Does the set $A$, of all permutations of $\{1,2,3,4\}$ that fixes $2$ and $4$ form a subgroup of $S_4$? Does the set $A$, of all permutations of $\{1,2,3,4\}$ that fixes $2$ and $4$ form a subgroup of $S_4$?
I’ve got as far,
The set $A$ will be the identity where all $4$ numbers map to themselves (as all fixed) and $(1, 3)$ as $2$ and $4$ are fixed. So the set $A =\{\operatorname{id}, (1 3)\}$
Now I need to show that this is a sub group of $S_4$, by identity closure and inverses.
Identity is in both so ok but I’m stuck on the other $2$.
Any help would be great
Kara x
Hint: Each such permutation has the form $\pi = \begin{pmatrix} 1 &amp; 2 &amp; 3 &amp; 4 \\ x_1 &amp; x_2 &amp; x_3 &amp;x_4 \end{pmatrix}$ with $x_2=2$ and $x_4=4$.
Show that the composition of two such permutations has the same shape.
Moreover show that the inverse permutation of such a permutation has the same format.
You're right about the structure of $A$, consisting of $(1)(2)(3)(4)$ and $(13)(2)(4)$.
You're correct: $A=\{{\rm id}, (13)\}$.
I will use the two-step subgroup test.
Observe that $\varnothing \neq A\subseteq S_4$.
Consider the multiplication table for $A$:
$$\begin{array}{c|cc}
\circ &amp; {\rm id} &amp; (13) \\
\hline
{\rm id} &amp; {\rm id} &amp; (13) \\
(13) &amp; (13) &amp; {\rm id}.
\end{array}$$
From the table, one can see that for any $a,b\in A$, we have $a\circ b\in A.$
Since $(13)^{-1}=(13)\in A$ and ${\rm id}^{-1}={\rm id}\in A$, $A$ is closed under inverses.
Hence $A\le S_4$.