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Can anyone explain me what would be the procedure for building a subgroup $H\leq S_4$ of order $8$?

I started obviously as $H=\{id$. Then I added two disjoint $2$-cycles $(1\ 2), (3\ 4)$ for they commute and they are equal to their inverse, that is, $$H=\{id, (1\ 2), (3\ 4), $$ then I added the product of this $2$-cycles, $$H=\{id, (1\ 2), (3\ 4), (1\ 2)(3\ 4).$$

Now I don't know what I should add because I have to worry about both with the products and inverses.

Is there a general guideline for building this kind of subgroup?

Thanks

Obs:

(i) $S_4$ is the group of the permutations of the first $4$ integers.

Thanks

PtF
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    Hint: Since $H$ has order 8, its elements must have order $1,2,4$ or $8$, by Lagrange's theorem. On the other hand, elements of $S_4$ have orders $1,2,3,4$. What is left if you discard from $S_4$ all the elements of order $3$? – MJD Oct 20 '14 at 15:43
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    The Klein $4$-group has order $4$! You might try starting with a $4$-cycle such as $(1,2,3,4)$ and attempt to build up a dihedral group. – James Oct 20 '14 at 15:43
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    Note also that such a subgroup, since it can't consist solely of even permutations ($8\nmid 12$) must have $4$ odd and $4$ even permutations, so other than $e$, there must be exactly $3$ even permutations. – rogerl Oct 20 '14 at 15:54
  • To your group you can add a 4-cycle such that its square is $(12)(34)$. Can you find such a 4-cycle? – Jyrki Lahtonen Oct 20 '14 at 19:58

2 Answers2

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A subgroup of order $8$ in $S_4$ is a $2$-Sylow subgroup, so it is the wreath product $$S_2\wr S_2= (S_2\times S_2)\rtimes S_2,$$ where the right most $S_2$ acts by permuting components of the first two. In other words, let the first $S_2$ be $\{1,(12)\}$, the second be $\{1,(34)\}$, then the third is $\{1,(13)(24)\}$.

Another way is to see that the subgroup $$\{1,(12)(34),(13)(24),(14)(23)\}$$ is normal in $S_4$, and then take any of the transpositions $(ij)$ as an extra generator.

Quang Hoang
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    What are the odds someone asking such a basic question has the faintest idea what a wreath product is? – Timbuc Oct 20 '14 at 16:26
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    @Timbuc Very likely. In fact I would guess that the OP didn't know about Sylow subgroups. That explains the explicit constructions. – Quang Hoang Oct 20 '14 at 16:30
  • @timbuc Answers on this site aren't only for the person asking the question. They are intended to form a permanent collection for perusal by future visitors. Some of those visitors may be interested in a solution involving wreath products. – MJD Oct 21 '14 at 13:10
  • @MJD. I'd agree with that as long as the answer also covers the question in the assumed level the OP has, as answering the question for the poster is the main goal of this site, I believe. – Timbuc Oct 21 '14 at 13:47
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Look at $S_4$ as the group of permutations of the vertices ABCD of a square.

The plane isometries fixing the square induce obvious permutations on the vertices, defining a map $$ D_4\hookrightarrow S_4. $$ Its image is your subgroup of order $8$.

You get all others by reshuffling the labels of the vertices, since this operation induces an inner automorphism of $S_4$ and you know that all subgroups of order $8$ of $S_4$, being its $2$-Sylows, are conjugated.

Andrea Mori
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