# Thread: Subnormal Series and Composition Series

1. I need some help with the proof of Lemma 21.2.4 in Eie & Chang's book: A Course on Abstract Algebra:

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  Reply With Quote

2.

3. 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?  Reply With Quote Originally Posted by mathbalarka 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:

So 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?

Peter

Note: 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.  Reply With Quote

5. 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 $bab^{-1} \in A$, $(bab^{-1})C \in A/C$. $\blacksquare$  Reply With Quote

6. Originally Posted by mathbalarka 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.  Reply With Quote

7. Originally Posted by Deveno
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 I admire it   Reply With Quote

8. Originally Posted by mathbalarka 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 I admire it 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.  Reply With Quote

+ Reply to Thread #### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
• 