Subnormal Series and Composition Series

In summary, Lemma 21.2.4 in Eie & Chang's book: A Course on Abstract Algebra states that it is possible to insert a normal subgroup H of H_{i+1} with H_i \subsetneq H \subsetneq H_{i+1} if and only if H/H_i is a proper nontrivial subgroup of H_{i+1}/H_i. This is because $H/H_i$ being a normal subgroup of $H_{i+1}/H_i$ is necessary for the insertion to take place, and this cannot happen if $H_{i+1}/H_i$ is a simple group. The proof involves the Correspondence Theorem for Groups and the concept of
  • #1
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I need some help with the proof of Lemma 21.2.4 in Eie & Chang's book: A Course on Abstract Algebra:

Lemma 21.2.4 reads as follows:View attachment 3322

My questions regarding the proof are as follows:

Question 1

Eie & Chang write:

"... ... It is possible to insert a normal subgroup \(\displaystyle H\) of \(\displaystyle H_{i+1}\) with \(\displaystyle H_i \subsetneq H \subsetneq H_{i+1} \) if and only if \(\displaystyle H/H_i\) is a proper nontrivial subgroup of \(\displaystyle H_{i+1}/H_i\). ... ... "

Can someone please give me a detailed explanation of exactly why this is the case?
Question 2

In the above text, Eie & Chang write:

" ... ... And this is true if and only if \(\displaystyle H_{i+1}/H_i\) is not simple. ... ... "

Can someone please give me a detailed explanation of exactly why this is the case?

Help will be much appreciated.

Peter
 
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  • #2
This is quite obvious. What you want to try is to insert a "normal component" in between $H_i$ and $H_{i+1}$, i.e., a group $H$ such that $H_i \triangleleft H \triangleleft H_{i+1}$.

Claim If $A$ is a normal subgroup of $B$, and $C$ is a normal subgroup of both, then $A/C$ is a normal subgroup of $B/C$.

Using the claim above, $H/H_i$ is a normal subgroup of $H_{i+1}/H_i$, but this cannot happen if $H_{i+1}/H_i$ is simple (by definition of simplicity).

Can you try to prove the claim above?
 
  • #3
mathbalarka said:
This is quite obvious. What you want to try is to insert a "normal component" in between $H_i$ and $H_{i+1}$, i.e., a group $H$ such that $H_i \triangleleft H \triangleleft H_{i+1}$.

Claim If $A$ is a normal subgroup of $B$, and $C$ is a normal subgroup of both, then $A/C$ is a normal subgroup of $B/C$.

Using the claim above, $H/H_i$ is a normal subgroup of $H_{i+1}/H_i$, but this cannot happen if $H_{i+1}/H_i$ is simple (by definition of simplicity).

Can you try to prove the claim above?

Thanks for the help Mathbalarka ... it is much appreciated!

I will try to prove the claim in terms of \(\displaystyle H_i, H \text{ and } H_{i+1}\) where \(\displaystyle H_i \triangleleft H \triangleleft H_{i+1}\) (instead of \(\displaystyle A, B \text{ and } C ) \)

It appears to me that the claim is essentially just an application of the Correspondence Theorem for Groups.

Eie and Chang's text gives the Correspondence Theorem for Groups as follows:View attachment 3323So in terms of \(\displaystyle H_i , H \text{ and } H_{i+1} \) we have that:

\(\displaystyle H_i\) is a normal subgroup of \(\displaystyle H_{i+1}\)

and we let

\(\displaystyle \pi \ : \ H_{i+1} \to H_i\)

and, further we let

\(\displaystyle \mathscr{A}\) be the family of subgroups of \(\displaystyle H_{i+1}\) containing \(\displaystyle H_i\)

and let

\(\displaystyle \mathscr{B}\) be the family of subgroups of \(\displaystyle H_{i+1} / H_i\)

Then there is an order-preserving one-to-one correspondence:

\(\displaystyle \Phi \ : \ \mathscr{A} \to \mathscr{B}\)

where

\(\displaystyle H \mapsto H / H_i \) \(\displaystyle \ \ \ \) (i.e. \(\displaystyle \Phi (H) = H / H_i\) )

[in the above, \(\displaystyle H\) is a subgroup of \(\displaystyle H_{i+1}\) containing \(\displaystyle H_i\)]

and (and here is what we want!)

\(\displaystyle H \triangleleft H_{i+1} \)

if and only if

\(\displaystyle H / H_i\) is a normal subgroup of \(\displaystyle H_{i+1}/ H_i \)

Can you please confirm that the above analysis is correct - or point out errors or shortcomings?

PeterNote: the mapping \(\displaystyle \Phi \ : \ \mathscr{A} \to \mathscr{B}\) is not only one-to-one but is also onto - so it is a bijection - and thus has an inverse, as indicated in the theorem.
 
  • #4
Eh, I don't know. I don't really care about big theorems here.

Recall the definition of normality : we are given that $A$ is normal in $B$. Then $bAb^{-1} = A$, i.e., $bab^{-1} \in A$ for all $a \in A$. We are also given that $C$ is normal in $A$ and $B$ both, thus $A/C$ and $B/C$ make sense.

We want to prove that $A/C$ is normal in $B/C$. Consider an element $aC \in A/C$. We want to prove that $(bC)(aC)(bC)^{-1} \in A/C$ for any $b \in B$. This is accomplished by elementary manipulations :

$$\begin{aligned}(bC)(aC)(bC)^{-1} & = (bC)(aC)(b^{-1}C) \\ &= (bC)(Ca)(b^{-1}C) \;\;\;\;\; \text{from normality of $C$ in $A$}\\ &= (bCa)(b^{-1}C) \\ &= (baC)(b^{-1}C) \;\;\;\;\; \text{from normality of $C$ in $A$} \\ &= (baCb^{-1}) \\ &= (bab^{-1})C \;\;\;\;\; \text{from normality of $C$ in $B$}\end{aligned}$$

As $bab^{-1} \in A$, $(bab^{-1})C \in A/C$. $\blacksquare$
 
  • #5
mathbalarka said:
Eh, I don't know. I don't really care about big theorems here.

Recall the definition of normality : we are given that $A$ is normal in $B$. Then $bAb^{-1} = A$, i.e., $bab^{-1} \in A$ for all $a \in A$. We are also given that $C$ is normal in $A$ and $B$ both, thus $A/C$ and $B/C$ make sense.

We want to prove that $A/C$ is normal in $B/C$. Consider an element $aC \in A/C$. We want to prove that $(bC)(aC)(bC)^{-1} \in A/C$ for any $b \in B$. This is accomplished by elementary manipulations :

$$\begin{aligned}(bC)(aC)(bC)^{-1} & = (bC)(aC)(b^{-1}C) \\ &= (bC)(Ca)(b^{-1}C) \;\;\;\;\; \text{from normality of $C$ in $A$}\\ &= (bCa)(b^{-1}C) \\ &= (baC)(b^{-1}C) \;\;\;\;\; \text{from normality of $C$ in $A$} \\ &= (baCb^{-1}) \\ &= (bab^{-1})C \;\;\;\;\; \text{from normality of $C$ in $B$}\end{aligned}$$

As $bab^{-1} \in A$, $(bab^{-1})C \in A/C$. $\blacksquare$

Given that $C \triangleleft A$, we can "skip a few steps" and write:

$(bC)(aC)(bC)^{-1} = (bab^{-1})C$

directly, by the definition of coset multiplication in $A/C$, in other words: the conjugate of the cosets is the coset of the conjugate (normality of $C$ is what allows us to say the product of cosets is the coset of the product).

The above is valid for any elements of $A$, and we only need to invoke normality of $B$ in $A$ as you did, the normality of $C$ in $B$ is not needed (of course $C$ is normal in any subgroup of $A$).

A word of caution, here: the relation "is normal in" is NOT transitive, for example we have:

$A = \{1,s\} \triangleleft B = \{1,r,s,sr\} \triangleleft D_4$ (since both subgroups are of index 2 in the next super-group), but it is NOT true that $A \triangleleft D_4$, for example:

$rAr^{-1} = \{1,rsr^{-1}\} = \{1,sr^2\} \neq A$.

Given the proper normality conditions, in a group $G/K$ the set $K$ in a "typical coset" $gK$ is pretty much "just along for the ride", its presence just obscures what is happening. One loses no actual information by using a notation like $[g]$ or $\overline{g}$ to indicate a homomorphic image under $\pi: G \to G/K$. In the integers modulo $n$, for example, often the modulus is taken to be "understood", and just the coset representatives are used (with an occasional (mod $n$) appended after calculations).

As another example, when one considers the (Euclidean) plane as the image of the projection map:

$\Bbb R^3 \to (\Bbb R^3)/(\{0\}\oplus\{0\}\oplus\Bbb R)$

one typically does NOT use "coset notation", we just ignore the third coordinate.
 
  • #6
Deveno said:
directly, by the definition of coset multiplication in A/C, in other words: the conjugate of the cosets is the coset of the conjugate (normality of C is what allows us to say the product of cosets is the coset of the product).

Fair enough, yes. Multiplication mod $N$ just works like usual group multiplication.

You always find a thing or two to add on every thread, don't you :p I admire it (Yes)
 
  • #7
mathbalarka said:
Fair enough, yes. Multiplication mod $N$ just works like usual group multiplication.

You always find a thing or two to add on every thread, don't you :p I admire it (Yes)

Not always. The quality of posts in this particular sub-forum has taken a quantum leap forward in recent days, and some posts by others such as yourself, Euge, ILikeSerena, and Opalg are virtually letter-perfect. It's encouraging to see the upswing in activity.
 

FAQ: Subnormal Series and Composition Series

What is a subnormal series?

A subnormal series is a sequence of subgroups within a group, where each subgroup is a normal subgroup of the previous one. This means that the quotient group formed by dividing each subgroup by the next is always a normal subgroup.

How is a subnormal series different from a normal series?

A normal series is a sequence of subgroups where each one is a normal subgroup of the previous one, but there is no guarantee that the quotient groups will also be normal. A subnormal series, on the other hand, ensures that the quotient groups are always normal.

What is a composition series?

A composition series is a subnormal series in which each of the quotient groups is simple, meaning it has no non-trivial normal subgroups. In other words, it is a series of subgroups that cannot be further broken down into smaller normal subgroups.

How are subnormal series and composition series related?

A composition series is a special type of subnormal series, where the quotient groups are all simple. However, not all subnormal series are composition series, as some may have quotient groups that are not simple.

Why are subnormal and composition series important in group theory?

Subnormal and composition series provide a way to break down a group into simpler subgroups, making it easier to understand and analyze. They also help in determining important properties of a group, such as its structure and symmetry. Additionally, these series have applications in other branches of mathematics, such as Galois theory and representation theory.

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