# A finite $p$-group has normal subgroup of index $p^2$

Someone suggested me to show by induction something more general is valid, i.e. if $|G|=p^n$ then $\forall\; 0\le i\le n\;\;\exists\; H_i\unlhd G$ s.t. $|H_i|=p^i$.

Lookin' at http://crazyproject.wordpress.com/2010/05/13/a-p-group-contains-subgroups-of-every-order-allowed-by-lagranges-theorem/ we can show that does exists subgroups of order $p^i\;\;\forall 0\le i\le n$, but the normality fails an half! I mean that I can prove the normality only considering subgroups of $G/Z(G)$; when we consider $Z(G)$ we can't apply the same argument since normality is NOT a transitive property! How can I do?

However if this is not true (someone suggested this to me, so I can't be 100% sure this is correct), I'm interested in showing what I wrote in the title: i.e. that a finite $p$-group has normal subgroup of index $p^2$ (and this is true for sure).

Thank you all

• It is not true if $|G|=p$ ;) – Servaes May 14 '14 at 9:16
• Yeah I meant $n\geq2$. Thanks – Joe May 14 '14 at 9:20

it is true.

Let $|G|=p^n$ use induction on $n$ then assume it is true for all $k<n$.

We know that $Z(G)$ is non trivial, so $p$ divides $|Z(G)|$ by Cauchy theorem there exist an element $a$ wit order $p$. Set $H=<a>$ then $H$ is normal in $G$.

Now, $G/H$ is a group of order $p^{n-1}$ by assumption $G/H$ has normal subgroup for every possible order. By correspondence theorem inverse image of normal is normal in $G$,so you have normal groups in $G$, with orders $p^2,p^3,...,p^n$ and we have $H$ so we are done.

Note: I showed that a finite $p$ group has normal subgroup of order $p^i$ for every possible $i$.

• It's almost all clear... but why is $H=\langle a\rangle$ normal in $G$? – Joe May 14 '14 at 9:44
• Notice that any subgroup of $Z(G)$ must be normal. – mesel May 14 '14 at 10:42
• Clear! Thanks! :-) – Joe May 14 '14 at 13:45
• @Joe: you are welcome :) – mesel May 14 '14 at 21:46