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I want to show that if $N$ is normal in $G$ then $G$ is solvable implies $G/N$ is solvable.

Now, $G$ is solvable implies there exists a chain
$\{e\}=G_0 \trianglelefteq G_1 \trianglelefteq G_2 \trianglelefteq G_3 \cdots \trianglelefteq G_s=G $, such that $G_i\trianglelefteq G_{i+1}$ and $G_{i+1}/G_i$ is abelian.

We can consider the chain $\overline{N} =G_0N/N \trianglelefteq G_1N/N \trianglelefteq G_2/N \trianglelefteq G_3N/N \cdots \trianglelefteq G_sN/N=G/N $ I want to show that

1. $G_iN/N \trianglelefteq G_{i+1}N/N $ which is equivalent to showing $G_iN\trianglelefteq G_{i+1}N$

2. and $\frac{G_{i+1}N/N} { G_{i}N/N }$ which is isomorphic to $\frac{G_{i+1}N}{G_iN}$ is abelian.

With a lot of brut force somehow I can prove the first part. But I am unable to prove the second part. Can someone suggest me an elegant proof of (1) and any proof of (2)?

I am including a proof of (1) which I have done Proof of (1)

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  1. follows immediately from the third isomorphism theorem. Consider the map $\phi: G_{i+1}N/N \to G_{i+1}N/G_iN$. It is easy to check that this is a homomorphism with kernel $G_iN/N$.

Having obtained $\frac{G_{i+1}N/N}{G_iN/N} \simeq G_{i+1}N/G_iN$ we need to show that the right-hand side is an abelian group. Elaborating on the same argument used here Quotients of Solvable Groups are Solvable:

consider the commutators $[xn, ym]$ for $x, y \in G_{i+1}$, $n,m \in N$. We want $[xn,ym] \in G_iN$, given that $[x,y] \in G_i$. Now, $[xn,ym]G_iN = ([xn,ym]N)G_i$, so it suffices to show that $[xn,ym]N = [x,y]N$. This is again brute force

$xnymn^{-1}x^{-1}m^{-1}y^{-1}N = xyx^{-1}y^{-1}N$

$\iff nymn^{-1}x^{-1}mN = yx^{-1}N$

$\iff (y^{-1}ny)(mn^{-1})(x^{-1}mx)N = N$, and this is true because $N$ is normal.

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  • $\begingroup$ Why is $G_i$ normal in $G_{i+1}N$? $\endgroup$ – Babai Mar 10 at 21:15
  • $\begingroup$ Sorry! I made a mistake and I removed it. I will have to think of another proof - the only ones I know of use brute force. $\endgroup$ – vxnture Mar 10 at 21:34
  • $\begingroup$ I can already show the isomorphism you have just shown above. What I am struggling to show is enough of those isomorphic quotients are abelian. $\endgroup$ – Babai Mar 10 at 21:54
  • $\begingroup$ math.stackexchange.com/questions/1339609/… there is a proof here (with commutators) $\endgroup$ – vxnture Mar 10 at 21:59
  • $\begingroup$ There is a gap in that proof too, I am not able to verify it $\endgroup$ – Babai Mar 11 at 5:24

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