Prove that $S_4$ does not have a normal subgroup of order 8, nor a normal subgroup of order 3. (Dummit/Foote 3.2.14)
Proof: Suppose $A$ and $B$ are normal subgroups of $S_4$ with orders 8 and 3, respectively.
(i) Since $A \trianglelefteq S_4$ then for any subgroup $X\leq S_4$ we have $AX \leq S_4$. Let $X= \{1,x\}$ such that $|x|=2$ and $x\notin A$. Since $|A|=8$, and $S_4$ contains more than 8 elements of order 2, then $x$ exists. By definition we therefore have $A\cap X = \{1\}$ and hence their product $AX$ has order
$$|AX|=\frac{|A||X|}{|A\cap X|}=\frac{8\cdot 2}{1}=16,$$
however by LaGrange's Theorem, if $AX\leq S_4$ then $|AX|$ divides $|S_4|$, but 16 does not divide 24, so $A$ cannot be a normal subgroup.
(ii) Similar to part (i), since $B \trianglelefteq S_4$ then for any subgroup $Y\leq S_4$ we will have $BY\leq S_4$. Let $Y= \{1, y, y^2\}$ such that $|y|=|y^2|=3$ and $y, y^2 \notin B$. Once again, this is possible since $S_4$ contains more than 3 elements of order 3. Then we will have,
$$|BY|=\frac{|B||Y|}{|B\cap Y|}=\frac{3\cdot 3}{1}=9,$$
but since 9 does not divide 24 therefore $BY$ is not a subgroup of $S_4$, so $B$ is not a normal subgroup.
Is this correct?
Best Answer
There is still another less involved method for showing that $S_4$ does not have any normal subgroups of order $3$ or $8$. The trick is that you observe that a normal subgroup must be the (disjoint) union of some of the conjugacy classes and that one of them is the class of the unity $1$ (since $1 \in$ normal subgroup).
The conjugacy classes are easy to figure out, since they correspond to cycle types; their sizes are $1, 3, 6, 6$, and $8$ (respectively the size of the conjugacy classes $[(1)], [(12)(34)], [(12)], [(1234)], [(123)]$).
Now you immediately see that there is no way of building a normal subgroup of order $3$ or $8$ out of these classes: for $3$, you are left with $2$ elements, impossible, for $8$ you are left with $7$, also impossible.
This is quite a cool way to check if something is a normal subgroup or not. With this method you can easily prove that $A_5$ is a simple group (that is without any non-trivial normal subgroups). I leave that to you.