MHB What Are the Composition Series for \(A_4\) and \(S_3 \times \mathbb{Z}_2\)?

[mathmari]

Hey!

I want to find all the composition series for $A_4$ and $S_3\times\mathbb{Z}_2$.

A composition series for $G$ is $$1=S_0\leq S_1\leq S_2\leq \cdots \leq S_k=G$$ with $S_i\trianglelefteq S_{i+1}$ and $S_{i+1}/S_i$ is a simple group, right? (Wondering)

Could you give me some hints how we could find all the composition series for $A_4$ and $S_3\times\mathbb{Z}_2$ ? (Wondering)

How can we find all the subgroups $S_i$ ? (Wondering)

 

[Deveno]

Step 1, identity normal subgroups.

 

[mathmari]

Step 1, identity normal subgroups.

So, for $G=A_4$, one composition series is $1\leq A_4$ and for $G=S_3\times\mathbb{Z}_2$ one composition series is $1\leq S_3\times\mathbb{Z}_2$, right? (Wondering)

We have to check if $A_4$ and $S_3\times\mathbb{Z}_2$ are simple, or not? (Wondering)

But when these groups are simple, then there are no other normal subgroups, are there? (Wondering)

 

[Deveno]

So, for $G=A_4$, one composition series is $1\leq A_4$ and for $G=S_3\times\mathbb{Z}_2$ one composition series is $1\leq S_3\times\mathbb{Z}_2$, right? (Wondering)

We have to check if $A_4$ and $S_3\times\mathbb{Z}_2$ are simple, or not? (Wondering)

But when these groups are simple, then there are no other normal subgroups, are there? (Wondering)

You have to check if the QUOTIENTS are simple. So far, neither series you have is a composition series. Refine them.

 

[mathmari]

You have to check if the QUOTIENTS are simple. So far, neither series you have is a composition series. Refine them.

Consider first the group $A_4$.

We have that one subgroup is $N=\{1, (12)(34), (13)(24), (14)(23)\}$.

Is the order of $A_4$ equal to $\frac{4!}{2}=12$ ? (Wondering)

If this is true, we have that $[A_4:N]=\frac{|A_4|}{|N|}=\frac{12}{4}=3$.

This implies that the quotient $A_4/N$ is simple since the orders of the subgroups can be $1$ and $3$. The trivial subgroup has order $1$ and the whole group has order $3$, so it has no non-trivial subgroups, right? (Wondering) $1\leq N\leq A_4$ is a composition series, if $N/1=N$ is also simple, right? (Wondering)
How can we check this? (Wondering)

 

[Deveno]

Consider first the group $A_4$.

We have that one subgroup is $N=\{1, (12)(34), (13)(24), (14)(23)\}$.

Is the order of $A_4$ equal to $\frac{4!}{2}=12$ ? (Wondering)

If this is true, we have that $[A_4:N]=\frac{|A_4|}{|N|}=\frac{12}{4}=3$.

This implies that the quotient $A_4/N$ is simple since the orders of the subgroups can be $1$ and $3$. The trivial subgroup has order $1$ and the whole group has order $3$, so it has no non-trivial subgroups, right? (Wondering) $1\leq N\leq A_4$ is a composition series, if $N/1=N$ is also simple, right? (Wondering)
How can we check this? (Wondering)

Well $N$ is abelian, but it's not cyclic of prime order. Maybe if you found a subgroup...

 

[mathmari]

Well $N$ is abelian, but it's not cyclic of prime order. Maybe if you found a subgroup...

You mean to take a subgroup of $N$ ? (Wondering)
Can we take the identity element and any of the other elements of $N$ ? Can we take for example $\tilde{N}= \{1, (12)(34)\}$ ? (Wondering)

Then we have that $[N:\tilde{N}]=\frac{|N|}{|\tilde{N}|}=\frac{4}{2}=2$.

Since the quotient $N/\tilde{N}$ is of prime order, it is simple, right? (Wondering)

The quotient $\tilde{N}/1=\tilde{N}$ is also of prime order, $2$, so it is also simple.

Therefore, a composition series for $A_4$ is $$1\leq \tilde{N}\leq N\leq A_4$$ right? (Wondering)

 

[Deveno]

You mean to take a subgroup of $N$ ? (Wondering)
Can we take the identity element and any of the other elements of $N$ ? Can we take for example $\tilde{N}= \{1, (12)(34)\}$ ? (Wondering)

Then we have that $[N:\tilde{N}]=\frac{|N|}{|\tilde{N}|}=\frac{4}{2}=2$.

Since the quotient $N/\tilde{N}$ is of prime order, it is simple, right? (Wondering)

The quotient $\tilde{N}/1=\tilde{N}$ is also of prime order, $2$, so it is also simple.

Therefore, a composition series for $A_4$ is $$1\leq \tilde{N}\leq N\leq A_4$$ right? (Wondering)

Yes, and the composition factor orders are (from right to left): 3,2,2.

Are there any others (for $A_4$)? Well, any other (non-trivial) subgroup of $N$ would yield an equivalent composition series (so we have at least 3), but if we tried to find a DIFFERENT $N$ with $N \lhd A_4$ and $|A_4/N| = 2$, we'd need for $|N| = 6$. You might want to show this cannot happen.

Try the other group, now. Will finding normal subgroups of $S_3$ or $\Bbb Z_2$ help?

 

[mathmari]

Are there any others (for $A_4$)? Well, any other (non-trivial) subgroup of $N$ would yield an equivalent composition series (so we have at least 3), but if we tried to find a DIFFERENT $N$ with $N \lhd A_4$ and $|A_4/N| = 2$, we'd need for $|N| = 6$. You might want to show this cannot happen.

Why can it be that there is a subgroup $N$ of order $6$ ? (Wondering)

Try the other group, now. Will finding normal subgroups of $S_3$ or $\Bbb Z_2$ help?

The only normal subgroups of $\mathbb{Z}_2$ are the trivial and the whole group, right? (Wondering)

A normal subgroup of $S_3$ is $A_3$, right? (Wondering)

 

[Deveno]

Why can it be that there is a subgroup $N$ of order $6$ ? (Wondering)

Yes, indeed, why can't there be? It's pretty obvious there's no cyclic subgroup of order 6 (there's no elements of order 6), but why isn't there a subgroup isomorphic to $S_3$? (hint: consider the subgroup generated by a 3-cycle and a double transposition).

The only normal subgroups of $\mathbb{Z}_2$ are the trivial and the whole group, right? (Wondering)

A normal subgroup of $S_3$ is $A_3$, right? (Wondering)

True...do you think that $S_3 \times \{0\}$ or $A_3 \times \Bbb Z_2$ might be normal? Why, or why not?

 

[mathmari]

Yes, indeed, why can't there be? It's pretty obvious there's no cyclic subgroup of order 6 (there's no elements of order 6), but why isn't there a subgroup isomorphic to $S_3$? (hint: consider the subgroup generated by a 3-cycle and a double transposition).

The possible cycle of $A_4$ are:
- the identity element (order $1$),
- double transpositions (order $2$)
- $3$-cycles (order $3$)

So, there is no element of order $6$. That means that $A_4$ has no cyclic subgroup of order $6$.

Assume that $A_4$ has a subgroup $H$ of order $6$, where $H \cong S_3$.
Do we know that $S_3$ has a subgroup generated by a $3$-cycle and a double transposition, or do we have to prove it? (Wondering)
So, $H$ must contain the identity element, a $3$-cycle and a double transposition. That means that $|H|=4$, a contradiction since we have supposed that $|H|=6$, right? (Wondering)

True...do you think that $S_3 \times \{0\}$ or $A_3 \times \Bbb Z_2$ might be normal? Why, or why not?

We have that $[S_3\times\mathbb{Z}_2:S_3\times\{0\}]=\frac{|S_3\times\mathbb{Z}_2|}{|S_3\times\{0\}|}=\frac{6\cdot 2}{6\cdot 1}=2$. So $S_3 \times \{0\}$ is a normal subgroup.

We have that $[S_3\times\mathbb{Z}_2:A_3 \times \Bbb Z_2]=\frac{|S_3\times\mathbb{Z}_2|}{|A_3 \times \Bbb Z_2|}=\frac{6\cdot 2}{3\cdot 2}=2$. So $A_3 \times \Bbb Z_2$ is also a normal subgroup.

Is this correct? (Wondering)

 

[Deveno]

The possible cycle of $A_4$ are:
- the identity element (order $1$),
- double transpositions (order $2$)
- $3$-cycles (order $3$)

So, there is no element of order $6$. That means that $A_4$ has no cyclic subgroup of order $6$.

Assume that $A_4$ has a subgroup $H$ of order $6$, where $H \cong S_3$.
Do we know that $S_3$ has a subgroup generated by a $3$-cycle and a double transposition, or do we have to prove it? (Wondering)
So, $H$ must contain the identity element, a $3$-cycle and a double transposition. That means that $|H|=4$, a contradiction since we have supposed that $|H|=6$, right? (Wondering)

We know that $S_3$ is generated by an element of order 3, and an element of order 2. Your argument that $|H| = 4$ isn't right. But look:

$(a\ b\ c)(a\ c)(b\ d) = (b d c)$ (this is just one example, the point is-what is the cycle type? Can you prove that the product of a 3-cycle and a double transposition is another 3-cycle which isn't a power of the generating 3-cycle?). Thus $\langle (a\ b\ c),(a\ c)(b\ d)\rangle$ has at least 4 elements of order 3, while $S_3$ only has two.

We have that $[S_3\times\mathbb{Z}_2:S_3\times\{0\}]=\frac{|S_3\times\mathbb{Z}_2|}{|S_3\times\{0\}|}=\frac{6\cdot 2}{6\cdot 1}=2$. So $S_3 \times \{0\}$ is a normal subgroup.

We have that $[S_3\times\mathbb{Z}_2:A_3 \times \Bbb Z_2]=\frac{|S_3\times\mathbb{Z}_2|}{|A_3 \times \Bbb Z_2|}=\frac{6\cdot 2}{3\cdot 2}=2$. So $A_3 \times \Bbb Z_2$ is also a normal subgroup.

Is this correct? (Wondering)

Yes, in a way, this example is almost "too easy", the order of our group is rather small, so we can find subgroups of index 2 without much difficulty.

here is a related problem:

If $N \lhd G$ and $K \lhd H$, is is true that $N \times K \lhd G \times H$?

In any case, the first subgroup $S_3 \times \{0\}$ leaves us with the (easier) problem of finding a composition series for $S_3$ (there's only one), while the second possibility is easy once you realize that $A_3 \times \Bbb Z_2 \cong \Bbb Z_3 \times \Bbb Z_2 \cong \Bbb Z_6$, and it's an *abelian* group (finding all the composition series for $\Bbb Z_6$ should be easy for you).

 

[mathmari]

the first subgroup $S_3 \times \{0\}$ leaves us with the (easier) problem of finding a composition series for $S_3$ (there's only one)

A composition series for $S_3$ is $1\leq A_3\leq S_3$, right? (Wondering)

the second possibility is easy once you realize that $A_3 \times \Bbb Z_2 \cong \Bbb Z_3 \times \Bbb Z_2 \cong \Bbb Z_6$, and it's an *abelian* group (finding all the composition series for $\Bbb Z_6$ should be easy for you).

Do we have to show that $A_3\cong \mathbb{Z}_3$ ? (Wondering)

 

[Deveno]

A composition series for $S_3$ is $1\leq A_3\leq S_3$, right? (Wondering)

The one and only, since $A_3$ is the ONLY non-trivial proper normal subgroup of $S_3$, and $A_3$ is simple.

Do we have to show that $A_3\cong \mathbb{Z}_3$ ? (Wondering)

What, it's not obvious?

 

[mathmari]

the second possibility is easy once you realize that $A_3 \times \Bbb Z_2 \cong \Bbb Z_3 \times \Bbb Z_2 \cong \Bbb Z_6$, and it's an *abelian* group (finding all the composition series for $\Bbb Z_6$ should be easy for you).

Are the composition series for $\Bbb Z_6$ the following? (Wondering)

$$1\leq \langle 2\rangle \leq \mathbb{Z}_6 \\ 1\leq \langle 3\rangle \leq \mathbb{Z}_6$$

 

[Deveno]

Are the composition series for $\Bbb Z_6$ the following? (Wondering)

$$1\leq \langle 2\rangle \leq \mathbb{Z}_6 \\ 1\leq \langle 3\rangle \leq \mathbb{Z}_6$$

Convince me.

 

[mathmari]

Convince me.

The prime factorization of $6$ is $2\cdot 3$.

The cyclic subgroup of order $2$ is $\langle 3\rangle=\{3, 1\}$ and the cyclic subgroup of order $3$ is $\langle 2\rangle=\{2, 4, 1\}$, right? (Wondering)

The order of the quotient group $\mathbb{Z}_6/\langle 3\rangle$ is $\frac{6}{2}=3$, a prime, so the quotient group is simple.
We have to check if $\langle 3\rangle$ is a normal subgroup of $\mathbb{Z}_6$, right? (Wondering)

The order of the quotient group $\mathbb{Z}_6/\langle 2\rangle$ is $\frac{6}{3}=2$, a prime, so the quotient group is simple, and we have that $\langle 2\rangle$ is a normal subgroup of $\mathbb{Z}_6$, right? (Wondering) So, are the composition series the ones I mentioned in post #15, or do we have to refine them? (Wondering)

 

[Deveno]

The prime factorization of $6$ is $2\cdot 3$.

So far, so good. Really, this is all you need, but let's continue.

The cyclic subgroup of order $2$ is $\langle 3\rangle=\{3, 1\}$ and the cyclic subgroup of order $3$ is $\langle 2\rangle=\{2, 4, 1\}$, right? (Wondering)

No, the subgroup of order 2 (remember the group operation in $\Bbb Z_6$ is ADDITION modulo $6$) is $\{0,3\}$, and the subgroup of order 3 is $\{0,2,4\}$.

The order of the quotient group $\mathbb{Z}_6/\langle 3\rangle$ is $\frac{6}{2}=3$, a prime, so the quotient group is simple.
We have to check if $\langle 3\rangle$ is a normal subgroup of $\mathbb{Z}_6$, right? (Wondering)

Repeat after me: all subgroups of an abelian group are normal.

The order of the quotient group $\mathbb{Z}_6/\langle 2\rangle$ is $\frac{6}{3}=2$, a prime, so the quotient group is simple, and we have that $\langle 2\rangle$ is a normal subgroup of $\mathbb{Z}_6$, right? (Wondering) So, are the composition series the ones I mentioned in post #15, or do we have to refine them? (Wondering)

You can't refine a composition series where the composition factors have prime order.