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  1. MHB Master
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    #1
    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

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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?

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    #3 Thread Author
    Quote Originally Posted by mathbalarka View Post
    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.

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

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    #5
    Quote Originally Posted by mathbalarka View Post
    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.

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    #6
    Quote 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

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    #7
    Quote Originally Posted by mathbalarka View Post
    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.

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