Finite p-group contains normal subgroups $H_i$, $|H_i| = p^i$

Question

Let $G$ be a finite group of order $p^n$, where $p$ is prime. Show that $G$ contains normal subgroups $H_i$ for $1 \leq H_i \leq n$ such that $|H_i| = p^i$ and $H_i < H_{i+1}$ for $1 \leq i < n$ .

Using induction:

$k = 0$ : $p^0$ is a normal subgroup of $G$, the trivial $\{e\}$.

$k = i $ : We assume that $G$ has a normal subgroup $H_i$ of order $p^i$, for some $i < n$ .

$k = i + 1$: There exists a normal subgroup $\tilde{H}$ of order $p$ in $Z[G/H_i]$ (Argument N0.$1$)

Thus, $H_{i + 1} = \gamma^{-1}[\tilde{H}]$ (where $\gamma$: normal homomorphism) is a normal subgroup of $G$ with order $p^{i+1}$ (Argument No.$2$)

Question:

It'd be very helpful to see an elaboration on Arg.$1$ and Arg.$2$.

Why $Z[G/H_i]$ has a normal subgroup of order $p$ and why does the inverse image of this subgroup gives the required normal subgroup of order $p^{i+1}$?

Answer 1

$p-$Groups (groups $G$ of order $p^n$) have non-trivial centre, this follows from class equation, $$ |G|=Z(G)+\sum_{distinct}(| \text{Orb}_G(x)|) $$

Where, $\text{Orb}_G(x)=\{ gxg^{-1} | g \in G \}$ with $ x\notin Z(G)$ and $(|\text{Orb}_G(x)|)=|G|/|\text{C}_{G}(x)|$ , (By Orbit-Stabilizer theorem)

$x \in Z(G) \text{ iff } 1=|\text{Orb}_G(x)|$

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Since $Z(G/H_i)$ is $p-$ group with order less than $G$ and greater than $1$, so we can use induction hypothesis (or Cauchy theorem) to infer that $Z(G/H_i)$ has subgroup $\overline{H_1}$ of oreder $p$, which is normal in $G/H_i$

Now consider , $f: G \to G/H_i$ be natural group homomorphism, then pullback $f^{-1}(\overline{H_1})$ is normal subgroup of $G$ of order $p^{i+1}$, which contains $H_i=\text{Ker}(f)$. (Correspondence theorem between groups).

Answer 2

You are using induction in the first argument. If $H\neq1$, then $|G/H|<|G|$. Assuming the claim holds for groups of order less than $p^n$, it holds for $G/H$, and so you get the existence of a normal subgroup of order $p$ of the quotient.

Next, if $h\in\gamma^{-1}(\tilde{H})$ and $g\in G$, then $\gamma(ghg^{-1})=\gamma(g)\gamma(h)\gamma(g)^{-1}\in \tilde{H}$, since $\tilde{H}$ is normal. This shows that $ghg^{-1}\in H_{i+1}$. Finally, $|\tilde{H}|=|H_{i+1}/H_i|=|H_{i+1}|/|H_i|$, from which you get $|H_{i+1}|=|\tilde{H}||H_i|=p^{i+1}$.