1. ## Soluble Groups....

Hi guys,

How can you prove that S3, S4 & A4 are soluble, & that A5 & S5 are not?

I know that every group of order less than 60 is soluble, but how would you proove it without using that fact in these cases? :-)

I have absolutely no idea where to start! :-s (our lecturer did not explain this bit very well AT ALL! :-( )

Any of the proofs, or even some explanation about what Soluble Groups actually mean & their significance, would be VERY MUCH appreciated! :-)

Many, many thanks. x

2. Originally Posted by AAM
Hi guys,

How can you prove that S3, S4 & A4 are soluble, & that A5 & S5 are not?

I know that every group of order less than 60 is soluble, but how would you proove it without using that fact in these cases? :-)

I have absolutely no idea where to start! :-s (our lecturer did not explain this bit very well AT ALL! :-( )

Any of the proofs, or even some explanation about what Soluble Groups actually mean & their significance, would be VERY MUCH appreciated! :-)

Many, many thanks. x
$A_3$ and $S_3/A_3$ are both solvable because they are abelian. therefore $S_3$ is solvable. $V,$ the klein-4 group, and $A_4/V$ are both abelian and hence solvable. therefore $A_4$

is solvable. also obviously $S_4/A_4$ is abelian and hence solvable. thus $S_4$ is solvable. $A_5',$ the commutator subgroup of $A_5,$ is a normal subgroup of $A_5.$ so either $A_5'=\{1\}$

or $A_5'=A_5$ because $A_5$ is simple. but $A_5' \neq \{1 \}$ because $A_5$ is not abelian. thus $A_5'=A_5$ and hence $A_5^{(k)}=A_5 \neq \{1 \},$ for all $k.$ so $A_5$ is not solvable which obviously

implies that $S_5$ is not solvable because every subgroup of a solvable group is solvable.

3. Originally Posted by AAM
Hi guys,

How can you prove that S3, S4 & A4 are soluble, & that A5 & S5 are not?

I know that every group of order less than 60 is soluble, but how would you proove it without using that fact in these cases? :-)

I have absolutely no idea where to start! :-s (our lecturer did not explain this bit very well AT ALL! :-( )

Any of the proofs, or even some explanation about what Soluble Groups actually mean & their significance, would be VERY MUCH appreciated! :-)

Many, many thanks. x
As NonCommAlg said you need to know that $A_5$ is a simple group - which means it has no proper non-trivial normal subgroup (so if $N\triangleleft A_5 \implies N = \{ e \}\text{ or }N=A_5$). If $A_5$ is simple then it certainly cannot be solvable. This is easy to see since you cannot write a subnormal series because you will not be able to form normal subgroups of $A_5$ other than the trivial and itself, hence $\{ e \}\triangleleft A_5$ is the only subnormal series for $A_5$. Since the factor group $A_5/\{e\}\simeq A_5$ is not abelian it means the group cannot be sovable. This means $S_5$ (in fact $S_n$) cannot be solvable because $A_5$ is a subgroup, and a subgroup of a solvable group is solvable. Thus, all we need to show is that $A_5$ is simple.
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Here is one method with orbits and stablizers. We will derive the class equation for $A_5$. First, we will work with $S_5$ to help figure out the conjugacy classes for $A_5$. For $S_5$ there are 7 conjugacy classes. These consist of: 1-cycle, 2-cycles, 3-cycles, 4-cycles, 5-cycles, (3-cycles)(2-cycles),(2-cycles)(2-cycles). The number of 1-cycles is just 1. The number of 2-cycles is ${5\choose 2} = 10$. The number of 3-cycles is $2!\cdot {5\choose 3} = 20$. The number of 4-cycles is $3!\cdot {5\choose 4} = 30$ and the number of 5-cycles is $4! = 24$. The number of (3-cycles)(2-cycles) is ${5\choose 3}=20$. The number of (2-cycles)(2-cycles) is $3{5\choose 4} = 15$. These are the conjugacy classes for $S_5$ and the number of elements in each class. What we will be interested in is the class equation for $A_5$.

Two elements that are conjugate in $A_5$ must be conjugate in $S_5$, obviously. However, elements that are conjugate in $S_5$ are not necessarily conjugate in $A_5$. Remember a result from group theory that the number of elements conjugate to $x\in A_5$ would be equal to the index of $C(x)$ (centralizer of $x$ in $A_5$) in $A_5$. The conjugacy classes for $A_5$ consist of: 5-cycles, 3-cycles, 1-cycle, (2-cycles)(2-cycles). We need to count the number of elements in each conjugacy class, we will use our work above to help answer this question. Let us start easy, 1-cycle, is just the identity, so there is just one element in that conjugacy class. Let $x=(12)(34)$, the number of conjugates to this in $S_5$ is $15$, but necessary in $A_5$. We will prove that it is $15$ for $A_5$ as well. Let $[(12)(34)]_{S_5}$ be the order of the conjugacy class of $(12)(34)$ in $S_5$, therefore $15=[(12)(34)]_{S_5} = 120/|C_{S_5}(x)| \implies |C_{S_5}(x)|=8$. We notice that (if you do the computation) $\left< (1324),(13)(24)\right> \subseteq C_{S_5}(x)$, but since $|\left<(1324),(13)(24)\right>| = 8$ it means $C_{S_5}(x) = \left<(1324),(13)(24)\right>$. Thus, $C_{A_5}(x) = A_5\cap C_{S_5}(x) = \{ e, (12)(34),(13)(24),(14)(23)\}$. Thus, $|C_{A_5}(x)| = 4$ which means $[(x)]_{A_5} = [A_5:C_{A_5}(x)] = 15$ which means the conjugacy class of $x=(12)(34)$ in $A_5$ is the entire conjugacy class for that element in $S_5$. Let $y=(123)$ then (all these arguments are similar) $|(123)|_{S_5} = 20$ so $|C_{S_5}(y)| = 6$, it can now be shown $C_{S_5}(y) = \left< (123),(45)\right>$ thus $C_{A_5}(y) = A_5\cap C_{S_5}(y) = \left< (123)\right>$. This means $[y]_{A_5} = 20$ and so the conjugacy class of $(123)$ is also all of conjugates in $S_5$. Now we get to $z=(12345)$ (the last remaining case to consider). Again (similar argument, steps excluded), $C_{S_5}(z) = \left< (12345)\right>$ and so $C_{A_5}(z) = \left< (12345)\right>$. However, this means $[(12345)]_{A_5} = 12$. Thus, conjugacy class for $(12345)$ in $A_5$ is not all of the ones in $S_5$ (which have $24$ of them). Repeating the same argument with the remaining 12 5-cycles we see (by the same argument as just now) that they consist of $12$ conjugate elements too. Thus, the conjugacy class equation for $A_5$ is $60 = 1 + 12 + 12 + 15 + 20$. It now follows that $A_5$ is simple since we cannot write a non-trivial proper divisor of $60$ as a summand of those numbers.