MHB Unlock Role of Correspondence Thm for Groups in Analysing Composition Series

[Math Amateur]

I have made two posts recently concerning the composition series of groups and have received considerable help from Euge and Deveno regarding this topic ... in particular, Euge and Deveno have pointed out the role of the Correspondence Theorem for Groups (Lattice Isomorphism Theorem for Groups) in analysing composition series ...

I am trying to fully understand the role of the Correspondence Theorem for Groupsin analysing composition series ... but need a bit more help ...

The issue I am focused on is the following:

Aluffi in his book, Algebra: Chapter 0 in the proof of the Jordan-Holder Theorem (pages 206 - 207) ... given a composition series:

$$G = G_0 \supsetneq G_1 \supsetneq G_2 \supsetneq \ ... \ ... \ \supsetneq G_n = \{e \} $$

states the following:

" ... ... there are no proper normal subgroups between $$G_1$$ and $$G$$ since $$G/G_1$$ is simple ... ... "

... so then more generally we have the following:

... there are no proper normal subgroups between $$G_{ i + 1}$$ and $$G_i$$ since $$G_i/G_{ i + 1}$$ is simple ... ... "Now ... ... in a previous post, Euge pointed out that this statement can be established through applying the Correspondence Theorem ... but how, exactly?To establish a notation, I am providing the statement of the Correspondence Theorem from Joseph J Rotman's undergraduate text, An Introduction to Abstract Algebra with Applications (Third Edition) ... as follows ... :View attachment 4919Now to restate the above in terms of our problem, we have:

$$G_{i + 1} \triangleleft G_i $$

Then $$\text{ Sub}( G_i ; G_{i + 1} )$$ is the family of all those subgroups $$S$$ of $$G_i$$ containing $$G_{i + 1}$$

and

$$\text{ Sub}( G_i / G_{i + 1} )$$ is the family of all subgroups of $$ G_i / G_{i + 1} $$
Now ... we need to show that

$$G_i / G_{i + 1}$$ is simple $$\Longrightarrow$$ there are no proper normal subgroups between $$G_i$$ and $$G_{i + 1}$$

... BUT ... ... how exactly do we do this ... ... ?Seems like we should use the Correspondence Theorem part (iii) ... but how exactly ... ?

Hope someone can help ...

Peter

 

[Euge]

Let's do it this way:

Lemma. Let $K$ be a normal subgroup of a group $G$. If $G/K$ is simple, then there are no normal subgroups $N$ of $G$ such that $K \subsetneq N \subsetneq G$.Proof. I'll prove the contrapositive, i.e., if there is a normal subgroup $N$ of $G$ such that $K\subsetneq N \subsetneq G$, then $G/K$ is not simple. Suppose such an $N$ exists, and consider the factor group $N/K$. Note that $N/K$ is the image of $N$ under the correspondence $\operatorname{Sub}(G;K) \to \operatorname{Sub}(G/K)$ in the Correspondence Theorem. By Proposition 2.123(iii), $N/K$ is normal in $G/K$. Furthermore, $N/K \neq G/K$ and $N/K \neq K$ (remember, $K = K/K$ is the identity of $G/K$), for otherwise Proposition 2.123(i) gives $N = G$ and $N = K$, respectively. These are contrary to assumption. Therefore, $N/K$ is a proper, nontrivial normal subgroup of $G/K$. This means that $G/K$ is not a simple group.

 

[Math Amateur]

Let's do it this way:

Lemma. Let $K$ be a normal subgroup of a group $G$. If $G/K$ is simple, then there are no normal subgroups $N$ of $G$ such that $K \subsetneq N \subsetneq G$.Proof. I'll prove the contrapositive, i.e., if there is a normal subgroup $N$ of $G$ such that $K\subsetneq N \subsetneq G$, then $G/K$ is not simple. Suppose such an $N$ exists, and consider the factor group $N/K$. Note that $N/K$ is the image of $N$ under the correspondence $\operatorname{Sub}(G;K) \to \operatorname{Sub}(G/K)$ in the Correspondence Theorem. By Proposition 2.123(iii), $N/K$ is normal in $G/K$. Furthermore, $N/K \neq G/K$ and $N/K \neq K$ (remember, $K = K/K$ is the identity of $G/K$), for otherwise Proposition 2.123(i) gives $N = G$ and $N = K$, respectively. These are contrary to assumption. Therefore, $N/K$ is a proper, nontrivial normal subgroup of $G/K$. This means that $G/K$ is not a simple group.

Thanks so so much for your help Euge ...

The proof is very clear and extremely helpful ...

Thanks again,

Peter

 

[Deveno]

Let's look at an example, because it's fun to play with our toys.

For our group $G$, we'll choose $D_6$ the symmetry group of the regular hexagon, of order $12$.

Explicitly, we'll write this as $\langle r,s: r^6 = s^2 = 1, sr = r^5s\rangle$, so that:

$G = \{1,r,r^2,r^3,r^4,r^5,s,rs,r^2s,r^3s,r^4s,r^5s\}$.

We can think of $r$ as being the linear transformation $\Bbb R^2 \to \Bbb R^2$ which has the matrix (in the standard basis):

$R = \begin{bmatrix} \frac{1}{2}&-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}&\frac{1}{2}\end{bmatrix}$

and $s$ being the linear transformation $\Bbb R^2 \to \Bbb R^2$ which has the matrix:

$S = \begin{bmatrix}1&0\\0&-1\end{bmatrix}$ (reflection about the $x$-axis, the line between vertex $1$ and $3$).

Now the subgroup $\langle r\rangle$ (the *rotation* subgroup) is of index $2$, so automatically normal, and the quotient, being of prime order, is necessarily simple.

So we can begin our composition series $D_6 \supsetneq \langle r\rangle$.

Now $\langle r\rangle$ is cyclic (of order $6$), but not of prime order, so it has a non-trivial subgroup (all of its subgroups are normal, since it is abelian)of order $d$ for any divisor $d$ of $6$. So let's pick $d = 3$, so we have the subgroup:

$\langle r^2\rangle = \{1,r^2,r^4\}$.

As before, we have $\langle r^2\rangle$ of index $2$ in $\langle r\rangle$. And $\langle r^2\rangle$ is simple (it is cyclic of prime order), so our composition series is the following:

$D_6 \supsetneq \langle r\rangle \supsetneq \langle r^2\rangle \supsetneq \{1\}$.

**********

Let's do this another way. First, we want to find a normal subgroup, let's use a different one than we used above. So we'll use $Z(D_6) = \{1,r^3\}$ (it's a fun exercise to prove this is indeed the center), which is always a normal subgroup.

Just for grins, let's investigate what $D_6/Z(D_6)$ is, and how the correspondence theorem plays out for it.

First, let's meet our cosets:

$Z(D_6) = \{1,r^3\}$
$rZ(D_6) = \{r,r^4\}$
$r^2Z(D_6) = \{r^2,r^5\}$
$sZ(D_6) = \{s,r^3s\}$ (recall $r^3$ commutes with $s$ so $sr^3 = r^3s$).
$rsZ(D_6) = \{rs, r^4s\}$
$r^2sZ(D_6) = \{r^2s,r^5s\}$

Clearly, $D_6/Z(D_6)$ has order $6$, so it is either cyclic of order $6$, or isomorphic to $S_3$. Which is it?

Note $(rZ(D_6))(sZ(D_6)) = (rs)Z(D_6)$, while:

$(sZ(D_6))(rZ(D_6)) = (sr)Z(D_6) = (r^5s)Z(D_6) = (r^2s)Z(D_6) \neq (rs)Z(D_6)$

so we conclude $D_6/Z(D_6) \cong S_3$ (since it is non-abelian).

For clarity's sake, let's abbreviate $gZ(D_6)$ as $[g]$. So the subgroups of the quotient $D_6/Z(D_6)$ are:

$\{[1],[r],[r^2],,[rs],[r^2s]\}$ -isomorphic to $S_3$
$\{[1],[r],[r^2]\}$ -isomorphic to $A_3$ (in this case the $3$-cycles and the identity).
$\{[1],\}$
$\{[1],[rs]\}$
$\{[1],[r^2s]\}$ - these correspond to the groups generated by a transposition
$\{[1]\}$

The correspondence theorem then tells us the subgroups of $D_6$ containing $Z(D_6)$ are:

$D_6$
$\langle r\rangle$ (this is the only subgroup of order $6$ containing $Z(D_6)$, it is cyclic.)
$\{1,r^3,s,r^3s\}$
$\{1,r^3,rs,r^4s\}$
$\{1,r^3,r^2s,r^5s\}$ (these three groups are isomorphic to $V_4$, the klein viergruppe)
$Z(D_6)$

Now let's use $S_3$ and $\Bbb Z_2$ to build a composition series for $D_6$:

First, the composition series for $S_3$:

$S_3,A_3,\{e\}$.

The composition series for $\Bbb Z_2$ is, of course, trivial, since $\Bbb Z_2$ is simple.

Pulling back the composition series for $S_3$ to $D_6$, and continuing with the composition series for $Z(D_6) \cong \Bbb Z_2$ we have:

$D_6, \langle r\rangle, \langle r^3\rangle,\{1\}$

Note this is a *different* composition series, but both series we have found have the same length, and the quotients that occur are both the same, just in a different order.

 

[Math Amateur]

Let's look at an example, because it's fun to play with our toys.

For our group $G$, we'll choose $D_6$ the symmetry group of the regular hexagon, of order $12$.

Explicitly, we'll write this as $\langle r,s: r^6 = s^2 = 1, sr = r^5s\rangle$, so that:

$G = \{1,r,r^2,r^3,r^4,r^5,s,rs,r^2s,r^3s,r^4s,r^5s\}$.

We can think of $r$ as being the linear transformation $\Bbb R^2 \to \Bbb R^2$ which has the matrix (in the standard basis):

$R = \begin{bmatrix} \frac{1}{2}&-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}&\frac{1}{2}\end{bmatrix}$

and $s$ being the linear transformation $\Bbb R^2 \to \Bbb R^2$ which has the matrix:

$S = \begin{bmatrix}1&0\\0&-1\end{bmatrix}$ (reflection about the $x$-axis, the line between vertex $1$ and $3$).

Now the subgroup $\langle r\rangle$ (the *rotation* subgroup) is of index $2$, so automatically normal, and the quotient, being of prime order, is necessarily simple.

So we can begin our composition series $D_6 \supsetneq \langle r\rangle$.

Now $\langle r\rangle$ is cyclic (of order $6$), but not of prime order, so it has a non-trivial subgroup (all of its subgroups are normal, since it is abelian)of order $d$ for any divisor $d$ of $6$. So let's pick $d = 3$, so we have the subgroup:

$\langle r^2\rangle = \{1,r^2,r^4\}$.

As before, we have $\langle r^2\rangle$ of index $2$ in $\langle r\rangle$. And $\langle r^2\rangle$ is simple (it is cyclic of prime order), so our composition series is the following:

$D_6 \supsetneq \langle r\rangle \supsetneq \langle r^2\rangle \supsetneq \{1\}$.

**********

Let's do this another way. First, we want to find a normal subgroup, let's use a different one than we used above. So we'll use $Z(D_6) = \{1,r^3\}$ (it's a fun exercise to prove this is indeed the center), which is always a normal subgroup.

Just for grins, let's investigate what $D_6/Z(D_6)$ is, and how the correspondence theorem plays out for it.

First, let's meet our cosets:

$Z(D_6) = \{1,r^3\}$
$rZ(D_6) = \{r,r^4\}$
$r^2Z(D_6) = \{r^2,r^5\}$
$sZ(D_6) = \{s,r^3s\}$ (recall $r^3$ commutes with $s$ so $sr^3 = r^3s$).
$rsZ(D_6) = \{rs, r^4s\}$
$r^2sZ(D_6) = \{r^2s,r^5s\}$

Clearly, $D_6/Z(D_6)$ has order $6$, so it is either cyclic of order $6$, or isomorphic to $S_3$. Which is it?

Note $(rZ(D_6))(sZ(D_6)) = (rs)Z(D_6)$, while:

$(sZ(D_6))(rZ(D_6)) = (sr)Z(D_6) = (r^5s)Z(D_6) = (r^2s)Z(D_6) \neq (rs)Z(D_6)$

so we conclude $D_6/Z(D_6) \cong S_3$ (since it is non-abelian).

For clarity's sake, let's abbreviate $gZ(D_6)$ as $[g]$. So the subgroups of the quotient $D_6/Z(D_6)$ are:

$\{[1],[r],[r^2],,[rs],[r^2s]\}$ -isomorphic to $S_3$
$\{[1],[r],[r^2]\}$ -isomorphic to $A_3$ (in this case the $3$-cycles and the identity).
$\{[1],\}$
$\{[1],[rs]\}$
$\{[1],[r^2s]\}$ - these correspond to the groups generated by a transposition
$\{[1]\}$

The correspondence theorem then tells us the subgroups of $D_6$ containing $Z(D_6)$ are:

$D_6$
$\langle r\rangle$ (this is the only subgroup of order $6$ containing $Z(D_6)$, it is cyclic.)
$\{1,r^3,s,r^3s\}$
$\{1,r^3,rs,r^4s\}$
$\{1,r^3,r^2s,r^5s\}$ (these three groups are isomorphic to $V_4$, the klein viergruppe)
$Z(D_6)$

Now let's use $S_3$ and $\Bbb Z_2$ to build a composition series for $D_6$:

First, the composition series for $S_3$:

$S_3,A_3,\{e\}$.

The composition series for $\Bbb Z_2$ is, of course, trivial, since $\Bbb Z_2$ is simple.

Pulling back the composition series for $S_3$ to $D_6$, and continuing with the composition series for $Z(D_6) \cong \Bbb Z_2$ we have:

$D_6, \langle r\rangle, \langle r^3\rangle,\{1\}$

Note this is a *different* composition series, but both series we have found have the same length, and the quotients that occur are both the same, just in a different order.

Hi Deveno ... thanks for the help ..

So good to have an example! ... I felt that I really needed an example on the Jordan-Holder Theorem as I did not have a good feel or sense of how/why the result held and how the result would look in a concrete case ... ... and being an obsessive collector of mathematics texts, especially algebra texts, I had searched my various texts for an example and found none ... yes, that is right, more than half a dozen texts and not one example! ... so a special thanks for your post!

I am now working through the details ...

I do find that examples help give a concrete sense of how and why a result holds ... maybe even more so than the proof ... especially where the proof is a bit abstract ...

Thanks again,

Peter