[Math] Without using Sylow: Group of order 28 has a normal subgroup of order 7

Question

Prove that a group of order 28 has a normal subgroup of order 7.

How can I prove this without using Sylow's theorem?

I know by Cauchy’s theorem, there exists an $x\in G$ with order 7, now I just need to prove it has a normal subgroup.

Answer 1

Another different proof: since as you said you know that $G$ has an element of order $7$, by Cauchy, you also know that there's at least one subgroup of order $7$, let's call it $H$.

Suppose that $K \leq G$ is another subgroup of order $7$, then we can consider the subset $HK$ that has order $|HK|=|H||K|/|H \cap K|$.

If $H$ and $K$ were distinct then $H \cap K$ should be a proper subgroup of both of them, but since they have order the prime $7$ this is possible iff $H \cap K=(\text{id})$ and so $|HK| =7\cdot 7 = 49$ which is clearly bigger then $28$.

We arrived to an absurdity, so we have to conclude that $H$ is the only subgroup of order $7$ and so it's characteristic, hence normal.

Edit (more details): let's consider a generic automorphism $\varphi \colon G \to G$, then by properties of homomorphisms $\varphi(H)$ is a subgroup of $G$ and since $\varphi$ is bijective $\varphi(H)$ should have order $7$.

Because as we have proved $H$ is the only subgroup of order $7$ it follows that $\varphi(H)=H$: so $H$ is fixed by all the automorphisms, i.e. is characteristic.

From this follows normality since a subgroup is normal iff is fixed by all inner automorphisms, i.e. is fixed by all the automorphisms of the form $$x \mapsto gxg^{-1}$$ for some $g \in G$.

Since $H$ is fixed by every automorphism it's fixed in particular by the inner automorphism and so it's normal.