Is $S_3$ isomorphic to anything other than $D_3$?

Question

Problem: Is $S_3$ isomorphic to anything other than $D_3$?

I know that $S_3$ is isomorphic to $D_3$. Is it isomorphic also to some $\mathbb{Z}_n$ or $\mathbb{Z}_m \times \mathbb{Z}_n$?

Answer 1

There are only two non-isomorphic groups of order $6$: $Z_6$ and $S_3\cong D_3$. Therefore, $S_3\cong S_3/Z(S_3)$ cannot be isomorphic to any product of abelian groups.

Answer 2

When we talk about groups, there isn't really a good definition of "different" besides "non-isomorphic." For example, I can define the group $G_1$ generated by some element $x$ satisfying $x^{10}=1$, and I can also define the group $G_2$ generated by some element $y$ satisfying $y^{10}=1$. On one hand, these groups are clearly the same thing -- if you rename $x$ to $y$, you get $G_2$ from $G_1$, and vice versa; equivalently, they are both the cyclic group of order $10$. On the other hand, their elements aren't actually equal: $x$ does not equal $y$; there is no element of $G_2$ that "is" $x$. Our intuition about $G_1$ and $G_2$ being "the same" actually means that they are isomorphic.

There are, of course, other ways of defining groups that are isomorphic to the cyclic group of order $10$ that are less obviously "the same." We can, if we want, define a group $G_3$ generated by two elements $a$ and $b$ for which $a$ and $b$ commute, $a^2=1$, and $b^5=1$. There are $10$ elements of this group, and they look like $$\{1,b,b^2,b^3,b^4,a,ab,ab^2,ab^3,ab^4\}.$$ It isn't that hard to check that $G_3$ is isomorphic to $G_1$ via $a\mapsto x^5$ and $b\mapsto x^2$. So, $G_3$ "is" the cyclic group of order $10$, which is to say that it is isomorphic to (any instance of) the cyclic group of order $10$.

So, for any group $G$, there are lots of different ways we can write groups that are isomorphic to $G$. For example, we can show that $S_3$ and $D_3$ are isomorphic, even though they do not "look the same." There are (probably) arbitrarily complex ways to write groups that are isomorphic to $S_3$, but in some sense, due to being isomorphic to $S_3$, they are actually $S_3$. One other example of a group that is "non-obviously" $S_3$ is $GL_2(\mathbb F_2)$, as noted by Qiaochu Yuan in the comments. In fact, because $|S_3|=6$ is a small number, it isn't too hard to show by hand that any group of order $6$ is either isomorphic to $\mathbb Z/6\mathbb Z$ or $S_3$ (we say that these are "the only groups of order $6$, up to isomorphism"). As a result, any group of order $6$ you can come up with will either be (isomorphic to) $\mathbb Z/6\mathbb Z$ or (isomorphic to) $S_3$.

One can, however, show that there are no groups in some particular classes of groups that are isomorphic to $S_3$. For example, the groups $\mathbb Z/m\mathbb Z$ and $\mathbb Z/m\mathbb Z\times \mathbb Z/n\mathbb Z$ are abelian groups. So, there is no way any of them are isomorphic to $S_3$.