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I wish to prove that $S_3$,$S_4$ (permutations on $3,4$ elements respectively) are solvable.

I know that $D_6,D_{24}$ ($D_n$=Dihedral group of order $n$) are solvable and if I could prove that $S_3$ is isomorphic to $D_6$, $S_4$ is isomorphic to $D_{24}$ it will do the trick.

With $S_3$ and $D_6$ I can do this 'the hard way' by defining the isomorphism, but with $S_4$ and $D_24$ proving that the function I define respects the multiplication of the group is too much work and doesn't seem like the 'right' way.

Maybe it is sufficient to prove that S4 is solvable, since there is a $1-1$ homomorphism from $S_3 \to S_4$ ?

I could use some help with this...

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+1. Well posed question. –  user21436 Mar 15 '12 at 9:30
    
The easy way is to simply verify that the factors in the normal series $1 \triangleleft V_4 \triangleleft A_4 \triangleleft S_4$ are abelian, where $V_4$ is the Klein 4 group, which is isomorphic to $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ (abelian). The other factors are too small to be nonabelian. –  William DeMeo Mar 15 '12 at 9:43
    
Why does it prove the claim ? |A4|=12, and there are groups of samller order that are not abelian, no ? –  Belgi Mar 15 '12 at 9:50
1  
What I mean is check that the factor groups are abelian. Perhaps your book isn't giving you this definition of solvability. If that's the case, you should review the definition here. So, you just need to check that $S_4/A_4$ is abelian, that $A_4/V_4$ is abelian, and that $V_4/1 = V_4$ is abelian. But this is obvious by considering the orders of these groups. –  William DeMeo Mar 15 '12 at 10:33
    
@WilliamDeMeo I have shown the OP completely how to do this. –  fpqc Mar 15 '12 at 11:38
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2 Answers

up vote 6 down vote accepted

First of all let me point out that $D_{24}$ cannot be isomorphic to $S_4$. $S_4$ does not have an element of order 12, where as $D_{24}$ does. This is because $D_{24}$ as a set is

$$\{1,r,r^2 ,\ldots r^{11}, sr, sr^2, \ldots sr^{11}\}$$

where the $r_i's$ are the rotations and the $sr^j$ the reflections; so that $r$ is your element of order 12. For the sake of knowledge though $S_4$ does have a subgroup of order 12 sitting inside it that is just $A_4$. Just to tell you though the subgroups of $S_4$ are (up to isomorphism):

$$\{e\}, \hspace{2mm} C_2, \hspace{2mm} C_3, \hspace{2mm} V_4, \hspace{2mm} C_4, \hspace{2mm} S_3, \hspace{2mm} D_8, \hspace{2mm} A_4, \hspace{2mm} S_4.$$

Actually it is not so hard to see directly that $S_3$ and $D_6$ are isomorphic, just list down the elements explicitly and if they obey exactly the same relations, their Cayley Tables are the same so you have your isomorphism.

By the way I think for your last line you want a surjective homomorphism from $S_4$ to $S_3$ to show that $S_3$ is solvable only if $S_4$ is. The way you do that is like this:

We get a homomorphism $\varphi$ from $S_4$ to $S_3$ by considering the action of $S_4$ on a finite set $Y = \{\Pi_1, \Pi_2, \Pi_3\}$ where $\Pi_1 = \left\{\{1,2\}, \{3,4\}\right\}$, $\Pi_2 = \left\{\{1,3\}, \{2,4\}\right\}$ and $\Pi_3 =\left\{ \{1,4\}, \{2,3\}\right\}$. Each $\Pi_i$ is one way of partitioning a set of 4 elements into two subsets of equal cardinality. $S_4$ acts on a $\Pi_i$ by permuting the numbers in each partition, so for example the cycle $(1234)$ interchanges $\Pi_1$ and $\Pi_3$ while keeping $\Pi_2$ fixed, so that $\varphi( (1234) ) = (13)(2)$. It is easy to see that $\ker \varphi$ is the Klein four- group $V_4= \{e, (12)(34), (14)(23), (13)(24)\}$ where $e$ denotes the identity in $S_4$. If you go further you should know that by the first isomorphism theorem the quotient group $S_4/V_4$ is isomorphic to the image of $\varphi$.

Now to see the surjectivity of $\varphi$, recall we have the counting formula

$|S_4| = |\ker \varphi||\operatorname{Im} \varphi|$. Therefore the order of the image is $24/4 = 6$. Now recall the image is a subgroup of $S_3$ so if we have a subgroup of order 6 sitting inside $S_3$, it must be the whole of $S_3$.

To answer your question on why $S_4$ is solvable, note that the only normal subgroups in $S_4$ are the trivial group, the Klein 4-group $V_4$, $A_4$ and $S_4$ itself.

So we have a subnormal series

$$\{e\} \triangleleft V_4 \triangleleft A_4 \triangleleft S_4.$$

Now you should know that $V_4/\{e\} \cong V_4$ that is abelian (all groups of order 4 are). You also know that $|A_4/V_4| = 3$ so that $A_4/V_4 \cong C_3$ which is abelian. Finally $|S_4/A_4| = 2$ so that $S_4/A_4 \cong C_2$ that is also abelian. So it remains to check that $A_4$ is normal in $S_4$ and $V_4$ is normal in $A_4$. The former you already know for the index of $A_4$ in $S_4$ is two. For the second I suggest listing out the conjugacy classes of $A_4$ and show that $V_4$ is a union of conjugacy classes. It should not be so hard to compute all the conjugacy classes of $A_4$ since there are only 12 elements. Once you've done all of this, you've proved that $S_4$ is solvable!

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Nice Answer. : ) –  user21436 Mar 15 '12 at 11:14
    
@KannappanSampath Thanks. –  fpqc Mar 15 '12 at 11:15
    
+1 A thorough explanation. –  Jyrki Lahtonen Mar 15 '12 at 11:32
    
In the list of subgroups I've replaces $S_6$ by the obviously intended $S_3$. However the "group" in braces is not a subgroup, and I didn't know what was intended by it, an explicit rendering of $V_4$? Or maybe $C_4$ which is missing from the list? Also note that some "subgroups" occur more than once: $C_2$ ($9=6+3$ times, two non-conjugate types), $C_3$ ($8$ times), $C_4$ ($3$ times), $S_3$ ($4$ times). –  Marc van Leeuwen Mar 15 '12 at 12:35
    
@MarcvanLeeuwen I have edited the mistakes. –  fpqc Mar 15 '12 at 13:10
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That $D_6$ is isomorphic to $S_3$ is easy: consider how $D_6$ acts one the corners of the triangle of which it is the symmetry group. However $D_{24}$ is not isomorphic to $S_4$ (consider elements of order $12$, or even of order $6$, which $S_4$ does not posess), so no amount of effort will suffice to prove it that way. The injective homomorphism (one of the many) $S_3\to S_4$ will not help you with $S_4$ either. However there is a surjective homomorphism $S_4\to S_3$ (think of a tetrahedron and the lines joining the midpoints of its sides) that will help you to prove that $S_4$ is solvable.

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Maybe I understand this wrong ? :"The groups S3,S4 and are isomorphic to the group of symmetries of the triangle and to the group of symmetries of the tetrahedron" - it's from a book I'm reading... –  Belgi Mar 15 '12 at 9:38
    
@Belgi The group of symmetries of the tetrahedron is not $D_{24}$. –  Alex Becker Mar 15 '12 at 9:45
    
@Belgi: Yes, but $D_{24}$ is not the symmetry group of a tetrahedron, but rather of a (plane) regular $12$-gon (pronounced dodecagon). –  Marc van Leeuwen Mar 15 '12 at 9:45
    
Oh, what is the notation for the symmetry group of a tetrahedron ? thanks for the help! –  Belgi Mar 15 '12 at 9:49
    
@Belgi: As you quoted yourself, the symmetry group of a tetrahedron is called (or isomorphic to) $S_4$. It is the group $D_{24}$ that is unrelated to your question. –  Marc van Leeuwen Mar 15 '12 at 12:25
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