# Is my solution correct? subgroups and quotients of solvable groups are solvable

I want to solve the following exercise from Dummit & Foote's Abstract Algebra:

Prove that subgroups and quotient groups of solvable groups are solvable

My attempt:

Let $G$ be a solvable group with a chain $1=G_0 \triangleleft G_1 \triangleleft \dots \triangleleft G_s=G$ such that all factors $G_i/G_{i-1}$ are abelian.

Let $H \leq G$ be a subgroup. We claim that the chain $1=H_0 \leq H_1 \leq \dots \leq H_s=H$ where $H_i=G_i \cap H$ does the trick. Firstly we prove that $H_i \triangleleft H_{i+1}$ for all $i$. Let $x \in H_{i+1}=G_{i+1} \cap H$. Since $x \in H$ we have $x H x^{-1}=H$. Since $x \in G_{i+1} \triangleright G_i$ we have $xG_i x^{-1}=G_i$. Combining these gives $x(H \cap G_i) x^{-1}=H \cap G_i$, as required.

Next, we prove that the quotients $H_{i+1}/H_i=(G_{i+1} \cap H)/(G_i \cap H)$ are abelian. Let $x,y \in G_{i+1} \cap H$. We would like to prove that $xy (G_i \cap H)=yx (G_i \cap H)$. Observe that using the fact the cosets of $G_i$ in $G_{i+1}$ can be interchanged, we find

$$\begin{split} &xy (G_i \cap H)=(xy G_i) \cap (xy H)=(x G_i) (y G_i) \cap H=(y G_i)(x G_i) \cap (yx H)\\ &=(yx G_i) \cap (yx H)=yx (G_i \cap H). \end{split}$$

Let $N \trianglelefteq G$ be a normal subgroup, and let $G/N$ be the corresponding quotient group. We suggest the following normal chain:

$$1=G_0/ N \triangleleft G_1/ N \triangleleft \dots \triangleleft G_s/ N =G/N.$$

We first prove that $G_i/N \triangleleft G_{i+1}/N$:

Let $gN \in G_{i+1}/N,aN \in G_i/N$. We have $(gN)(aN)(gN)^{-1}=(gag^{-1})N$. Since $G_i \trianglelefteq G_{i+1}$ the element $gag^{-1} \in G_i$, which gives $(gN)(G_i/N)(gN)^{-1} \in G_i/N$ for all $gN \in G_{i+1}/N$.

The quotients (or factors) are

$$(G_{i+1}/N) \big/ (G_i/N),$$ and we wish to prove they are abelian. Indeed, let $g_1N(G_iN),g_2 N(G_iN) \in (G_{i+1}/N) \big/ (G_i N)$ (i.e. $g_1,g_2 \in G_{i+1}$). These two arbitrary elements commute iff

$$[g_2Ng_1N]^{-1}[g_1Ng_2N] \in G_i/N$$ which is equivalent to the condition $g_1^{-1}g_2^{-1}g_1g_2 \in G_i$. Since we have that the quotient $G_{i+1}/G_i$ is abelian we know that $(g_1G_i)(g_2G_i)=(g_2G_i)(g_1 G_i)$ which gives precisely this condition. We conclude that quotient groups of solvable groups are solvable.

• Looks fine. Alternatively to show that $H_{i+1}/H_i$ is abelian, note that for $x,y\in(G_{i+1}\cap H)$, their commutator $xyx^{-1}y^{-1}$ is both in $G_i$ (because $G_{i+1}/G_i$ is abelian) and in $H$ – Hagen von Eitzen Jun 3 '14 at 6:47
• Looks good user1337. Alternatively, you could use the fact that $G$ is solvable if and only if the derived series of $G: G', G'', \dots$ ends after a finite number of steps. This can be easily proved from the definition of solvability. Solvability of subgroups and quotients can be concluded from that with less effort. – Nicky Hekster Jun 3 '14 at 11:02
• I'm not sure your proof makes sense. What is $G_0 /N = 1/N$? We can't assume $G_i > N$... – Eric Auld Dec 15 '14 at 5:42
• Yeah, instead maybe we should consider the chain with factors $\frac{G_iN}{N}$ – Mathematical Mushroom Jun 11 at 18:41

As Eric Auld mentioned in the comment, the statement that $G_0/N=1/N$ is incorrect. To fix this, you may use the normal chain $$$$1=G_0N/ N \triangleleft G_1N/ N \triangleleft \dots \triangleleft G_sN/ N =G/N.$$$$

To show subgroup of solvable group is solvable. .. Let H is subgroup of G where G is solvable $G^k=\{e\}$. Claim: $H^n=G ^n$ for all $n \geq 1$. We will prove the claim by applying induction.

For $n=1$

$H^1=[H, H]$ is a subgroup of $[G, G] =G^1$.

Suppose , result is true for $n-1$.

Now, $H^n=[H^{(n-1)}, H^{(n-1)}]$ is subgroup of $[G^{(n-1)}, G^{(n-1)}]=G^n$

In particular, $H^k$ is subgroup of $G^k=\{e\}$

$H^k={e}$ and so , $H$ is solvable