[proofplan] We treat the two parts separately, each using the same strategy: inherit a solubility series from $G$ and verify that the induced chain in the subgroup (resp. quotient) again has cyclic successive quotients. For Part 1 (subgroups), given $H \le G$ and a solubility series $\{1\} = G_r \trianglelefteq \cdots \trianglelefteq G_0 = G$, we intersect each $G_i$ with $H$ to form $H_i := H \cap G_i$. The [Second Isomorphism Theorem](/theorems/???) provides an injection $H_i / H_{i+1} \hookrightarrow G_i / G_{i+1}$, so each quotient is isomorphic to a subgroup of a cyclic group, hence cyclic. For Part 2 (quotients), given $N \trianglelefteq G$, we project each $G_i$ to $G/N$ via $\overline{G}_i := G_i N / N$. The [Third Isomorphism Theorem](/theorems/???) identifies consecutive quotients $\overline{G}_i / \overline{G}_{i+1}$ as quotients of $G_i / G_{i+1}$, hence cyclic. [/proofplan] [step:Intersect the solubility series with $H$ to build a chain of subgroups] Let $G$ be a soluble group with solubility series \begin{align*} \{1\} = G_r \trianglelefteq G_{r-1} \trianglelefteq \cdots \trianglelefteq G_1 \trianglelefteq G_0 = G, \end{align*} where each quotient $G_i / G_{i+1}$ is cyclic. Let $H \le G$ be a subgroup. For each $0 \le i \le r$, define \begin{align*} H_i := H \cap G_i. \end{align*} Then $H_0 = H \cap G = H$ and $H_r = H \cap \{1\} = \{1\}$, so the chain \begin{align*} \{1\} = H_r \subset H_{r-1} \subset \cdots \subset H_1 \subset H_0 = H \end{align*} runs from $\{1\}$ to $H$. We verify that $H_{i+1} \trianglelefteq H_i$ for each $0 \le i \le r-1$. Let $h \in H_i$ and $x \in H_{i+1}$. Then $h \in G_i$ and $x \in G_{i+1}$, so $hxh^{-1} \in G_{i+1}$ (since $G_{i+1} \trianglelefteq G_i$). Also $hxh^{-1} \in H$ (since $H$ is a subgroup and $h, x \in H$). Therefore $hxh^{-1} \in H \cap G_{i+1} = H_{i+1}$, confirming $H_{i+1} \trianglelefteq H_i$. [guided] The idea is to "restrict" the given solubility series of $G$ to $H$ by intersecting. We need to check two things: that the resulting chain is a subnormal series in $H$ (i.e., each $H_{i+1}$ is normal in $H_i$), and that the successive quotients are cyclic. This step handles the first task. Define $H_i := H \cap G_i$ for $0 \le i \le r$. The endpoints are correct: $H_0 = H \cap G_0 = H \cap G = H$ and $H_r = H \cap G_r = H \cap \{1\} = \{1\}$. The inclusion $H_{i+1} \subset H_i$ holds because $G_{i+1} \subset G_i$ implies $H \cap G_{i+1} \subset H \cap G_i$. For normality, we must show that conjugation by elements of $H_i$ preserves $H_{i+1}$. Take $h \in H_i = H \cap G_i$ and $x \in H_{i+1} = H \cap G_{i+1}$. The conjugate $hxh^{-1}$ lies in $G_{i+1}$ because $G_{i+1} \trianglelefteq G_i$ and $h \in G_i$, $x \in G_{i+1}$. The conjugate also lies in $H$ because $H$ is closed under products and inverses and $h, x \in H$. Therefore $hxh^{-1} \in H \cap G_{i+1} = H_{i+1}$. Note that we are using normality of $G_{i+1}$ in $G_i$ (not in all of $G$). The solubility series provides normality only in consecutive terms, and that is sufficient because we conjugate by elements of $H_i \subset G_i$. [/guided] [/step] [step:Apply the Second Isomorphism Theorem to show each $H_i / H_{i+1}$ is cyclic] Fix $0 \le i \le r-1$. We apply the [Second Isomorphism Theorem](/theorems/???) inside the group $G_i$, with the subgroup $H_i = H \cap G_i$ and the normal subgroup $G_{i+1} \trianglelefteq G_i$. The Second Isomorphism Theorem states: if $A$ is a subgroup of a group $\Gamma$ and $B \trianglelefteq \Gamma$, then $A \cap B \trianglelefteq A$ and \begin{align*} A / (A \cap B) \cong AB / B. \end{align*} We apply this with $\Gamma := G_i$, $A := H_i$, and $B := G_{i+1}$. We verify the hypotheses: - $H_i \le G_i$: this holds since $H_i = H \cap G_i \subset G_i$. - $G_{i+1} \trianglelefteq G_i$: this is part of the given solubility series. The theorem yields \begin{align*} H_i / (H_i \cap G_{i+1}) \cong H_i G_{i+1} / G_{i+1}. \end{align*} Now $H_i \cap G_{i+1} = (H \cap G_i) \cap G_{i+1} = H \cap G_{i+1} = H_{i+1}$ (using $G_{i+1} \subset G_i$). Therefore \begin{align*} H_i / H_{i+1} \cong H_i G_{i+1} / G_{i+1}. \end{align*} The right-hand side $H_i G_{i+1} / G_{i+1}$ is a subgroup of $G_i / G_{i+1}$. Since $G_i / G_{i+1}$ is cyclic by hypothesis, every subgroup of $G_i / G_{i+1}$ is cyclic (subgroups of cyclic groups are cyclic). Therefore $H_i / H_{i+1}$ is cyclic. Since this holds for every $0 \le i \le r-1$, the chain $\{1\} = H_r \trianglelefteq \cdots \trianglelefteq H_0 = H$ is a solubility series for $H$, proving that $H$ is soluble. This completes Part 1. [guided] The crucial observation is that the Second Isomorphism Theorem provides not just a quotient isomorphism, but an *injection* of $H_i / H_{i+1}$ into $G_i / G_{i+1}$. We need this because solubility requires the successive quotients to be cyclic, and we know $G_i / G_{i+1}$ is cyclic but have no direct information about $H_i / H_{i+1}$. We apply the [Second Isomorphism Theorem](/theorems/???) in the ambient group $\Gamma = G_i$, taking $A = H_i$ (a subgroup of $G_i$) and $B = G_{i+1}$ (a normal subgroup of $G_i$). The hypotheses are satisfied: - $H_i = H \cap G_i$ is a subgroup of $G_i$ (as an intersection of a subgroup with $G_i$). - $G_{i+1} \trianglelefteq G_i$ is given by the solubility series of $G$. The theorem gives $H_i / (H_i \cap G_{i+1}) \cong H_i G_{i+1} / G_{i+1}$. We simplify the left-hand side. Since $G_{i+1} \subset G_i$, we have \begin{align*} H_i \cap G_{i+1} = (H \cap G_i) \cap G_{i+1} = H \cap (G_i \cap G_{i+1}) = H \cap G_{i+1} = H_{i+1}. \end{align*} So the isomorphism reads $H_i / H_{i+1} \cong H_i G_{i+1} / G_{i+1}$. Now, $H_i G_{i+1} / G_{i+1}$ is a subgroup of $G_i / G_{i+1}$: it consists of those cosets $g G_{i+1}$ where $g \in H_i G_{i+1} \subset G_i$. Since $G_i / G_{i+1}$ is cyclic (by the solubility series hypothesis), and every subgroup of a cyclic group is cyclic, $H_i G_{i+1} / G_{i+1}$ is cyclic. Via the isomorphism, $H_i / H_{i+1}$ is cyclic. Why do we use "subgroup of a cyclic group is cyclic" rather than "quotient of a cyclic group is cyclic"? Because the Second Isomorphism Theorem embeds $H_i / H_{i+1}$ as a *subgroup* of $G_i / G_{i+1}$ (specifically, the subgroup $H_i G_{i+1} / G_{i+1}$), not as a quotient. Both statements are true for cyclic groups, but the subgroup statement is the one that applies here. [/guided] [/step] [step:Project the solubility series to $G/N$ to build a chain in the quotient] Now let $N \trianglelefteq G$. For each $0 \le i \le r$, define \begin{align*} \overline{G}_i := G_i N / N, \end{align*} the image of $G_i$ under the canonical projection $\pi \colon G \to G/N$, $g \mapsto gN$. Then $\overline{G}_0 = G_0 N / N = GN/N = G/N$ (since $N \subset G$) and $\overline{G}_r = G_r N / N = N/N = \{eN\}$. The chain \begin{align*} \{eN\} = \overline{G}_r \subset \overline{G}_{r-1} \subset \cdots \subset \overline{G}_1 \subset \overline{G}_0 = G/N \end{align*} runs from the identity in $G/N$ to $G/N$ itself. We verify $\overline{G}_{i+1} \trianglelefteq \overline{G}_i$ for each $0 \le i \le r-1$. Since $\pi$ is a surjective group homomorphism and $G_{i+1} \trianglelefteq G_i$, the image $\pi(G_{i+1}) = G_{i+1}N/N$ is normal in $\pi(G_i) = G_i N/N$. (This follows because for any $g \in G_i$ and $x \in G_{i+1}$, the conjugate $\pi(g) \pi(x) \pi(g)^{-1} = \pi(gxg^{-1}) \in \pi(G_{i+1})$, using $gxg^{-1} \in G_{i+1}$.) [guided] The strategy for Part 2 mirrors Part 1, but with projection replacing intersection. Given $N \trianglelefteq G$ and the solubility series of $G$, we push each term $G_i$ forward through the quotient map $\pi \colon G \to G/N$. Define $\overline{G}_i := \pi(G_i) = G_i N / N$ for each $0 \le i \le r$. The endpoints work out: - $\overline{G}_0 = G_0 N / N = GN / N = G/N$, since $G_0 = G$ and $N \subset G$ implies $GN = G$. - $\overline{G}_r = G_r N / N = \{1\} \cdot N / N = N/N = \{eN\}$, the identity element of $G/N$. For the normality of $\overline{G}_{i+1}$ in $\overline{G}_i$: the canonical projection $\pi$ is a group homomorphism, so it preserves normality. Concretely, let $\bar{g} \in \overline{G}_i$ and $\bar{x} \in \overline{G}_{i+1}$. Write $\bar{g} = gN$ for some $g \in G_i$ and $\bar{x} = xN$ for some $x \in G_{i+1}$. Then \begin{align*} \bar{g}\, \bar{x}\, \bar{g}^{-1} = (gN)(xN)(g^{-1}N) = (gxg^{-1})N = \pi(gxg^{-1}). \end{align*} Since $G_{i+1} \trianglelefteq G_i$ and $g \in G_i$, $x \in G_{i+1}$, we have $gxg^{-1} \in G_{i+1}$, so $\pi(gxg^{-1}) \in \pi(G_{i+1}) = \overline{G}_{i+1}$. This confirms $\overline{G}_{i+1} \trianglelefteq \overline{G}_i$. [/guided] [/step] [step:Apply the Third Isomorphism Theorem to show each $\overline{G}_i / \overline{G}_{i+1}$ is cyclic] Fix $0 \le i \le r-1$. We identify the quotient $\overline{G}_i / \overline{G}_{i+1}$ using the [Third Isomorphism Theorem](/theorems/???). The Third Isomorphism Theorem states: if $A \trianglelefteq B$ are both normal subgroups of a group $\Gamma$, then $B/A$ is a normal subgroup of $\Gamma/A$, and \begin{align*} (\Gamma / A) / (B / A) \cong \Gamma / B. \end{align*} We apply this differently. Consider the group $G_i N$ and its subgroups. The quotient $\overline{G}_i / \overline{G}_{i+1}$ can be identified directly as follows: \begin{align*} \overline{G}_i / \overline{G}_{i+1} = (G_i N / N) / (G_{i+1} N / N) \cong G_i N / G_{i+1} N, \end{align*} where the isomorphism is by the [Third Isomorphism Theorem](/theorems/???), applied with the ambient group $G_i N$, and the normal subgroups $G_{i+1} N \trianglelefteq G_i N$ (noting that $N \subset G_{i+1} N$ since $1 \in G_{i+1}$). The hypotheses are satisfied: $G_{i+1} N / N \trianglelefteq G_i N / N$ was verified in the previous step, and $N \subset G_{i+1}N$, so the Third Isomorphism Theorem applies. It remains to show that $G_i N / G_{i+1} N$ is cyclic. Define the surjective homomorphism \begin{align*} \varphi \colon G_i &\to G_i N / G_{i+1} N \\ g &\mapsto g \cdot G_{i+1} N. \end{align*} This is well-defined and surjective: every element of $G_i N$ has the form $gn$ with $g \in G_i$, $n \in N$, and $gn \cdot G_{i+1}N = g \cdot G_{i+1}N$ (since $n \in N \subset G_{i+1}N$). The kernel satisfies $\ker(\varphi) \supset G_{i+1}$ (since $g \in G_{i+1}$ implies $g \cdot G_{i+1}N = G_{i+1}N$). Therefore $\varphi$ factors through $G_i / G_{i+1}$, yielding a surjection $G_i / G_{i+1} \twoheadrightarrow G_i N / G_{i+1} N$. Since $G_i / G_{i+1}$ is cyclic (by the solubility series hypothesis) and $G_i N / G_{i+1} N$ is a quotient of a cyclic group, $G_i N / G_{i+1} N$ is cyclic (quotients of cyclic groups are cyclic). Therefore \begin{align*} \overline{G}_i / \overline{G}_{i+1} \cong G_i N / G_{i+1} N \end{align*} is cyclic. Since this holds for every $0 \le i \le r-1$, the chain $\{eN\} = \overline{G}_r \trianglelefteq \cdots \trianglelefteq \overline{G}_0 = G/N$ is a solubility series for $G/N$, proving that $G/N$ is soluble. This completes Part 2. [guided] The key step is to recognise $\overline{G}_i / \overline{G}_{i+1}$ as a *quotient* of the cyclic group $G_i / G_{i+1}$, which is the dual of the subgroup argument used in Part 1. The [Third Isomorphism Theorem](/theorems/???) (sometimes called the "cancellation" or "correspondence" isomorphism theorem) tells us that for nested normal subgroups $A \trianglelefteq B \trianglelefteq \Gamma$, the quotient $(B/A)$ is normal in $(\Gamma/A)$ and $(\Gamma/A)/(B/A) \cong \Gamma/B$. We use this to simplify $\overline{G}_i / \overline{G}_{i+1}$. Write $\overline{G}_i = G_i N/N$ and $\overline{G}_{i+1} = G_{i+1}N/N$. Both are subgroups of $G/N$ containing $N/N$, and we verified $\overline{G}_{i+1} \trianglelefteq \overline{G}_i$ in the previous step. Since $N \subset G_{i+1}N \subset G_i N$, the Third Isomorphism Theorem (applied within the group $G_i N / N$ with quotient by $G_{i+1}N/N$) gives \begin{align*} \overline{G}_i / \overline{G}_{i+1} = (G_i N / N) \,\big/\, (G_{i+1} N / N) \cong G_i N / G_{i+1} N. \end{align*} Now we must show $G_i N / G_{i+1} N$ is cyclic. The strategy is to exhibit it as a quotient of the cyclic group $G_i / G_{i+1}$. Consider the composite map $G_i \hookrightarrow G_i N \twoheadrightarrow G_i N / G_{i+1} N$. This sends $g \in G_i$ to the coset $g \cdot G_{i+1}N$. It is surjective: any element of $G_i N$ is of the form $gn$ with $g \in G_i$ and $n \in N$, and $gn \cdot G_{i+1}N = g \cdot G_{i+1}N$ because $n \in N \subset G_{i+1}N$. The kernel of this composite contains $G_{i+1}$: if $g \in G_{i+1}$, then $g \in G_{i+1}N$, so $g \cdot G_{i+1}N = G_{i+1}N$ is the identity coset. By the first isomorphism theorem, the map factors as \begin{align*} G_i / G_{i+1} \twoheadrightarrow G_i N / G_{i+1} N. \end{align*} This is a surjective group homomorphism from the cyclic group $G_i / G_{i+1}$ onto $G_i N / G_{i+1} N$. Every quotient of a cyclic group is cyclic: if $C = \langle c \rangle$ is cyclic and $\pi \colon C \to Q$ is a surjection, then $Q = \langle \pi(c) \rangle$ is generated by a single element. Therefore $G_i N / G_{i+1} N$ is cyclic, and so is $\overline{G}_i / \overline{G}_{i+1}$. Note the contrast with Part 1: there we used that *subgroups* of cyclic groups are cyclic (to handle the injection $H_i / H_{i+1} \hookrightarrow G_i / G_{i+1}$), while here we use that *quotients* of cyclic groups are cyclic (to handle the surjection $G_i / G_{i+1} \twoheadrightarrow G_i N / G_{i+1} N$). Both properties hold for cyclic groups, but the direction of the map determines which one is needed. [/guided] [/step]