# Thread: Soluble Groups....

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
$\displaystyle A_3$ and $\displaystyle S_3/A_3$ are both solvable because they are abelian. therefore $\displaystyle S_3$ is solvable. $\displaystyle V,$ the klein-4 group, and $\displaystyle A_4/V$ are both abelian and hence solvable. therefore $\displaystyle A_4$

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

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

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

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