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Let $G$ is a group with order $p^n$ where $p$ is prime and $n \geq 3$. By Sylow's Thm, we know that $G$ has a subgroup with order $p^2$.

But, I wonder to proof without Sylow's Thm.

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An interesting fact: The number of such subgroups is congruent to 1 modulo $p$. The proof I learned requires the Sylow theorems, though. – Brett Frankel Apr 19 '12 at 3:00
@DimitriSurinx This is not one of the Sylow theorems: The subgroup in question does not have maximal $p$-power order. The theorem I am quoting is that if $p^n$ divides the order of a $G$, then $G$ has 1 mod $p$ subgroups of order $p^n$, even if $p^{n+1}$ also divides the order of $G$. See, for instance, Burnside's Theory of Groups of Finite Order. – Brett Frankel Apr 19 '12 at 16:28
up vote 4 down vote accepted

Theorem. Let $G$ be a group of order $p^n$, where $p$ is a prime and $n\gt 0$. Then $G$ has subgroups of order $p^i$ for each $i$, $0\leq i\leq n$.

Proof. Induction on $n$. The result is true if $n=1$ or if $n=2$ by Cauchy's Theorem.

Assume the result holds for groups of order $p^k$, $k\lt n$. If $i=0$, we can take the trivial subgroup. If $i=1$, the result follows from Cauchy's Theorem. So assume $i\gt 1$.

Since $G$ is a $p$-group, $Z(G)\neq\{1\}$ by the class formula. Hence, $Z(G)$ has an element of order $p$. Let $g\in Z(G)$ generate a subgroup of order $p$. Then $G/\langle g\rangle$ has order $p^{n-1}$, and so has a subgroup $\overline{H}$ of order $p^{i-1}$ by the induction hypothesis. By the isomorphism theorem, $\overline{H}$ corresponds to a subgroup $H$ of $G$ that contains $\langle g\rangle$. Thus, $$|H| = [H:\langle g\rangle][\langle g\rangle:1] = \left|\frac{H}{\langle g\rangle}\right|p = |\overline{H}|p = p^{i-1}p = p^i.$$ Thus, $G$ has a subgroup of order $p^i$. $\Box$

Alternatively, the result holds for abelian groups. If $p^i\leq |Z(G)|$, we can find a subgroup of $Z(G)$ of order $p^i$; otherwise, find a subgroup of order $p^j$ of $G/Z(G)$, where $p^j|Z(G)| = p^i$.

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Since $G$ is a $p$-group, $G$ has a nontrivial center by the class equation. Either the center has order at least $p^2$, and so we can find the desired subgroup inside the center by the classification of finitely generated abelian groups, or else the center has order $p$. In that case, take any other nontrivial element of $G$. If that element has order at least $p^2$, it generates a cyclic subgroup which itself has a subgroup of order $p^2$. If it has order $p$, this element will generate the desired subgroup along with the center.

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Well, we could apply the fact that the center of such a group $G$ is nontrivial (proof). Since the center is nontrivial, it either has order $p$ or $p^m$ for $1<m\leq n$. In the former case, $G/Z(G)$ is of order $p^{n-1}$ so $Z(G/Z(G))$ is nontrivial, hence has order a power of $p$. Since $Z(G/Z(G))$ is abelian, it is easy to find such a subgroup (i.e. using the Fundamental Theorem of Finite Abelian Groups). The latter case is similarly easy, as $Z(G/Z(G))$ is an abelian group of $p^m$ for $m>1$.

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Let's try a direct approach assuming the weaker condition $G$ finite with $p^2\mid |G|\quad$ (i.e., $G$ not necessarily a $p$-group):

By Cauchy, there exists an element $x \in G$ of order $p$.

Case 1: $p$ divides the index of the normalizer $\mathop{N}_G(\langle x \rangle)$ in $G$

$\langle x \rangle$ acts by right multiplication on the set of cosets $\mathop{N}_G(x)\backslash G = \{\mathop{N}_G(\langle x \rangle)g : g\in G\}\quad$. All orbits of $\langle x \rangle$ have order $1$ or $p$, implying that the number of fixed points of $\langle x \rangle$ equals the index of $\mathop{N}_G(\langle x \rangle)$ in $G$ modulo $p$, which is $0$ modulo $p$ in this case. As $\mathop{N}_G(\langle x \rangle)$ is a fixed point of $\langle x \rangle$, there has to be another fixed point, say $\mathop{N}_G(\langle x \rangle)g = \mathop{N}_G(\langle x \rangle)gx\quad$, which is equivalent to $gxg^{-1} = x^{g^{-1}} \in \mathop{N}_G(\langle x \rangle)\quad$. As $g\not\in\mathop{N}_G(\langle x \rangle)\quad$, $x^{g^{-1}}$ is an element of order $p$ not contained in $\langle x \rangle$ that normalizes $\langle x \rangle$. Hence $\langle x, x^{g^{-1}} \rangle$ is a subgroup of order $p^2$ of $G$.

Case 2: The index of the normalizer $\mathop{N}_G(\langle x \rangle)$ in $G$ is coprime to $p$

Then the quotient group $\mathop{N}_G(\langle x \rangle)/\langle x \rangle\quad$ has order divisible by $p$. By Cauchy, it contains a subgroup of order $p$, whose preimage has order $p^2$.

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The question assumes $G$ to be a finite $p$-group. The Sylow theorems (existence, conjugacy and index $= 1 \bmod p$) do not give any nontrivial information in this case! – meinereiner Apr 19 '12 at 13:39
There is a version of Sylow's First Theorem that states: "If $G$ is a finite group, and $p^n$ divides $|G|$, then $G$ has subgroups of order $p^i$, $0\leq i\leq n$." This implies the existence of Sylow subgroups, and does give nontrivial information as per the statement. – Arturo Magidin Apr 19 '12 at 15:00
@ArturoMagidin: I prefer to think of Sylow subgroups as maximal $p$-subgroups, and the Sylow theorem(s) as the statement that all the maximal $p$-subgroups are conjugate and of maximal possible order. The version of Sylow's first theorem you mention can be proved by minimally adapting my answer. – meinereiner Apr 19 '12 at 16:10
It can be proven in many different ways, of course. My point is that the version I mention (which appears, e.g., in Hungerford's Algebra as "Sylow's First Theorem") would make your claim that "The Sylow theorems do not give any nontrivial information in this case" incorrect. The version you prefer may not give any nontrivial information, but there are other, equivalent versions that do. – Arturo Magidin Apr 19 '12 at 16:16

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