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 $p^0$.