Prove $S_4$ has no normal subgroup of order 8, or of order 3?

Question

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?

Answer 1

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.

Answer 2

Here's another low calculation approach that relies on properties of quotient groups instead of on conjugacy classes.

I show that $S_4$ can't have a normal subgroup with order $8$ by contradiction. Suppose $H \unlhd S_4$ is a normal subgroup with order $8$. Then $S_4 / H$ is a quotient group with order $3$. Since all groups with order $3$ are cyclic, $S_4 / H$ is cyclic and it's two non-identity cosets $C_1$ and $C_2$ are generators of $S_4 / H$ and have order $3$. So, for example, $(C_1)^3 = H$, since $H$ is the identity. Observe that $S_4$ has only $8$ elements with order $3$. Since the two non-identity cosets require $16$ elements between them, at least one of them, say $C_1 \in S_4 / H$, must contain elements that have order $2$ or $4$ (since the elements of $S_4$ have orders $1$, $2$, $3$, and $4$ and the order $1$ element is in $H$--the identity coset,and we run out of elements with order $3$). If $\sigma \in C_1$ has order $2$, then $(C_1)^3 = \sigma^3 H = \sigma H \ne H$ which is a contradiction. If $\sigma \in C_1$ has order $4$, then $(C_1)^3 = \sigma^3 H \ne H$ which also is a contradiction. Therefore, $H$ is not normal and $S_4$ cannot have any normal subgroups with order $8$.

I also show that that $S_4$ can't have a normal subgroup with order $3$ by contradiction. Suppose $H \unlhd S_4$ is a normal subgroup with order $3$. Then $S_4 / H$ is a quotient group with order $8$ (i.e. with $8$ cosets). Since $H$ only uses two of the order $3$ elements in $S_4$, at least one of the non-identity cosets in $S_4 / H$, say $C_1$, contains an element with order $3$. Let $\sigma \in C_1$ with $|\sigma| = 3$. Then $(C_1)^3 = \sigma^3 H = H$. But that means that $<C_1>$ is a subgroup of $S_4 / H$ with order $3$. That's a contradiction since the order of subgroups of $S_4 / H$ must divide $8$. Therefore, $H$ is not a normal subgroup and $S_4$ cannot have any normal subgroups with order $3$.