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?


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.

  • $\begingroup$ I am really impressed esp. about your idea in part 1. Thank you very much. Just one question: as it's also obvious from the comparison you suggested in part 1, the book has chosen $r^ns$ in the meaning of 'first $r^n$ then $s$ happen'; is this (weird) choice (instead of $sr^n$) a standard (i.e. in every textbook) notation or it's just this book's choice? Thank you. $\endgroup$ – L.G. Jul 24 '15 at 16:43
  • $\begingroup$ ... e.g. in clockwise sense, 1,3,5 is $id$ and 5,3,1 is $r^2s$ (in bijective 'relation'), i.e. in the bijection to go from 1,3,5 to 5,3,1 replacement first must $r^2$ happen then $s$ but purely in $S_3$ first must $s$ happen then $r^2$! $\endgroup$ – L.G. Jul 24 '15 at 17:14
  • $\begingroup$ There is no consistency. Different textbooks use different conventions. $\endgroup$ – Gerry Myerson Jul 24 '15 at 23:50
  • $\begingroup$ Your solution in the first part, made a question for me which I asked here1 and here2. Nobody is agree with what I said, would you please help me with that, because I think my question is reasonable and I am pretty sure you will understand what I mean. Thank you. $\endgroup$ – L.G. Jul 28 '15 at 7:05

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).

  • $\begingroup$ Part 1 of your answer is not answering proof of isomorphism of $S_3$ and $K$. $\endgroup$ – L.G. Jul 24 '15 at 12:16
  • 2
    $\begingroup$ $K$ has $6$ elements, there are (up to isomorphism) only $2$ groups of order $6$, namely $S_3$ and $\mathbb{Z}/6\mathbb{Z}$. $K$ is not abelian and of order $6$, and is therefore isomorphic to $S_3$. $\endgroup$ – Krijn Jul 24 '15 at 12:19
  • $\begingroup$ @Krijn - would you please tell me a book or link to learn about "K has 6 elements, there are (up to isomorphism) only 2 groups of order 6, namely S_3 and Z/6Z"; I am sorry but it is absolutely unfamiliar for me. Thank you $\endgroup$ – L.G. Jul 24 '15 at 16:16
  • 1
    $\begingroup$ Hmm, I could not think of a book immediatly (and a link such as en.wikipedia.org/wiki/List_of_small_groups would not explain this very well, I think) but you could prove it by writing out the multiplication tabels of a group of $6$ elements. You will see that there are only two possible multiplication tables, one of them corresponding to $S_3$ and one to $\mathbb{Z}/6\mathbb{Z}$. This is however quite some work. Most of the time it is assumed that one knows all groups up to order $14$ or so, as there are not that many. $\endgroup$ – Krijn Jul 24 '15 at 17:00
  • $\begingroup$ An outline of a proof there're only two non-isomorphic groups of order $6$: Let $G$ be a non-cyclic group of order $6$. $G$ has no elements of order $6$, and cannot have all (non-identity) elements of order $2$, or else $G$ is abelian, and thus would contain a subgroup of order $4$ generated by two order $2$ elements. But $4$ does not divide $6$. Thus $G$ contains an element $a$ of order $3$, and an element $b$ of order $2$. If $a,b$ commute, $ab$ has order $6$, and so it can be shown that $ba$ can only be $a^2b$. Then $a \mapsto (1\ 2\ 3)$ and $b \mapsto (1\ 2)$ is the desired isomorphism. $\endgroup$ – David Wheeler Jul 24 '15 at 23:31

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