Prove using First Isomorphism Theorem 
Prove that $N=\{ e,(1\ 2)(3\ 4),(1\ 3)(2\ 4),(1\ 4)(2\ 3)\}$ is a normal subgroup of $S_4$ and $N\subseteq A_4$ such that $S_4/N \cong S_3$ and $A_4/N \cong \Bbb Z_3$.
Hint: Use that $$(i\ j)=(1\ i)(1\ j)(1\ i)$$ $$(1\ j)=(1\ j-1)(j-1\ j)(1\ j-1)$$
and the First Isomorphism Theorem.

We've tried to prove that $N\lhd S_4$, but we can't see how to expand an element of $S_4$ using the hint. Help.
 A: Let $R$ be a rectangle that is not a square. The easiest way to see that $N$ is a subgroup of $S_4$ is to let $G$ be the group of symmetries of $R$. Label the vertices of $R$ sequentially with the numbers $1$, $2$, $3$, $4$. Then $G$ will act on the set $\{1,2,3,4\}$. This gives us a homomorphism $\phi:G\to S_4$. $N$ is a subgroup of $S_4$, since $N=\phi(G)$.
To see that $N$ is normal, first note that for any $\sigma,\pi\in S_n$, if $\sigma$ is a $k$-cycle, i.e.
$$\sigma=(a_1a_2\ldots a_k)$$
for some $a_1,\ldots,a_k\in\{1,\ldots,n\}$, then it follows that
$$\pi\sigma\pi^{-1}=(\pi(a_1)\pi(a_2)\ldots \pi(a_k)).$$
In other words, whenever we conjugate a $k$-cycle, we get another $k$-cycle. Note that $N$ consists of the identity and all the elements of $S_4$ that are product of disjoint $2$-cycles. Hence $N\lhd S_4$.
Note that $S_4/N$ has order $6$ so $S_4/N\cong S_3$ or $S_4/N\cong\Bbb{Z}_6$. Since $S_4$ has no elements of order $6$, we must have $S_4/N\cong S_3$. Similarly, $A_4/N$ has order $3$ so $A_4/N\cong\Bbb{Z}_3$.
