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?

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.)

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 $$ S_4/V = S_3 V / V \cong S_{3}/(S_{3}\cap V) = S_3/\{1\} \cong S_3. $$