
MHB Master
#1
September 28th, 2014,
02:14
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:
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

September 28th, 2014 02:14
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MHB Journeyman
#2
September 28th, 2014,
04:00
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?

MHB Master
#3
September 28th, 2014,
05:53
Thread Author
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 orderpreserving onetoone 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 onetoone but is also onto  so it is a bijection  and thus has an inverse, as indicated in the theorem.

MHB Journeyman
#4
September 28th, 2014,
06:23
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 $bab^{1} \in A$, $(bab^{1})C \in A/C$. $\blacksquare$

MHB Master
#5
September 29th, 2014,
05:30
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 supergroup), 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.

MHB Journeyman
#6
September 29th, 2014,
06:39
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

MHB Master
#7
September 29th, 2014,
07:34
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 subforum has taken a quantum leap forward in recent days, and some posts by others such as yourself, Euge, ILikeSerena, and Opalg are virtually letterperfect. It's encouraging to see the upswing in activity.