What is Aluffi's notation for composition series in Algebra: Chapter 0?

[Math Amateur]

I am reading Paolo Aluffi's book, Algebra: Chapter 0 ... I am currently focused on Chapter 4, Section 3: Composition Series and Solvability ...

I need help with Exercise 3.3 on page 213, which reads as follows:View attachment 4909

I hope someone can help ... and in so doing use Aluffi's notation ...

So that MHB readers will understand Aluffi's basic definitions and notation in the area of composition series I am now providing Aluffi's introduction to Chapter 4, Section 3: Composition Series and Solvability ...View attachment 4910
View attachment 4911Help using Aluffi's notation will be much appreciated ...

Peter

 

[Euge]

Hi Peter,

As mentioned in the text, the first part of the exercise can be proven by induction on the order of the group. Suppose $G$ is your finite group. If $|G| = 1$, then $G = \{e\}$ is your composition series (of length zero). Now assume $|G| > 1$ and the result holds for all groups of order strictly less than the order of $G$. If $G$ is simple, then $G \supsetneq \{e\}$ is a composition series for $G$. If $G$ is not simple, there exists a nontrivial normal subgroup $N\triangleleft G$. Since $N$ and the factor group $G/N$ have orders strictly less than the order of $G$, use the induction hypothesis on $G/N$, then $N$, to prove that there is a composition series for $G$. Use the lattice isomorphism theorem to get a series $G = G_0 \supsetneq G_1 \supsetneq G_2\supsetneq\cdots \supsetneq G_k = N$ from the composition series for $G/N$ obtained by the induction hypothesis.

 

[Deveno]

(snip)

...For the second part, first note that every subgroup of $\Bbb Z$ is of the form $m\Bbb Z$ for some integer $m$. Now if a composition series for $\Bbb Z$ exists, then its length is $> 1$ (since $\Bbb Z$ is not simple). Suppose $\Bbb \ell(G) = n$. The first group $G_1$ must be simple...

Actually, the first *quotient* must be simple (this is $G/G_1$, which in this case must be $\Bbb Z/p\Bbb Z$, for some prime $p$).

Again, since $G_1/G_2$ is simple, it must be that $G_2 = q(p\Bbb Z) = pq\Bbb Z$ for some (not necessarily distinct) prime $q$.

Continuing in this way, we have $G_k = p_1p_2\cdots p_k\Bbb Z$, for any $k$, where each $p_i$ is a positive prime integer (again, these need not be distinct).

Since each $p_i > 0$ so is their product, so that $G_k \cong \Bbb Z$, for any $k$ (the isomorphism being $m \mapsto p_1p_2\cdots p_km$). Since this is not simple (since $\Bbb Z$ is not simple), we never have $G_k/\{0\}$ simple, for any $k$, so we can never have a composition series of finite length.

(I *do* hope this doesn't count as "trampling", I normally would refrain from posting, but this part of Euge's post simply isn't correct).

 

[Euge]

Actually, the first *quotient* must be simple

Yes that's right, my apologies. And no, you weren't trampling :D Please do if you see something off, as I sometimes I write posts when I'm tired (Wasntme)

Anyway, since you've given the correct proof of the second part, I removed it from my post.

Edit: Funny thing is, in my notes, the proof I wrote for the second part is the same as Deveno's, but I wrote something totally off here. More of an indication that I need rest. (Bigsmile)

 

[Math Amateur]

Yes that's right, my apologies. And no, you weren't trampling :D Please do if you see something off, as I sometimes I write posts when I'm tired (Wasntme)

Anyway, since you've given the correct proof of the second part, I removed it from my post.

Edit: Funny thing is, in my notes, the proof I wrote for the second part is the same as Deveno's, but I wrote something totally off here. More of an indication that I need rest. (Bigsmile)

Thanks to Euge and Deveno for the help ...

It is much appreciated ...

Still puzzling over exactly how the Principle of Induction actually works in this exercise ... but ... will reflect some more ... then I will post a question if I still do not fully understand how the Principle of Induction works in this exercise ...

Thanks again to you both for your help ...

Peter

 

[Deveno]

Let's twist Euge's proof into what I like to call "The Devil's proof".

Suppose the theorem were false. Then for some finite $G$, it does *not* have a composition series.

Let $n = \min\{|G|: G \text{ does not have a composition series}\}$.

If $G$ were simple, it would have the composition series $G_0 = G \supsetneq \{e\}$.

(in other words we pick $G$ to be a minimal counter-example).

Since $G$ is not simple, it has a normal nontrivial proper subgroup, say $N$.

Since $|N| < |G|$ (it is a proper subgroup of a finite group), we must have that $N$ has a composition series, say:

$N \supsetneq N_1 \supsetneq \cdots \supsetneq N_k = \{e\}$.

Since $N$ is nontrivial, and normal, we have:

$|G/N| = |G|/|N| < |G|/1 = |G|$, so $G/N$ has a composition series, say:

$H_0 = G/N \supsetneq H_1 \supsetneq \cdots \supsetneq H_{k'} = \{e_{G/N}\} = \{N\}$.

Let $\pi_N: G \to G/N$ be the canonical homomorphism $g \mapsto gN$.

Set $K_j = \pi_N^{-1}(H_j)$ for $j = 0,1,\dots,k'$ (the pre-image of the $j$-th subgroup in the composition series for $G/N)$.

It ought to be "obvious" that $K_j$ is a subgroup of $G$, for each $j$, but we can prove it, to be perfectly explicit.

Suppose $x,y \in K_j$. Note this means that $\pi_N(x)$ and $\pi_N(y)$ lie within $H_j$. Since $H_j$ is a subgroup of $G/N$ we have that:

$\pi_N(x)\pi_N(y) \in H_j$ ($H_j$ is closed under multiplication).

Since $\pi_N$ is a homomorphism, $\pi_N(x)\pi_N(y) = \pi_N(xy) \in H_j$.

But this means $\pi_N$ maps $xy$ into $H_j$, so $xy \in \pi_N^{-1}(H_j) = K_j$, so $K_j$ is closed under multiplication.

Since $G$ is finite, and $K_j$ is a subset of $G$ closed under multiplication, $K_j$ is a subgroup if it is non-empty.

But if $n \in N$, we have $\pi_N(n) = nN = N \in H_j$ (since every subgroup of $G/N$ contains the identity of $G/N$).

This shows that not only is $K_j$ non-empty, it also contains all of $N$ (as a subgroup), and since $N \lhd G$, it is true "all the more" that $N \lhd K_j$ (for any $j$).

Now we have: $K_j/K_{j+1} \cong (K_j/N)/(K_{j+1}/N)$. I claim $K_j/N = H_j$. For if $x \in K_j$, then $\pi_N(x) \in H_j$ (by the way we defined $K_j$), which is to say $xN \in H_j$. This shows that $K_j/N \subseteq H_j$.

On the other hand, if $yN \in H_j$, then $\pi_N(y) \in H_j$ (by the definition of $\pi_N$), whence $y \in \pi_N^{-1}(H_j) = K_j$, so that $yN \in K_j/N$. Since $H_j/H_{j+1}$ is simple (since we have a composition series for $H_0 = G/N$), it follows that $K_j/K_{j+1}$ is likewise simple, via the isomorphism above.

Thus we have:

$G = G_0 = K_0 \supsetneq K_1 \supsetneq \cdots \supsetneq K_{k'} = N \supsetneq N_1 \supsetneq N_k = \{e\}$

is a composition series for $G$, contradiction.

Thus no counterexample exists, and the theorem is true.

********************

This kind of proof is sort of "sneaky", one winds up wondering "what does induction have to do with this"? The pivotal step is in assuming the minimality of the order of $G$. Finite groups have orders which are natural numbers, and any non-empty set of natural numbers has a least element (this is called the *well-ordered* property of the natural numbers, and is equivalent to the principle of induction). If there are ANY finite groups that do not possesses composition series then the set of orders of such groups forms a non-empty subset of the natural numbers (there is a slight subtlety, here-the collection of all finite groups is "too big" to be a set, but the collection of all orders of those groups is indeed a set-the natural numbers (that this set is *all* of the natural numbers is evidenced by the existence of $\Bbb Z_n$, for any natural number $n$)).

Technically, Euge's proof is a "strong induction proof", and what I have written above is induction "turned inside-out".

 

[Math Amateur]

Let's twist Euge's proof into what I like to call "The Devil's proof".

Suppose the theorem were false. Then for some finite $G$, it does *not* have a composition series.

Let $n = \min\{|G|: G \text{ does not have a composition series}\}$.

If $G$ were simple, it would have the composition series $G_0 = G \supsetneq \{e\}$.

(in other words we pick $G$ to be a minimal counter-example).

Since $G$ is not simple, it has a normal nontrivial proper subgroup, say $N$.

Since $|N| < |G|$ (it is a proper subgroup of a finite group), we must have that $N$ has a composition series, say:

$N \supsetneq N_1 \supsetneq \cdots \supsetneq N_k = \{e\}$.

Since $N$ is nontrivial, and normal, we have:

$|G/N| = |G|/|N| < |G|/1 = |G|$, so $G/N$ has a composition series, say:

$H_0 = G/N \supsetneq H_1 \supsetneq \cdots \supsetneq H_{k'} = \{e_{G/N}\} = \{N\}$.

Let $\pi_N: G \to G/N$ be the canonical homomorphism $g \mapsto gN$.

Set $K_j = \pi_N^{-1}(H_j)$ for $j = 0,1,\dots,k'$ (the pre-image of the $j$-th subgroup in the composition series for $G/N)$.

It ought to be "obvious" that $K_j$ is a subgroup of $G$, for each $j$, but we can prove it, to be perfectly explicit.

Suppose $x,y \in K_j$. Note this means that $\pi_N(x)$ and $\pi_N(y)$ lie within $H_j$. Since $H_j$ is a subgroup of $G/N$ we have that:

$\pi_N(x)\pi_N(y) \in H_j$ ($H_j$ is closed under multiplication).

Since $\pi_N$ is a homomorphism, $\pi_N(x)\pi_N(y) = \pi_N(xy) \in H_j$.

But this means $\pi_N$ maps $xy$ into $H_j$, so $xy \in \pi_N^{-1}(H_j) = K_j$, so $K_j$ is closed under multiplication.

Since $G$ is finite, and $K_j$ is a subset of $G$ closed under multiplication, $K_j$ is a subgroup if it is non-empty.

But if $n \in N$, we have $\pi_N(n) = nN = N \in H_j$ (since every subgroup of $G/N$ contains the identity of $G/N$).

This shows that not only is $K_j$ non-empty, it also contains all of $N$ (as a subgroup), and since $N \lhd G$, it is true "all the more" that $N \lhd K_j$ (for any $j$).

Now we have: $K_j/K_{j+1} \cong (K_j/N)/(K_{j+1}/N)$. I claim $K_j/N = H_j$. For if $x \in K_j$, then $\pi_N(x) \in H_j$ (by the way we defined $K_j$), which is to say $xN \in H_j$. This shows that $K_j/N \subseteq H_j$.

On the other hand, if $yN \in H_j$, then $\pi_N(y) \in H_j$ (by the definition of $\pi_N$), whence $y \in \pi_N^{-1}(H_j) = K_j$, so that $yN \in K_j/N$. Since $H_j/H_{j+1}$ is simple (since we have a composition series for $H_0 = G/N$), it follows that $K_j/K_{j+1}$ is likewise simple, via the isomorphism above.

Thus we have:

$G = G_0 = K_0 \supsetneq K_1 \supsetneq \cdots \supsetneq K_{k'} = N \supsetneq N_1 \supsetneq N_k = \{e\}$

is a composition series for $G$, contradiction.

Thus no counterexample exists, and the theorem is true.

********************

This kind of proof is sort of "sneaky", one winds up wondering "what does induction have to do with this"? The pivotal step is in assuming the minimality of the order of $G$. Finite groups have orders which are natural numbers, and any non-empty set of natural numbers has a least element (this is called the *well-ordered* property of the natural numbers, and is equivalent to the principle of induction). If there are ANY finite groups that do not possesses composition series then the set of orders of such groups forms a non-empty subset of the natural numbers (there is a slight subtlety, here-the collection of all finite groups is "too big" to be a set, but the collection of all orders of those groups is indeed a set-the natural numbers (that this set is *all* of the natural numbers is evidenced by the existence of $\Bbb Z_n$, for any natural number $n$)).

Technically, Euge's proof is a "strong induction proof", and what I have written above is induction "turned inside-out".

Thanks for a helpful and very interesting proof, Deveno ...

Will work through this proof carefully and in detail in the morning ...

Thanks so much for the help,

Peter

 

[Math Amateur]

Hi Peter,

As mentioned in the text, the first part of the exercise can be proven by induction on the order of the group. Suppose $G$ is your finite group. If $|G| = 1$, then $G = \{e\}$ is your composition series (of length zero). Now assume $|G| > 1$ and the result holds for all groups of order strictly less than the order of $G$. If $G$ is simple, then $G \supsetneq \{e\}$ is a composition series for $G$. If $G$ is not simple, there exists a nontrivial normal subgroup $N\triangleleft G$. Since $N$ and the factor group $G/N$ have orders strictly less than the order of $G$, use the induction hypothesis on $G/N$, then $N$, to prove that there is a composition series for $G$. Use the lattice isomorphism theorem to get a series $G = G_0 \supsetneq G_1 \supsetneq G_2\supsetneq\cdots \supsetneq G_k = N$ from the composition series for $G/N$ obtained by the induction hypothesis.

Thanks again to Euge and Deveno for helping me with this exercise from Aluffi on composition series ... ...

I am going through the proofs by Euge and Deveno again to ensure that I understand them fully ...

In Euge's proof of the fact that every finite group has a composition series, we read the following:

" ... ... Since $N$ and the factor group $G/N$ have orders strictly less than the order of $G$, use the induction hypothesis on $G/N$, then $N$, to prove that there is a composition series for $G$. ... ... "It is very clear that \(\displaystyle N\) and \(\displaystyle G/N\) exist and have orders less than \(\displaystyle G\) ... ... so by the induction hypothesis, they both have composition series ... BUT ... how exactly do we put these results together to prove that there is a composition series for $G$. ... ... AND FURTHER ... how do we use the Lattice Isomorphism Theorem in this proof ... ... Hope one of you can help ... ...

Peter

 

[Euge]

In Euge's proof of the fact that every finite group has a composition series, we read the following:

" ... ... Since $N$ and the factor group $G/N$ have orders strictly less than the order of $G$, use the induction hypothesis on $G/N$, then $N$, to prove that there is a composition series for $G$. ... ... "It is very clear that \(\displaystyle N\) and \(\displaystyle G/N\) exist and have orders less than \(\displaystyle G\) ... ... so by the induction hypothesis, they both have composition series ... BUT ... how exactly do we put these results together to prove that there is a composition series for $G$. ... ... AND FURTHER ... how do we use the Lattice Isomorphism Theorem in this proof ... ...

By the induction hypothesis, $G/N$ has a composition series

$$\bar{G}_0 = G/N \supsetneq \bar{G}_1 \supsetneq \cdots \supsetneq \bar{G}_k = N\tag{1}$$

By the lattice isomorphism theorem, the above chain corresponds to the chain

$$G_0 = G \supsetneq G_1 \supsetneq \cdots \supsetneq G_k = N,\tag{2}$$

where $G_i$ is the inverse image of $\bar{G}_i$ under the canonical projection $\pi : G \to G/N$. For each $i$, $G_i/G_{i+1}$ is isomorphic to $\bar{G}_i/\bar{G}_{i+1}$, which is simple by construction of the composition series (1). So each factor group $G_i/G_{i+1}$ is simple.

Now using the induction hypothesis on $N$, we obtain a composition series for $N$:

$$G_k = N \supsetneq G_{k+1} \supsetneq \cdots \supsetneq G_{k+s} = \{e\} \tag{3}$$

In particular, for $k \le i \le k + s -1$, $G_i/G_{i+1}$ is simple. Combining (2) and (3), we get a composition series for $G$:

$$G_0 = G \supsetneq G_1 \supsetneq \cdots \supsetneq G_k \supsetneq \cdots \supsetneq G_{k+s} = \{e\}.$$

 

[Math Amateur]

By the induction hypothesis, $G/N$ has a composition series

$$\bar{G}_0 = G/N \supsetneq \bar{G}_1 \supsetneq \cdots \supsetneq \bar{G}_k = N\tag{1}$$

By the lattice isomorphism theorem, the above chain corresponds to the chain

$$G_0 = G \supsetneq G_1 \supsetneq \cdots \supsetneq G_k = N,\tag{2}$$

where $G_i$ is the inverse image of $\bar{G}_i$ under the canonical projection $\pi : G \to G/N$. For each $i$, $G_i/G_{i+1}$ is isomorphic to $\bar{G}_i/\bar{G}_{i+1}$, which is simple by construction of the composition series (1). So each factor group $G_i/G_{i+1}$ is simple.

Now using the induction hypothesis on $N$, we obtain a composition series for $N$:

$$G_k = N \supsetneq G_{k+1} \supsetneq \cdots \supsetneq G_{k+s} = \{e\} \tag{3}$$

In particular, for $k \le i \le k + s -1$, $G_i/G_{i+1}$ is simple. Combining (2) and (3), we get a composition series for $G$:

$$G_0 = G \supsetneq G_1 \supsetneq \cdots \supsetneq G_k \supsetneq \cdots \supsetneq G_{k+s} = \{e\}.$$

Thanks for the help, Euge ... just working through your post now ... ...

Just a quick point, though ... how do we now that the composition series for \(\displaystyle G/N\) ends in \(\displaystyle N\) ...?

I imagine it is because the end of a composition series is the group containing only the identity element of the group concerned and \(\displaystyle N = 0N \text{ or } 0 +N\) is the identity element in \(\displaystyle G/N\) ... ... Is that right ... ?

Thanks again,

Peter

 

[Euge]

Yes Peter, $N$ is the identity of $G/N$, and therefore a composition series for $G/N$ must end with $N$.