# Every group with order prime power $p^n$ then it has subgroup of order $p^k$ for $0\leq k \leq n$

Every group with order prime power $$p^n$$ then it has subgroup of order $$p^k$$ for $$0\leq k \leq n$$

My Attempt:
By Mathematical Induction (Strong),
As Every subgroup of order $$p$$ , it is cyclic and it is obvious .
For group of order $$p^2$$
Case 1: If cyclic , done
Case 2: If not then by $$G/Z(G)$$ theorem , its $$Z(G)$$ has order $$p$$.
So done

Assume for any group of order $$p^n$$ it is true. And assume it also true for any $$k\leq n$$

TO prove for group of order $$p^{n+1}$$
AS from application of conjugacy class, its centre is nontrivial SO its order must be $$p^a$$ And by strong mathematical induction we have a subgroup of order $$p^t$$ for $$0\leq t \leq a$$

Now I can not able to show this for $$n+1\geq k>a$$
I thought to use fact that $$Z(G)$$ is normal so $$G/Z(G)$$ is the group but I did not get much .

Any Help will be appreciated

Assume the statement is true for any group of order $$p^{n-1}$$ when $$n\geq 2$$ and let $$G$$ be a group of order $$p^n$$. As you said the center is not trivial. So by Cauchy's theorem we know there is an element $$x\in Z(G)$$ of order $$p$$. $$x$$ is in the center so $$g\langle x\rangle g^{-1}=\langle x \rangle$$ for all $$g\in G$$. Hence $$\langle x \rangle \triangleleft G$$ and we can look on the quotient group $$G/\langle x \rangle$$ which has order $$p^{n-1}$$. By induction $$G/\langle x \rangle$$ has subgroups $$H_0, H_1,...H_{n-1}$$ when $$H_k$$ has order $$p^k$$.
Now let's define the projection homomorphism $$\pi:G\to G/\langle x \rangle$$ by $$\pi(g)=g\langle x\rangle$$. For each $$0\leq k\leq {n-1}$$ we have $$\pi^{-1}(H_k)\leq G$$. What is the order of $$\pi^{-1}(H_k)$$? Well, by our assumption $$H_k$$ has $$p^k$$ left cosets of $$\langle x \rangle$$, and we know that each coset contains $$p$$ elements. So there are $$p^{k+1}$$ elements $$g\in G$$ that satisfy $$g\langle x\rangle\in H_k$$. Hence $$|\pi^{-1}(H_k)|=p^{k+1}$$. So that way we found subgroups of $$G$$ of orders $$p,p^2,p^3,...,p^n$$ and of course there is also the trivial group of order $$p^0$$.