we just saw.


I may be overenthusiastic, but this simple remark gives much cachet to this innocent-looking isomorphism. (Another peculiarity of the same kind is that the only interesting action of $\mathfrak S(n)$ on $n+1$ tokens is an action of $\mathfrak S(5)$ on 6 tokens, coming from another exceptional behaviour of the symmetric group.)
A: Hints (complete the argumentation):
(i) $\,|S_4|=4!=24\,\,,\,\,|V|=4\,$ and $\,V\triangleleft S_4\,$ , so $\,\left|S_4/V\right|=6=|S_3|\,$ . Now just prove that there can't be an element $\,\sigma\in S_4\,\,\,s.t.\,\,\,ord_{S_4/V}(\sigma V)=6\,$ (further hint: otherwise, what'd be $\,ord_{S_4}(\sigma)\,$ ?)
(ii) By definition
$$D_7:=\langle r,s\;\;;\;\;s^2=r^7=1\;\;,\;\;srs=r^{-1}=r^6\rangle$$
Now remember that for any homomorphism $\,\phi: D_7\to C_7\;\;,\;\;\phi(D_7)\le C_7\,$ (Further hint: %$\,7\,$ is a prime number...)
A: For the first one, note that there are precisely $2$ groups of order $6$ (up to isomorphism)--namely, $C_6$ and $S_3$. All you have to do is show that $S_4/V$ isn't abelian (or, perhaps more simply, isn't cyclic).
For the second, note that $D_7=\langle x,y\mid x^7=y^2=xyxy=1\rangle$, that a homomorphism is completely determined by where it maps a group's generators, and that if $\phi:G\to H$ is a homomorphism, then the order of $\phi(g)$ divides the order of $g$ for each $g\in G$. This should be enough to let you completely determine the homomorphisms $D_7\to C_7$.
A: We see that $S_4/V$ cannot be abelian, because otherwise $S_4' = A_4 \leq V$ which is definitely not true. Thus $S_4/V$ must be nonabelian of order $6$, which means that $S_4/V \cong S_3$.
