# Solvable normal subgroup that contains all other solvable normal subgroups

Suppose $H_1,H_2,...,H_n$ are solvable normal subgroups of a group $G$. Prove that there exists a solvable normal subgroup $F$ of $G$ containing all $H_i$.

I am struggling with how to come up with such an $F$. My idea is to let $G$ be solvable of degree $n$ and then since $G$ is solvable of degree $n$, each $H_i \leq G$ is also solvable of degree $n$. Then we have that ${e} = H_{11} \trianglelefteq H_{12} \trianglelefteq...\trianglelefteq H_{1n} = H_1$ and for each $H_i, {e} = H_{i1} \trianglelefteq H_{i2} \trianglelefteq...\trianglelefteq H_{in} = H_i$. We also know that since each of these subgroups $H_i$ is normal, $G/H_i$ is also solvable of degree $n$ and we know that each $H_{ij}/H_{ij-1}$ is abelian. Now I think that there is a way to put them all together and still having all quotients be abelian.

Maybe this is the completely wrong way and I should just consider the union of all $H_i$ then since each element is a subgroup of $G$, we know $F = \cup_{i=1}^{n} H_i \leq G$ so all that remains is to show that $F$ is normal in $G$ and also solvable.

Which of these is correct and how can I finish the proof?

• Hint: Suppose $F$ is the subgroup generated by all the $H_i$'s. Can you prove $F$ is normal? Write an element of $F$ as a finite product of $h$'s, then to conjugate by $g$ insert a $gg^{-1}$ between each $h$. – Michael Burr May 18 '18 at 0:35
• @MichaelBurr so you are saying the second method I mentioned? Would such an element look like $\{h_1, h_2, ..., h_m\} \in F$? Then we would know that for each of these $h_i$ we have $\forall g\in G gh_ig^{-1}\in H_j$ for some $H_j$ (whichever $h_i$ came from) so since $H_j \leq F, gh_ig^{-1} \in F$? – user562567 May 18 '18 at 0:44
• Is the union of subgroups a subgroup? – Michael Burr May 18 '18 at 0:46
• Ooh I thought it was, but I see that its not. So how is such an $F$ constructed? I see you said let $F$ be generated by all the $H_i's$, but can you show what an element would look like? – user562567 May 18 '18 at 0:49
• It's all finite words in the $H$'s. – Michael Burr May 18 '18 at 0:54

Sketch:

• Let $F$ be the subgroup of $G$ generated by the $H_i$'s. Then an element of $F$ is $$h_1h_2\dots h_k,$$ where $h_j\in H_{i_j}$.

• $F$ is normal because $$g(h_1h_2\dots h_k)g^{-1}=gh_1g^{-1}gh_2g^{-1}\dots gh_kg^{-1},$$ and each $H_i$ is normal.

• $F$ is solvable because if $F'$ is the subgroup generated by $H_1,\dots,H_{n-1}$, then, by the argument above, $F'$ is normal in $F$ and the quotient is a subgroup of $H_n$, and therefore solvable. Then proceed by induction.

• Could you tell me if this is where your induction step is going? I want to say that then $F^{k}$ generated by $H_1...H_{n-k}$ is normal in $F^{k}$ and the quotient is a subgroup of $H_n...H_{n-k}$ and is therefore solvable and then I want to show that the same thing holds for $k+1$ or is that not where you are trying to lead me? – user562567 May 18 '18 at 1:16
• TThe induction is on the number of subgroups generating $F$, so it looks like your index $H_{n-k}$ should be $n-k-1$. The quotient would be a subgroup of $H_{n-k}$, not all the larger groups. There is a condition that a group $K$ is solvable iff there is a normal subgroup $N$ so that $N$ and $K/N$ are both solvable. – Michael Burr May 18 '18 at 1:22
• So the endgame is to get that $F^{n-1} \triangleleft F^{n-2}$ and we know $F^{n-1}$ is generated by just $H_1$ so it is just $H_1$ so its solvable and $F^{n-2}/F^{n-1}$ is solvable would have been shown in the previous step so now we know $F^{n-2}$ is solvable and then we work our way back up until we reach $F$? – user562567 May 18 '18 at 1:37
• I also don't see why it would be a subgroup of just $H_{n-k}$. Could you explain? – user562567 May 18 '18 at 1:39
• Since $F^k$ is generated by $H_1,\dots,H_{n-k}$, and you're quotienting out by $F^{k+1}$, which is generated by $H_1,\dots,H_{n-k-1}$, all that you're left with is $H_{n-k}$. Note that $H_{n-k+1},\dots,H_n$ might not even be subsets of $F^k$. Your endgame sounds right (although the induction allows you to hide the endgame details). – Michael Burr May 18 '18 at 1:54