# Subgroups of the Permutations on Three Elements

Today we will find all the subgroups of the group of the permutations on a three-element set. Our goal here will be to find the subgroups, determine the left and right cosets of each of these subgroups, find which subgroups are normal and, for the normal subgroups, find the group modulo this normal subgroup.

## The Group of Permutations

Before determining each of our subgroups, we will first need to find the elements in our group. Since $$S_{3}$$ is the group of permutations on any three-element set, we will assume our set is $$\left\{1,2,3\right\}$$ so that we can denote these. Then, we can denote these elements as
\begin{align*}
\begin{pmatrix}
1 & 2 & 3 \\
1 & 2 & 3
\end{pmatrix},
\begin{pmatrix}
1 & 2 & 3 \\
2 & 3 & 1
\end{pmatrix},
\begin{pmatrix}
1 & 2 & 3 \\
3 & 1 & 2
\end{pmatrix}, \\
\begin{pmatrix}
1 & 2 & 3 \\
2 & 1 & 3
\end{pmatrix},
\begin{pmatrix}
1 & 2 & 3 \\
3 & 2 & 1
\end{pmatrix},
\begin{pmatrix}
1 & 2 & 3 \\
1 & 3 & 2
\end{pmatrix}
\end{align*}
or, we can use cycle notation and have these be $$(1), (123), (132), (12), (13)$$ and $$(23)$$ respectively. With these elements, we can now make a Cayley table given the composition of these permutations. We then arrive at

 (1) (123) (132) (12) (13) (23) (1) (1) (123) (132) (12) (13) (23) (123) (123) (132) (1) (13) (23) (12) (132) (132) (1) (123) (23) (12) (13) (12) (12) (23) (13) (1) (132) (123) (13) (13) (12) (23) (123) (1) (132) (23) (23) (13) (12) (132) (123) (1)

While we won’t make use of it in this post, we can compare this table to the one given in Symmetric group of an equilateral triangle. If we do so, we will note that we can find an isomorphism between the two groups. Therefore, if we wanted to approach the permutations of a three element set visually, we could do so by picturing them as the symmetries of an equilateral triangle.

### Subgroups

Now that we’ve constructed our Cayley table and know what $$S_{3}$$ looks like, we can now start finding the subgoups of $$S_{3}$$. In order to keep track of what we’ve done, we will start by finding the cyclic subgroups generated by each of the elements. Then we will move on to finding subgroups generated by two or more elements.

### $$\langle (1) \rangle$$

We begin by noting that $$(1)$$ is the identity element of this group. Therefore, the group generated by $$(1)$$ will precisely be the trivial group consisting of only the identity. Furthermore, for every $$x \in S_{3}$$, we will have that $$x \langle (1) \rangle =\left\{x\right\}=\langle (1) \rangle x$$. That is, the left and right cosets for each element are precisely the sets consisting of only that element. Since the left and right cosets are the same, this will be a normal subgroup, and the $$S_{3}/\langle (1) \rangle \cong S_{3}$$.

### $$\langle (123) \rangle$$

We will now look at the subgroup generated by $$(123)$$. To do this, we find the $$(123) \circ (123)= (132)$$ and $$(123) \circ (132) =(132) \circ (123) =(1)$$. Therefore, the group generated by $$(123)$$ will be $$\left\{(1), (123), (132) \right\}$$. This also implies that the order of $$(123)$$ is 3.

In order to find the cosets of this subgroup, we note that $$x \langle (123) \rangle=\langle (123) \rangle$$ for any $$x \in \langle (123)\rangle$$ since it is a group. Therefore, we need only find the left and right cosets for the elements not in this subgroup. We will, therefore, start with $$(12)$$. We then find that
\begin{align*}
(12) \circ (1) =(12) \qquad (12) \circ (123) = (23) \qquad (12) \circ (132)=(13) \\
\end{align*}
Therefore, $$(12) \langle (123) \rangle= \left\{(12), (13), (23) \right\}=\langle (123) \rangle (12)$$.

If we continue to check the compositions with $$(13$$ and $$(23)$$, we will indeed find that $$x \langle (123) \rangle=\left\{(12),(13),(23)\right\}= \langle (123) \rangle x$$ for each $$x \in \left\{ (12),(13),(23)\right\}$$. Because the left and right cosets are the same for every element in $$S_{3}$$ we see that $$\langle (123)\rangle$$ is also a normal subgroup of $$S_{3}$$.

#### $$S_{3} / \langle (123) \rangle$$

Now, if we let $$R=\left\{(1),(123),(132)\right\}$$ and $$F=\left\{(12),(13),(23)\right\}$$, we find that any element in $$R$$ composed with any element in $$R$$ is again in $$R$$, any element in $$R$$ when composed with an element in $$F$$ (in either direction) is in $$F$$ and any element in $$F$$ composed with any element in $$F$$ is in $$R$$. This gives us the Cayley Table

 $$R$$ $$F$$ $$R$$ $$R$$ $$F$$ $$F$$ $$F$$ $$R$$

Extra Note

If we look at the isomorphism from $$S_{3}$$ to the symmetries of an equilateral triangle, we see that that the elements in $$R$$ would be mapped to rotations, whereas the elements in $$F$$ would be mapped to flips. This is foreshadowed by the use of $$R$$ and $$F$$. Looking at this specific case, we do indeed see that a rotation and rotation is again a rotation, a flip and flip is a rotation and one of each gives us a flip. It is particularly nice to see this is the specific case of permutations on the three element set since we proved that this was true for all finite groups of symmetries in Da Vinci’s Theorem.

Furthermore, we should note that following the same steps, we will find that $$\langle (132)\rangle =\langle (123)\rangle$$.

### $$\langle (12) \rangle$$

We begin to find the group generated by $$(12)$$ by finding that $$(12) \circ (12) =(1)$$. Since we have already arrived by at the identity, we will note that $$(12)$$ has order $$2$$, so $$\langle (12) \rangle =\left\{(1),(12)\right\}$$.

Now that we know the subgroup generated by $$(12)$$ we can find the left and right cosets of this subgroup. We know the left and right cosets of $$(1)$$ and $$(12)$$ will be this subroup. However, we will find the other cosets by taking the compositions with these elements. This gives us
\begin{align*}
(123) \langle (12) \rangle &=\left\{(123),(13) \right\} \\
\langle (12) \rangle (123)&= \left\{(123),(23) \right\} \\
(132) \langle (12) \rangle &=\left\{(132),(23) \right\} \\
\langle (12) \rangle (132)&= \left\{(132),(13) \right\} \\
(13) \langle (12) \rangle &=\left\{(123),(13) \right\} \\
\langle (12) \rangle (13)&= \left\{(132),(13) \right\} \\
(23) \langle (12) \rangle &=\left\{(132),(23) \right\} \\
\langle (12) \rangle (23)&= \left\{(123),(23) \right\}
\end{align*}

In this case, we see that the left cosets and right cosets are not equal. In particular, $$(123) \langle (12) \rangle =\left\{(123),(13) \right\} \neq \left\{(123),(23) \right\}=\langle (12) \rangle (123)$$. Therefore, this is not a normal subgroup, so we cannot find $$S_{3}$$ modulo this subgroup.

Extra Note

A similar process will give you the groups generated by $$(13)$$ and $$(23)$$. In each case, the order of the subgroups will be 2 and the subgroups will not be normal subgroups.

### Non-cyclic subgroups

Instead of working through all of the compositions to find another subgroup, we will look at the order of the elements and recall that the size a subgroup must divide the size of the group. Since $$|S_{3}|=6$$, this implies that any subgroup must have size $$1,2,3$$ or $$6$$. Now that we have found all the cyclic subgroups (some of each with order 1, 2 and 3), we can look at any groups not generated by one element.

In this case, suppose that we have a group generated by two of the permutations $$(12),(13)$$ or $$(23)$$. Then, note that $$2$$ would have to divide the order of the resulting group. Furthermore, there will be at least 3 elements (the two we chose and the identity) so the only option left is that the order of the group would be 6. That is, we would have the entire group.

We have already seen if we use two, or three, elements from the set $$(1),(123),(132)$$, we will get the cyclic group generated by $$(123)$$ or $$(132)$$. Therefore, using a combination of these elements will not give us a new subgroup.

Finally, if we choose $$(123)$$ and $$(132)$$ along with one of the $$(12),(13)$$ or $$(23)$$, we will note that the size of the group must be divisible by both 2 and 3. Therefore, the order of the group must be 6. That is, we will again have all of $$S_{3}$$.

Therefore, the only subgroups of $$S_{3}$$ will be $$S_{3}$$ itself and the cyclic subgroups we’ve already found. Putting these into a subgroup lattice, we find

### Homomorphic Images

Now that we’ve found that the normal subgroups of $$S_{3}$$ are $$S_{3}$$ itself, $$\langle (123) \rangle$$ and the trivial group $$\left\{e\right\}$$, we can find all homomorphic images of $$S_{3}$$. In particular, since $$S_{3}/\left\{e\right\} \cong S_{3}$$, $$S_{3}/ \langle (123) \rangle \cong \mathbb{Z}_{2}$$ and $$S_{3}/S_{3} \cong \left\{e\right\}$$, we have that all homomorphic images of $$S_{3}$$, up to isomorphism, are $$S_{3}, \mathbb{Z}_{2}$$ and $$\left\{e\right\}$$.

## Conclusion

Thank you for reading and as we worked through finding all the subgroups of $$S_{3}$$. I hope this helped you with finding subgroups, cosets, normal subgroups and group modulo normal subgroups. If it did, be sure to check out the other posts at Algebraic Structures and our YouTube channel.

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