[proofplan] We prove by strong induction on $|G|$ that $G$ contains a subgroup of order $p^k$ for every $0 \le k \le a$. The key dichotomy is whether $p$ divides $|Z(G)|$. If it does, [Cauchy's Theorem](/theorems/797) produces an element of order $p$ in the centre, and we quotient by it and apply the inductive hypothesis to the smaller group $G / \langle z \rangle$, lifting subgroups back via the correspondence theorem. If $p \nmid |Z(G)|$, the [Class Equation](/theorems/3242) forces the existence of a conjugacy class whose size is not divisible by $p$, meaning the centraliser of a non-central element has the full $p$-power $p^a$ dividing its order but is strictly smaller than $G$, so the inductive hypothesis applies directly to that centraliser. [/proofplan] [step:Establish the induction framework and the base case] We proceed by strong induction on $|G|$. The statement to be proved is: **$P(n)$:** Every finite group $G$ with $|G| = n$ and $p^a \mid n$, $p^{a+1} \nmid n$, contains a subgroup of order $p^k$ for each $0 \le k \le a$. **Base case.** If $|G| = 1$, then $a = 0$ and the only value of $k$ is $k = 0$. The trivial subgroup $\{e\}$ has order $p^0 = 1$. **Inductive step.** Assume $P(j)$ holds for all $j < n$. Let $G$ be a group with $|G| = n = p^a m$, $p \nmid m$, $a \ge 1$. We must produce a subgroup of order $p^k$ for each $0 \le k \le a$. The case $k = 0$ is the trivial subgroup, so assume $1 \le k \le a$. [guided] We want to show that every finite group $G$ with $p^a \mid |G|$ has subgroups of every $p$-power order up to $p^a$. The natural strategy is strong induction on $|G|$: if we can reduce to a strictly smaller group that still carries enough $p$-power in its order, we can invoke the inductive hypothesis and lift the resulting subgroup back to $G$. The two cases in the inductive step correspond to two different ways of finding such a smaller group — one via the centre, one via a centraliser forced by the Class Equation. **Base case.** If $|G| = 1$, then $p^0 = 1$ divides $|G|$, so $a = 0$ and the only admissible value of $k$ is $k = 0$. The trivial subgroup $\{e\}$ has order $1 = p^0$, and $P(1)$ holds. **Inductive step.** Fix $n \ge 2$ and assume $P(j)$ holds for every positive integer $j < n$. Let $G$ be a finite group with $|G| = n = p^a m$ where $a \ge 1$ and $p \nmid m$. We must produce, for each $0 \le k \le a$, a subgroup of $G$ of order $p^k$. The case $k = 0$ is handled by the trivial subgroup $\{e\}$, so assume $1 \le k \le a$. We now split into two cases depending on whether $p$ divides $|Z(G)|$, the order of the centre of $G$. [/guided] [/step] [step:Apply the Class Equation to split into two cases based on whether $p \mid |Z(G)|$] Let $Z(G) = \{g \in G : gx = xg \text{ for all } x \in G\}$ denote the centre of $G$. The [Class Equation](/theorems/3242) states \begin{align*} |G| = |Z(G)| + \sum_{i=1}^{r} [G : C_G(g_i)], \end{align*} where $g_1, \ldots, g_r$ are representatives of the distinct conjugacy classes of size greater than $1$, and $C_G(g_i) = \{h \in G : hg_i = g_ih\}$ is the centraliser of $g_i$ in $G$. Each index $[G : C_G(g_i)] > 1$ divides $|G|$ by [Lagrange's Theorem](/theorems/782). We split into two cases: $p \mid |Z(G)|$ and $p \nmid |Z(G)|$. [guided] The Class Equation is the central tool. It partitions $G$ into the centre and the non-trivial conjugacy classes: \begin{align*} |G| = |Z(G)| + \sum_{i=1}^{r} [G : C_G(g_i)]. \end{align*} Here $g_1, \ldots, g_r$ are representatives of distinct conjugacy classes with more than one element, and $C_G(g_i) = \{h \in G : hg_i = g_ih\}$ is the centraliser. By the [Orbit-Stabiliser Theorem](/theorems/796) applied to the conjugation action $G \curvearrowright G$ defined by $(g, x) \mapsto gxg^{-1}$, the size of the conjugacy class of $g_i$ equals $[G : C_G(g_i)]$, which divides $|G|$ by [Lagrange's Theorem](/theorems/782). Since $p \mid |G|$, two things can happen: either $p$ divides $|Z(G)|$ (the centre absorbs some $p$-divisibility), or $p$ does not divide $|Z(G)|$ (meaning $p$ divides every non-trivial conjugacy class size, and hence some centraliser is strictly smaller than $G$ but still has $p^a$ dividing its order). These two cases lead to two different constructions. [/guided] [/step] [step:Case 1 — When $p \mid |Z(G)|$, apply Cauchy's Theorem to the centre and quotient] Suppose $p \mid |Z(G)|$. Since $Z(G)$ is an abelian group, [Cauchy's Theorem](/theorems/797) (applied to the abelian group $Z(G)$, which has order divisible by $p$) produces an element $z \in Z(G)$ with $\operatorname{ord}(z) = p$. Since $z \in Z(G)$, the cyclic subgroup $N := \langle z \rangle$ is normal in $G$: for any $g \in G$, $gzg^{-1} = z$ because $z$ commutes with every element, so $gNg^{-1} = N$. We have $|N| = p$. Form the quotient group $\overline{G} := G / N$. Its order is \begin{align*} |\overline{G}| = \frac{|G|}{|N|} = \frac{p^a m}{p} = p^{a-1} m. \end{align*} Since $|\overline{G}| = p^{a-1} m < |G|$ and $p \nmid m$, the inductive hypothesis $P(|\overline{G}|)$ applies. For any $1 \le k \le a$, consider: - If $k = 1$: the subgroup $N = \langle z \rangle$ has order $p = p^1$. - If $2 \le k \le a$: since $a - 1 \ge k - 1 \ge 1$, the inductive hypothesis gives a subgroup $\overline{H} \le \overline{G}$ with $|\overline{H}| = p^{k-1}$. In the second case, let $\pi: G \to G/N$ denote the canonical projection $\pi(g) = gN$. By the [Correspondence Theorem for Groups](/theorems/854), the preimage $H := \pi^{-1}(\overline{H})$ is a subgroup of $G$ containing $N$, and $|H| = |\overline{H}| \cdot |N| = p^{k-1} \cdot p = p^k$. [guided] Suppose $p \mid |Z(G)|$. The centre $Z(G)$ is abelian, and its order is divisible by $p$. We apply [Cauchy's Theorem](/theorems/797), which states: if a finite group has order divisible by a prime $p$, then the group contains an element of order $p$. The hypothesis is satisfied because $Z(G)$ is a finite group with $p \mid |Z(G)|$. This yields $z \in Z(G)$ with $\operatorname{ord}(z) = p$. Why is $N := \langle z \rangle$ normal in $G$? Because $z \in Z(G)$ commutes with every element of $G$: for any $g \in G$ and any $z^j \in N$, we have $g z^j g^{-1} = (gzg^{-1})^j = z^j \in N$. So $gNg^{-1} = N$ for all $g \in G$, i.e., $N \trianglelefteq G$. Now form $\overline{G} = G/N$. We have $|\overline{G}| = |G|/|N| = p^a m / p = p^{a-1}m$, which is strictly less than $|G|$. The highest power of $p$ dividing $|\overline{G}|$ is $p^{a-1}$. For $k = 1$, the subgroup $N$ itself has order $p^1$ and we are done. For $2 \le k \le a$, we need $k - 1 \le a - 1$, which holds. By the inductive hypothesis applied to $\overline{G}$ (which has order $p^{a-1}m < n$), there exists $\overline{H} \le \overline{G}$ with $|\overline{H}| = p^{k-1}$. How do we lift $\overline{H}$ back to a subgroup of $G$? The [Correspondence Theorem for Groups](/theorems/854) states that there is an order-preserving bijection between subgroups of $G/N$ and subgroups of $G$ containing $N$, given by $\overline{H} \mapsto \pi^{-1}(\overline{H})$ where $\pi: G \to G/N$ is the canonical projection. Setting $H = \pi^{-1}(\overline{H})$, we get $H \le G$ with \begin{align*} |H| = |\overline{H}| \cdot |N| = p^{k-1} \cdot p = p^k. \end{align*} This uses the fact that $\pi|_H: H \to \overline{H}$ is surjective with kernel $N$, so $|H| = |\overline{H}| \cdot |\ker(\pi|_H)| = |\overline{H}| \cdot |N|$ by [Lagrange's Theorem](/theorems/782). [/guided] [/step] [step:Case 2 — When $p \nmid |Z(G)|$, find a centraliser with full $p$-power and apply the inductive hypothesis] Suppose $p \nmid |Z(G)|$. Since $|G| = p^a m$ and $p \mid |G|$, the Class Equation \begin{align*} |G| = |Z(G)| + \sum_{i=1}^{r} [G : C_G(g_i)] \end{align*} implies that not every conjugacy class size $[G : C_G(g_i)]$ can be divisible by $p$: if $p \mid [G : C_G(g_i)]$ for all $i$, then $p \mid |G| - |Z(G)|$ and $p \mid |G|$ together give $p \mid |Z(G)|$, contradicting our assumption. Choose $g_j$ such that $p \nmid [G : C_G(g_j)]$. Write $|C_G(g_j)| = |G| / [G : C_G(g_j)] = p^a m / [G : C_G(g_j)]$. Since $p \nmid [G : C_G(g_j)]$, the full $p$-power $p^a$ divides $|C_G(g_j)|$: writing $[G : C_G(g_j)] = d$ with $\gcd(d, p) = 1$, we have $|C_G(g_j)| = p^a (m/d)$, so $p^a \mid |C_G(g_j)|$. Since $g_j \notin Z(G)$, there exists some $x \in G$ with $g_j x \neq x g_j$, so $x \notin C_G(g_j)$. Therefore $C_G(g_j) \subsetneq G$, i.e., $|C_G(g_j)| < |G|$. By the inductive hypothesis applied to $C_G(g_j)$ (which has order $< n$ with $p^a \mid |C_G(g_j)|$), for every $0 \le k \le a$ the group $C_G(g_j)$ has a subgroup of order $p^k$. Since $C_G(g_j) \le G$, this subgroup is also a subgroup of $G$. [guided] Suppose $p \nmid |Z(G)|$. We use the Class Equation to find a proper subgroup of $G$ whose order is still divisible by $p^a$. The Class Equation reads \begin{align*} p^a m = |G| = |Z(G)| + \sum_{i=1}^{r} [G : C_G(g_i)]. \end{align*} We claim that there exists some $g_j$ with $p \nmid [G : C_G(g_j)]$. Suppose for contradiction that $p \mid [G : C_G(g_i)]$ for all $1 \le i \le r$. Then $p$ divides the sum $\sum_{i=1}^r [G : C_G(g_i)]$. Since $p \mid |G|$, subtracting gives $p \mid |Z(G)|$, contradicting our assumption $p \nmid |Z(G)|$. So fix $g_j$ with $p \nmid [G : C_G(g_j)]$. What does this tell us about $|C_G(g_j)|$? By [Lagrange's Theorem](/theorems/782), $|G| = [G : C_G(g_j)] \cdot |C_G(g_j)|$, so \begin{align*} |C_G(g_j)| = \frac{|G|}{[G : C_G(g_j)]} = \frac{p^a m}{[G : C_G(g_j)]}. \end{align*} Since $p \nmid [G : C_G(g_j)]$, the entire factor $p^a$ remains in $|C_G(g_j)|$. Writing $|C_G(g_j)| = p^a m'$ where $m' = m / [G : C_G(g_j)]$ and $p \nmid m'$. Is $C_G(g_j)$ a proper subgroup of $G$? Yes: since $g_j$ is a representative of a non-trivial conjugacy class, $g_j \notin Z(G)$, so there exists $x \in G$ with $xg_jx^{-1} \neq g_j$. This $x$ does not belong to $C_G(g_j)$, so $C_G(g_j) \subsetneq G$ and $|C_G(g_j)| < |G|$. Now the inductive hypothesis applies to $C_G(g_j)$: it is a finite group with $|C_G(g_j)| < n$ and $p^a \mid |C_G(g_j)|$, so it contains a subgroup of order $p^k$ for every $0 \le k \le a$. Any subgroup of $C_G(g_j)$ is a subgroup of $G$, completing this case. [/guided] [/step] [step:Conclude the induction] In both cases, for every $0 \le k \le a$, we have produced a subgroup of $G$ of order $p^k$. This completes the inductive step. By the principle of strong induction, $P(n)$ holds for all $n \ge 1$. In particular, taking $k = a$ yields a subgroup of order $p^a$, which is a Sylow $p$-subgroup of $G$. [guided] We have shown that the inductive step holds in both cases. Whether $p \mid |Z(G)|$ (Case 1) or $p \nmid |Z(G)|$ (Case 2), every finite group $G$ with $|G| = p^a m$, $p \nmid m$, $a \ge 1$, contains a subgroup of order $p^k$ for each $0 \le k \le a$. The principle of strong induction — which allows us to assume $P(j)$ for all $j < n$ when proving $P(n)$ — applies because our base case $P(1)$ holds. In both cases of the inductive step we applied the hypothesis only to groups of strictly smaller order: $|G/N| < |G|$ in Case 1, $|C_G(g_j)| < |G|$ in Case 2. Therefore $P(n)$ holds for all $n \ge 1$. Taking $k = a$ in the conclusion of $P(n)$, every finite group $G$ with $p^a \mid |G|$ and $p^{a+1} \nmid |G|$ has a subgroup of order $p^a$. This subgroup is by definition a Sylow $p$-subgroup of $G$, completing the proof of Sylow's First Theorem. [/guided] [/step]