# Is there an elegant way to determine which subgroups of $S_3$ are normal?

I have a homework problem which reads

List all subgroups of $S_3$ and determine which subgroups are normal and which are not normal.

I understand the definitions of subgroup and normal subgroup, and so I can answer the question easily enough by brute force: find all of the subgroups, and then for each $n$ in a subgroup $N$, make sure that $sns^{-1} \in N$ for all $s \in S_3$. With a group of order six this is even doable. I feel like there has to be a more elegant way, though. Any hints? (Just an answer of the form “you should read about $x$” would be fine.)

• Hint: $H$ is a normal subgroup if $[G:H] = 2$. – user9413 Sep 4 '11 at 6:33

## 3 Answers

The keyword is "cycle structure". (I will leave it as an exercise for you to discover the meaning of "cycle structure of a permutation".)

Exercise 1: If $n$ is a positive integer and if $\alpha,\beta\in S_n$, then $\alpha$ and $\beta$ are conjugate in $S_n$ if and only if $\alpha$ and $\beta$ have the same cycle structure.

There are three different possible cycle structure for elements of $S_3$ (if one includes the cycle structure of the identity element of $S_3$). The two elements in $S_3$ of order $3$ share one cycle structure, the three elements in $S_3$ of order $2$ share a different cycle structure, and finally the identity element in $S_3$ has the trivial cycle structure.

Exercise 2: Prove that a subgroup of $S_3$ of order $2$ is never a normal subgroup of $S_3$. How many subgroups of $S_3$ have order $2$? (Hint: Exercise 1 is relevant.)

Exercise 3: Prove that there is exactly one subgroup of $S_3$ of order $3$ and that this subgroup of $S_3$ is a normal subgroup of $S_3$.

Exercise 4: Prove that in addition to the trivial subgroup and the entire group $S_3$, you have determined all subgroups of $S_3$ in Exercise 2 and Exercise 3. (Hint: Lagrange's theorem.)

Challenge Exercise: If $n$ is a positive integer factor of $24$, then determine the number of normal subgroups of $S_4$ of order $n$. (Hint: a basic knowledge of group actions is very helpful and a knowledge of Sylow theory probably trivializes the question.)

I hope this helps!

• In the first exercise, when you say “$\alpha$ and $\beta$ are conjugate in $S_n$” does that mean that $\alpha \beta \alpha^{-1} = \beta$? – bdesham Sep 5 '11 at 3:38
• @bdesham No. We have $\alpha\beta\alpha^{-1}=\beta$ if and only if $\alpha$ and $\beta$ commute in $S_n$. We say that $\alpha$ and $\beta$ are conjugate in $S_n$ if and only if there exists $\gamma\in S_n$ such that $\gamma^{-1}\alpha\gamma=\beta$. – Amitesh Datta Sep 5 '11 at 5:05
• @bdesham Exercise 5: We say that $\alpha$ is conjugate to $\beta$ if there exists $\gamma\in S_n$ such that $\gamma^{-1}\alpha\gamma=\beta$. In this case, we write $\alpha\equiv \beta$. Prove that $\equiv$ is an equivalence relation on $S_n$. – Amitesh Datta Sep 5 '11 at 5:10

Conjugating by a permutation amounts to renumbering the numbers in the conjugated permutation. For instance, $(132)(12)(132)^{-1}=(31)$, since the conjugating permutation renumbers $1$ to $3$ and $2$ to $1$ in the conjugated permutation. So you only have to check whether the subgroup is invariant under arbitrary renumberings, which is pretty straightforward.

For instance, for $S_4$, the identity and the three permutations of the form $(12)(34)$ form a subgroup, and this is normal since it's invariant under renumberings. More generally, a normal subgroup has to consist of entire conjugacy classes, and in a symmetric group a conjugacy class corresponds to a cycle type, so a subgroup is normal if and only if for each cycle type it contains either all or none of the permutations of that type.

Consider $S_3$ as isometries of an equilateral triangle with vertices labelled $\{1,2,3\}$. Then reflections in the lines passing through a vertex and center of opposite edge are orientation reversing isometries; have order two and so a reflection together with identity forms a subgroup of order $2$.

If we consider isometries which preserve the orientation, then they form a normal subgroup of $S_3$. [In fact, if $G$ is a group(not necessarily full group) of isometries of $R^n$ and $H$ is the subgroup of $G$, of orientation preserving isometries, then $H\triangleleft G$. For a geometric way to look at (some) groups, see Algebra-Artin, and Basic Notions of Algebra- Shafarevich]