I am to trying to prove this theorem: A finite $p$-group cannot be simple unless it has order $p$.
I have this:
Let $G = P$ and $|G|=p$; then there exists $N$, a normal subgroup of $G$ by Lagrange's theorem, such that $|N| \mid |G|$ so, $|N|= |e| = 1$ or $|N| = p$
If $|N|=p$ so $N=G$.
But I don't know how I can prove the part in the case of $|N|=|e| = 1$
Is my start correct?