Prime power order implies subgroups of all orders dividing the group order

Statement
Suppose $$G$$ is a group of prime power order $$p^n$$. Suppose $$0 \le m \le n$$. Then, $$G$$ has a subgroup of order $$p^m$$.

Congruence condition on the number of subgroups

 * Congruence condition on number of subgroups of given prime power order: This states that for any finite group $$G$$, and any prime power $$p^m$$ dividing the order of $$G$$, the number of subgroups of $$G$$ of order $$p^m$$ is congruent to $$1$$ modulo $$p$$. The special case of this (which is used as a first step in the proof) is where the order of $$G$$ is itself a power of $$p$$.

Other related facts

 * Finite group has subgroups of all prime power orders dividing its order
 * Finite nilpotent iff normal subgroups of all orders dividing the group order
 * Subgroups of all orders dividing the group order implies solvable

Facts used

 * 1) uses::Prime power order implies not centerless

Proof
Given: A group $$G$$ of prime power order $$p^n$$. An integer $$m$$ with $$0 \le m \le n$$.

To prove: $$G$$ has a subgroup of order $$p^m$$.

Proof by moving upward on $$m$$
Proof: We prove the result by induction on $$n$$. The base case, $$n = 0$$, is trivial. Suppose $$n \ge 1$$. Note also that if $$m = 0$$, the trivial subgroup works, so we can consider $$m \ge 1$$.


 * 1) $$G$$ has a normal subgroup $$N$$ of order $$p$$: Since $$G$$ is nontrivial, fact (1) tells us that its center $$Z(G)$$ is nontrivial. Pick any non-identity element $$x \in Z(P)$$. The order of $$x$$ is $$p^j$$, with 1 \le j \le n . Pick $$y = x^{p^{j-1}}$$. $$N = \langle y \rangle$$ has order $$p$$, and since $$N \le Z(G)$$, $$N$$ is normal.
 * 2) $$G/N$$ has a subgroup of order $$p^{m-1}$$: $$G/N$$ has order $$p^{n-1}$$, so the induction hypothesis applies to yield that it has a subgroup, say $$M/N$$, of order $$p^{m-1}$$.
 * 3) The subgroup $$M$$ has order $$p^m$$: This is a direct consequence of Lagrange's theorem, and what we know about the orders of $$N$$ and $$M/N$$.

Proof by moving downward on $$m$$
Proof: This uses the idea that every maximal subgroup is normal, and has index $$p$$, hence we can find a subgroup of the required order inside the maximal subgroup.