Prove $K\cong S_3$

Question

In the book Abstract Algebra by Thomas W. Judson, Ch. 9 Example 13 -

Example 13. The dihedral group $D_6$ is an internal direct product of its two subgroups $H={\{id,r^3}\}$ and $K={\{id,r^2,r^4,s,r^2s,r^4s}\}$.

It can easily be shown that $K\cong S_3$; consequently, $D_6\cong Z_2×S_3$.

My questions:

1- How to prove $K\cong S_3$? Obviously one way is to write all elements of S_3 and find a bijection between its elemnts and K's elements, which is a lot way to go. Is there any shorter way?

2- Up to isomorphism Cartesian direct product and internal direct product seems to be in coincidence. Is it true in general case?

Answer 1

Some other ways to tell that $K$ is $S_3$:

1. Label the vertices of the hexagon 1 through 6. Forget about 2, 4, and 6, and see what $K$ does to 1, 3, and 5. You will see that the 6 elements of $K$ are precisely the 6 permutations of 1, 3, 5, thus, precisely $S_3$.

2. Group presentations: note that $S_3$ is generated by $a=(12)$ and $b=(123)$ subject to the relations $a^2=1$, $b^3=1$, and $ba=ab^2$. Note that $K$ is generated by $a=s$ and $b=r^2$, subject to the relations $a^2=1$, $b^3=1$, and $ba=ab^2$. Conclude that the two groups are isomorphic.

Answer 2

1) There are only two groups with six elements. It is enough to check that $K$ is not abelian.

2) Yes. In fact, one definition of internal direct product could be that the product of the two subgroups inside the ambient group behaves like a direct product (i.e. is isomorphic to the Cartesian direct product).