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

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?

• Look fine, and nice, to me. Commented Feb 18, 2022 at 12:39
• $A$ contains at most 7 elements of order 2 and $B$ contains only 2 elements of order 3. So the existences of $X$ and $Y$ are marginally easier than you make them out to be. Not that it matters much for the correctness. Commented Feb 18, 2022 at 12:54

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.

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 $$$$ is a subgroup of $$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$$.