Is $S_3 \times S_3 \cong D_{18}$?

Question

Here, $D_n$ denotes the dihedral group of degree $n$ (order $2n$).

I know that it is possible that there exists a isomorphism by the sole fact that they have the same order. By listing the elements of $S_3$ I noticed that we have 3 elements of order $2$ and $2$ elements of order $3$, and thus highest order of an element in $S_3 \times S_3$ is 6. For example, $((12),(123))$ has order $6$.

Are all my conclusions correct?

You are correct that highest possible order of an element in $S_3 \times S_3$ is 6. Do you know why this proves that $S_3 \times S_3$ cannot be isomorphic to $D_{18}$? — manthanomen, Dec 01 '16 at 05:58
To expound a bit further, what is the order of a generator of the rotation subgroup of $D_{18}$? — David Wheeler, Dec 01 '16 at 06:00
@manthanomen Not sure. I am just making observations. By your comment, I assume that there does not exist an element (or as many) in $D_{18}$ of order 2,3, or 6? — Quarternion, Dec 01 '16 at 06:02
All you need to do now is show that there exists an element in $D_{18}$ whose order is bigger than 6. Specifically, show there's an element of order 18. — manthanomen, Dec 01 '16 at 06:04
@DavidWheeler 18. I was confusing as to how many degrees we rotate by, which in this case is 20. — Quarternion, Dec 01 '16 at 06:09
@manthanomen well, that would be r, where r is the rotation of the n-gon by 20 degrees. — Quarternion, Dec 01 '16 at 06:09
So, one group has an element of order $18$, the other does not. Problem solved! — David Wheeler, Dec 01 '16 at 06:11

score 2 · Answer 1 · answered May 09 '25 at 10:12

I present two similar ways to see this.

One way to see this, as others have observed, is to look at the orders of elements in the two groups. Since $S_3 \times S_3$ is a direct product, the order of an element $(g,h) \in S_3 \times S_3$ will be equal to the lowest common multiple of $g$ and $h$. Each element in $S_3$ has order $1,2$ or $3$, hence an element in $S_3 \times S_3$ has order at most $6$. But $D_{18}$ has elements of order $18$. Since two isomorphic groups must have the same orders for their elements, we conclude $S_3 \times S_3 \not \cong D_{18}$.

Note also that since $S_3 \times S_3$ is a direct product, we can construct an injective homomorphism

$$\rho: S_3 \times S_3 \to S_6$$

defined by

$$\rho((g,h))(i) = \begin{cases} \rho(g)(i) & i=1,2,3 \\ \rho(h)(i-3) & i=4,5,6 \end{cases}$$

where a permutations $\sigma \in S_n$, $\sigma(i)$ denotes the image of $i$ under the permutation $\sigma$, for $i=1,\dots,n$.

Basically what I am saying is the following. Due to how in direct products the different parts of the tuples behave independently from each other, when we are given two permutations $g, h \in S_3$ acting on the elements $\{1, 2, 3 \}$, we can relabel $h$ so that it instead acts on $\{4, 5, 6 \}$. So, for example

$$\rho((1 \, 2 \, 3), ((1 \, 2 \, 3))) = (1 \, 2 \, 3) (4 \, 5 \, 6).$$

What this means is that $S_3 \times S_3$ can act faithfully on a set of size $6$, i.e. it occurs as a subgroup of $S_6$.

However, $D_{18}$ can't do this. If we were to embed $D_{18}$ as the subgroup of a symmetric group, we'd need to map one of its elements to an element of order $18$, which $S_6$ doesn't have. Indeed, $D_{18}$ naturally acts on a set of $18$ points, namely the vertices of an $18$-gon.

Kan't · Answer 2 · 2025-05-09T14:04:20.857

2

You are correct (I take for granted that you know that $D_{18}$ has an element of order $18$, which you didn't mention in the OP). Or also: $S_3\times S_3$ is centerless, whereas $D_{18}$ is not.

edited May 09 '25 at 14:04

answered May 09 '25 at 13:39

Kan't

4,819

Is $S_3 \times S_3 \cong D_{18}$?

2 Answers2