If $p> \sqrt{|G|}$ divides $|G|$, then $G$ contains a normal subgroup of order $p$

Question

Let $G$ be a finite group and let $p$ be a prime dividing the size $|G|$ of $G$ such that $p > \sqrt{|G|}$. Prove that $G$ contains a normal subgroup of order $p$.

I tried generalizing the proof of Cauchy's theorem (that there is a subgroup of order $p$, not necessarily normal), but didn't seem to work. Any help appreciated!

(Feel free to use group actions, but please avoid e.g. outsorcing to Sylow theorems, if possible.)

Answer 1

Let $H$ be a subgroup of order $p$ (Cauchy guarantees its existence , $p | |G||$) and assume $H$ is not normal. Then there exists a $g \in G$ with $H^g:=g^{-1}Hg \neq H$, whence (using Lagrange and $p$ being prime) $H \cap H^g=1$. Observe that $g \notin H$ and this implies that the set $HH^g$ is a proper subset of $G$ (if $G=HH^g$, then $g=hg^{-1}kg$ $(h,k \in H)$, from which $g=kh \in H$ follows). But then $|G| \gt |HH^g|= \frac{|H||H^g|}{|H \cap H^g|}= p^2$. This yields $p \lt \sqrt{|G|}$, a contradiction. So $H$ must be normal.

Answer 2

Promoting the comments to an answer as the OP worked it out to some extent. See Nicky Hekster's answer for an even lower tech argument.

---

Let $|G|=mp$, so $m<p$. Let $H$ be a subgroup of order $p$ (Cauchy's theorem promises the existence). Let $X=G/H$ be the set of left cosets of $H$. Then

• $G$ acts (transitively) on the set $X$ by left multiplication, this gives a group homomorphism $f:G\to \mathrm{Sym}(X)\simeq S_m$.
• The stabilizer of the coset $1H=H\in X$ is obviously $H$, so we can conclude that $\mathrm{Ker}(f)\le H$.
• On the other hand, because $m<p$, the group $\mathrm{Sym}(X)$ has no elements of order $p$. Therefore elements of order $p$ must be mapped to the identity permutation (here it is crucial that $p$ is a prime), and we can deduce that $H\le \mathrm{Ker}(f)$.
• Therefore $\mathrm{Ker}(f)=H$. Consequently $H\unlhd G$ as the kernel of a homomorphism.