7
$\begingroup$

I just proved that subgroups of solvable groups are solvable. So given that $G$ is solvable there is $1=G_0 \unrhd G_1 \unrhd \cdots \unrhd G_s=G$ where $G_{i+1}/G_i$ is abelian and for $N$ a normal subgroup we know it is solvable so there is $1=N_0 \unrhd N_1 \unrhd \cdots \unrhd N_{r}=N$ where $N_{i+1}/N_i$ is abelian.

I'm trying to construct the chain for the entire group $G/N$ using the Lattice Isomorphism Theorem but am stuck. Is it possible to do this?

$\endgroup$
  • 1
    $\begingroup$ You have a run-on sentence starting with "So given that". It would be clearer if it were at least two sentences. $\endgroup$ – Michael Hardy Jun 26 '15 at 1:13
  • $\begingroup$ Oops didn't see that. Idk how that happened. $\endgroup$ – TuoTuo Jun 26 '15 at 1:16
5
$\begingroup$

The "standard" proof:

Consider, for each $i$, the subgroups $G_iN/N$ of $G/N$. It is straight-forward to show that $G_iN$ is normal in $G_{i+1}N$ (and thus $G_iN/N$ is normal in $G_{i+1}N/N$).

Now $\dfrac{G_{i+1}N/N}{G_iN/N} \cong G_{i+1}N/G_iN$.

Taking any $x,y \in G_{i+1}$, and $n,n' \in N$, we see that:

$[xn(G_iN),yn'(G_iN)] = [xn,yn']G_iN$, and since $N \lhd G$, $[xn,yn'] \in [x,y]N$, so that:

$[xn,yn']G_iN = [xn,yn']NG_i = ([xn,yn']N)G_i = $

$([x,y]N)G_i = [x,y]NG_i = [x,y]G_iN = ([x,y]G_i)N$.

Since $G_{i+1}/G_i$ is abelian, $[x,y]G_i = G_i$ so $[xn,yn']G_iN = G_iN$,so that

$G_{i+1}N/G_iN$ is abelian, as desired.

(here, $[x,y] = xyx^{-1}y^{-1}$ and we are using the fact that for a group $G$, $G$ is abelian iff $[x,y] = e$ for all $x,y \in G$).

It should be noted that some of the quotients $G_{i+1}N/G_iN$ may be trivial, resulting in a shorter subnormal series for $G/N$.

$\endgroup$
  • $\begingroup$ Can you explain why $([xn,yn']NG_i)=[x,y]NG_i$? $\endgroup$ – Babai Mar 10 at 22:19
5
$\begingroup$

(Corrected) Note that since the chain starts at $1<N$, there exists a maximun $i$ such that $G_i<N$. Multiplying and quotienting out that part of the chain gives a chain of quotients $1\lhd NG_i/N\lhd \cdots \lhd G/N$ and $(NG_{i+1}/N)/(NG_i/N)\simeq NG_{i+1}/NG_i\hookrightarrow G_{i+1}/G_i$ is abelian. The argument is essentially the same as the following: suppose that $\eta:G\to H$ is onto and $G$ is solvable. Then $H$ is solvable.

There's another part to the story: if $G/N$ and $N$ are solvable, show that $G$ is solvable too.

$\endgroup$
  • $\begingroup$ I think there might be an issue with the first term in your chain. In particular, this requires that $(G_{i}/N)/1 \cong G_{i}/N$ be abelian. But how do we know this is abelian? It is true that $(G_{i+1}/N)/(G_{i}/n) \cong G_{i+1}/G_{i}$ for all of them in the chain except for the first one (unless the minimum $G_{i}$ containing $N$ happens to be $G_{i}=N$. $\endgroup$ – TuoTuo Jun 26 '15 at 3:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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