How many elements of $S_5$ send the set $\{2,4\}$ into the set $\{2,4\} ?$

Question

How many elements of $S_5$ send the set $\{2,4\}$ into the set $\{2,4\} ?$

My attempt :let $\sigma\in S_5 $ be a permutation which send the set $\{2,4\}$ into the set $\{2,4\} $.Since $\sigma$ is injective so we have

$\sigma(2)=4$

$\sigma(4)=2$

$\sigma(4)=4$

$\sigma(2)=2$

$\implies$ there are $2$ elements of $S_5 $ which send the set $\{2,4\}$ into the set $\{2,4\} $

Answer 1

Okay, so we need a bijection from $\{1,2,3,4,5\}$ to itself. Also, importantly, we need to send $2$ and $4$ to $2$ and $4$, and we don't care about where $1,3,5$ are sent. You can send $2$ to $2$ or you can send $2$ to $4$, and in these cases you have to send $4$ to $4$ or $2$ respectively. There are $3!$ ways to arrange $1,3,5$, and hence $3!$ bijections which do the rearrangement. So, $2$ possibilities for where $2$ and $4$ go, and $3!$ possibilities for where $1,3,5$ go. In total that is $2\cdot 3!=12$ total bijections.

Answer 2

Let $H$ be the subgroup of $S_5$ of elements that fix the elements of $\{1,3,5\}$. Let $K$ be the subgroup of $S_5$ of elements that fix the elements of $\{2,4\}$. (These are clearly subgroups by transfer of structure onto $S_2$ and $S_3$ respectively.)

We have:

• $H\cap K=\{e\}$.
• For all $h\in H, k\in K$, $$hk=kh.$$

Thus $H\times K$ is isomorphic to an internal direct product $H\otimes K$ in $S_5$.

The elements of $H\otimes K$ contain all and only those elements of $S_5$ that send $\{2,4\}$ to $\{2,4\}$; $H$, in particular, clearly does this, and elements of $K$ leave $\{2,4\}$ untouched.

Thus the number you are after is

$$\begin{align} |H\times K|&=|H||K|\\ &=|S_2||S_3|\\ &=2\times 3!\\ &=12. \end{align}$$