Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am interested in how far further than the Jordan-Holder theorem we can go.

Say $1\triangleleft A\triangleleft B\triangleleft G$ and $1\triangleleft C\triangleleft D\triangleleft G$ are two subnormal series of a group $G$ such that

$$\frac{A}{1}\cong\frac{C}{1},\quad\frac{B}{A}\cong\frac{D}{C},\quad\frac{G}{B}\cong\frac{G}{D}.$$

That is, the succesive factors are isomorphic in parallel. Is it possible for $B\not\cong D$?

What if we consider subnormal, normal, composition, or chief series of given length $n$; if the factor groups are isomorphic in parallel then do any of the terms (other than two directly over $1$) need to be isomorphic, and can any given ordered sequence of $\cong$ and $\not\cong$s be realized by a pair of series? If we impose the constraint that some chosen subset of the terms in each series is isomorphic can we force the remaining ones to also be isomorphic?

share|improve this question

2 Answers 2

Suppose you have the group $G = C_4 \times C_2$. Then there exist subnormal series $1 \subset C_2 \subset C_4 \subset G$ and $1 \subset C_2 \subset C_2 \times C_2 \subset G$. The corresponding quotients are all isomorphic, but $C_4$ is not isomorphic to $C_2 \times C_2$.

In general, you could take $G = C_{2^n} \times C_2$. There exist composition series $$1 \subset C_2 \subset C_4 \subset C_8 \subset \ldots \subset C_{2^n} \subset G$$ and $$1 \subset C_2 \subset C_2 \times C_2 \subset C_4 \times C_2 \subset \ldots \subset C_{2^{n-1}} \times C_2 \subset G.$$ Then the factor groups are all $C_2$, but none of the corresponding groups other than the initial 1, the second group $C_2$, and the final $G$, are isomorphic.

share|improve this answer
    
Excellent. I was able adapt this family of examples to answer the generalized question. –  blue Nov 15 '12 at 6:22
up vote 0 down vote accepted

More generally: given any ordered sequence of $\cong$s and $\not\cong$s, we can realize a pair of composition series whose terms are related by this sequence. In particular, if $s$ denotes the number of switches from $\cong$ to $\not\cong$ in the sequence, and $\ell$ the length of the sequence, then for any prime $p$ there are pairs of composition series of $C(p^\ell)\oplus C(p^s)$ for which this holds.

Let $f:\{1,2,\cdots,\ell\}\to \{\cong,\not\cong\}$ represent an ordered sequence of $\cong$s and $\not\cong$s, with the condition that $f(1),f(\ell)$ are both $\cong$. Let $s$ be the number of 'breaks,' or indices $1\le i< \ell$ such that $f(i)$ is the $\cong$ symbol and $f(i+1)$ is the $\not\cong$ symbol. Given a prime $p$, define $G=C(p^{\ell-s})\oplus C(p^s)$ to be a direct sum of cyclic groups, which will be a finite abelian $p$-group. We construct a pair of series $H_i$ and $K_i$ for $G$ such that $H_i\,f(i)\,K_i$ for each $1\le i\le\ell$ through a recursive alternating process.

To wit: set $H_1=C(p)\oplus C(1)=K_1$. For each $1<i\le\ell$,

  • if $\displaystyle\begin{cases}f(i)~~~~~~~\text{ is }\cong \\ f(i+1)\text{ is }\cong\\H_i,K_i=C(p^\alpha)\oplus C(p^\beta) \end{cases}$ then $\displaystyle\begin{cases}H_{i+1}=C(p^{\alpha+1})\oplus C(p^\beta)\\ K_{i+1}=C(p^{\alpha+1})\oplus C(p^\beta)\end{cases}$;

  • if $\displaystyle\begin{cases}f(i)~~~~~~~\text{ is }\cong \\ f(i+1)\text{ is }\not\cong\\H_i,K_i=C(p^\alpha)\oplus C(p^\beta) \end{cases}$ then $\displaystyle\begin{cases}H_{i+1}=C(p^\alpha)\oplus C(p^{\beta+1}) \\ K_{i+1}=C(p^{\alpha+1})\oplus C(p^\beta)\end{cases}$;

  • if $\displaystyle\begin{cases}f(i)~~~~~~~\text{ is }\not\cong \\ f(i+1)\text{ is }\not\cong\\H_i=C(p^\alpha)\oplus C(p^{\beta+1}) \\ K_i=C(p^{\alpha+1})\oplus C(p^\beta) \end{cases}$ then $\displaystyle\begin{cases}H_{i+1}=C(p^{\alpha+1})\oplus C(p^{\beta+1}) \\ K_{i+1}=C(p^{\alpha+2})\oplus C(p^\beta)\end{cases}$;

  • if $\displaystyle\begin{cases}f(i)~~~~~~~\text{ is }\not\cong \\ f(i+1)\text{ is }\cong\\H_i=C(p^\alpha)\oplus C(p^{\beta+1}) \\ K_i=C(p^{\alpha+1})\oplus C(p^\beta) \end{cases}$ then $\displaystyle\begin{cases}H_{i+1}=C(p^{\alpha+1})\oplus C(p^{\beta+1}) \\ K_{i+1}=C(p^{\alpha+1})\oplus C(p^{\beta+1})\end{cases}$.

In other words, we create two paths in $\Bbb N^2$ starting at $(1,0)$: $\cong\to\cong$ and $\not\cong\to\not\cong$ correspond to adding the vector $(1,0)$ to both paths; $\cong\to\not\cong$ corresponds to adding $(0,1)$ and $(1,0)$ to the paths resp. and $\not\cong\to\cong$ corresponds to adding $(1,0)$ and $(1,0)$ to the paths resp.

As an example, for $p=2$ and $(f(i))_{i=1}^4=(\cong,\not\cong,\not\cong,\cong)$, we have for $C_8\oplus C_2$,

$$\begin{array}{|c|c|c|c|c|}\hline i & 1 & 2 & 3 & 4 \\ \hline H_i & C_2\oplus C_1 & C_2\oplus C_2 & C_4\oplus C_2 & C_8\oplus C_2 \\ \hline K_i & C_2\oplus C_1 & C_4\oplus C_1 & C_8\oplus C_1 & C_8\oplus C_2 \\ \hline \end{array}.$$

It is then easy to see, setting $H_0=C_1=K_0$ for convenience, that $$\frac{H_4}{H_3}\cong\frac{H_3}{H_2}\cong\frac{H_2}{H_1}\cong \frac{H_1}{H_0}\cong C_2\cong\frac{K_1}{K_0}\cong\frac{K_2}{K_1}\cong\frac{K_3}{K_2}\cong\frac{K_4}{K_3}, \\ H_1\cong K_1,\quad H_2\not\cong K_2,\quad H_3\not\cong K_3,\quad H_4\cong K_4.$$

Note that, for every positive integer and divisor, $d\mid n$, we view $C_d$ as a subgroup of $C_n$.

This extends Ted's examples so that every question is answered: every sequence can be realized, no subset of terms can be set isomorphic in order to force the other terms to be isomorphic, and this applies to chief series (since all subgroups of abelian groups are normal) which are examples of normal, subnormal, and composition series as well. All with finite abelian $p$-groups of two factors.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.