[proofplan]
We first verify directly that $V$ is a subgroup of $S_4$ by checking that the product of any two of its nonidentity elements is the third nonidentity element, while each nonidentity element is its own inverse. To prove normality, we show that conjugation by an arbitrary $\sigma \in S_4$ sends each element of $V$ back into $V$. The key point is that conjugation relabels the entries in cycle notation, so a double transposition is sent to a double transposition; in $S_4$ there are exactly the three double transpositions listed in $V$.
[/proofplan]
[step:Verify that the listed set is a subgroup of $S_4$]
Let
\begin{align*}
a:=(12)(34), \qquad b:=(13)(24), \qquad c:=(14)(23).
\end{align*}
Each of $a,b,c$ belongs to $S_4$, and $e$ is the identity element of $S_4$. Since each transposition has order $2$ and the transpositions in each displayed product are disjoint, we have
\begin{align*}
a^2=b^2=c^2=e.
\end{align*}
A direct evaluation on the set $\{1,2,3,4\}$ gives
\begin{align*}
ab=c, \qquad bc=a, \qquad ca=b.
\end{align*}
Because $a,b,c$ all have order $2$, reversing each product gives the same identities:
\begin{align*}
ba=c, \qquad cb=a, \qquad ac=b.
\end{align*}
Thus $V$ contains the identity element, is closed under products, and contains the inverse of each of its elements. Hence $V \le S_4$.
[/step]
[step:Conjugate each double transposition by an arbitrary permutation]
Fix an arbitrary permutation $\sigma \in S_4$. By [citetheorem:8541], [conjugation relabels cycle notation](/theorems/8541). Therefore
\begin{align*}
\sigma(12)(34)\sigma^{-1}=(\sigma(1)\ \sigma(2))(\sigma(3)\ \sigma(4)).
\end{align*}
Similarly,
\begin{align*}
\sigma(13)(24)\sigma^{-1}=(\sigma(1)\ \sigma(3))(\sigma(2)\ \sigma(4)),
\end{align*}
and
\begin{align*}
\sigma(14)(23)\sigma^{-1}=(\sigma(1)\ \sigma(4))(\sigma(2)\ \sigma(3)).
\end{align*}
Since $\sigma$ is a bijection $\{1,2,3,4\} \to \{1,2,3,4\}$, the four symbols $\sigma(1),\sigma(2),\sigma(3),\sigma(4)$ are pairwise distinct. Hence each displayed conjugate is a product of two disjoint transpositions in $S_4$.
[guided]
Fix an arbitrary permutation $\sigma \in S_4$. To prove normality, we must understand the elements of the conjugate set
\begin{align*}
\sigma V\sigma^{-1}:=\{\sigma v\sigma^{-1}:v\in V\}.
\end{align*}
The identity element causes no difficulty, because
\begin{align*}
\sigma e\sigma^{-1}=e.
\end{align*}
So it remains to track the three nonidentity elements of $V$.
The useful mechanism is that conjugation by $\sigma$ merely changes the labels in cycle notation. By [citetheorem:8541], for every cycle $(x_1\ \cdots\ x_k)$ in $S_4$ we have
\begin{align*}
\sigma(x_1\ \cdots\ x_k)\sigma^{-1}=(\sigma(x_1)\ \cdots\ \sigma(x_k)).
\end{align*}
Applying this to the two disjoint transpositions in $(12)(34)$ gives
\begin{align*}
\sigma(12)(34)\sigma^{-1}=(\sigma(1)\ \sigma(2))(\sigma(3)\ \sigma(4)).
\end{align*}
The same relabelling gives
\begin{align*}
\sigma(13)(24)\sigma^{-1}=(\sigma(1)\ \sigma(3))(\sigma(2)\ \sigma(4)),
\end{align*}
and
\begin{align*}
\sigma(14)(23)\sigma^{-1}=(\sigma(1)\ \sigma(4))(\sigma(2)\ \sigma(3)).
\end{align*}
Why are these still double transpositions? Because $\sigma$ is a bijection from $\{1,2,3,4\}$ to itself. Therefore the four values $\sigma(1),\sigma(2),\sigma(3),\sigma(4)$ are pairwise distinct. Consequently each displayed expression is a product of two transpositions with disjoint supports, hence is a double transposition in $S_4$.
[/guided]
[/step]
[step:List all double transpositions in $S_4$]
A double transposition in $S_4$ is determined by a partition of the set $\{1,2,3,4\}$ into two unordered pairs. These partitions are exactly
\begin{align*}
\{\{1,2\},\{3,4\}\}, \qquad \{\{1,3\},\{2,4\}\}, \qquad \{\{1,4\},\{2,3\}\}.
\end{align*}
They correspond respectively to
\begin{align*}
(12)(34), \qquad (13)(24), \qquad (14)(23).
\end{align*}
Thus the only double transpositions in $S_4$ are $a,b,c$, and every double transposition in $S_4$ lies in $V$.
[/step]
[step:Conclude that $V$ is invariant under conjugation]
From the preceding steps, for every $\sigma \in S_4$ and every $v \in V$, the element $\sigma v\sigma^{-1}$ lies in $V$. Hence
\begin{align*}
\sigma V\sigma^{-1}\subset V.
\end{align*}
Applying the same inclusion with $\sigma^{-1}$ in place of $\sigma$ gives
\begin{align*}
\sigma^{-1}V\sigma\subset V.
\end{align*}
Multiplying this inclusion on the left by $\sigma$ and on the right by $\sigma^{-1}$ yields
\begin{align*}
V\subset \sigma V\sigma^{-1}.
\end{align*}
Therefore
\begin{align*}
\sigma V\sigma^{-1}=V
\end{align*}
for every $\sigma \in S_4$. Since $V \le S_4$ and $V$ is invariant under conjugation by every element of $S_4$, we conclude that $V\trianglelefteq S_4$.
[/step]
