i'm studying, and tried the following exercise
Prove that a group of order 28 has a normal subgroup of order 7.
Which i found a solution, but i don't know which theorem or how to deduce something, the proof is
Proof: Note that $o(G)=28=2^2\cdot 7$, therefore there exists $x\in G$ such that $o(x)=7$, by Cauchy theorem, consider $\left<x\right>$ of order $7$, we know that $\left<x\right>\leq G$, therefore we know
$$|G|=28\nmid 24=4!=[G:\left<x\right>]!$$
then $\left<x\right>$ must contain a non-trivial normal subgroup of $G$.
Therefore, $\left<x\right>$ is the normal subgroup of order $7$ of $G$.
What i don't get, is the result in bold, is there a theorem that says, that? or how do you deduce it from there? thanks.
Here is one way to think about it. If $H\subset G$ is a subgroup of $G$, then $G$ acts on the cosets of $H$ in $G$ via the rule for $g\in G$, $aH\in G/H$:
$$g\cdot aH = gaH$$
Now, if the index of $H$ in $G$ is $n$, then there are $n$ cosets of $H$ in $G$ so the group action gives us a group homomorphism
$$\phi:G\longrightarrow S_n$$
where $S_n$ denotes the symmetric group on $n$ elements. The group homomorphism is where the $4!$ comes in in your question. If $|G|$ does not divide $|S_n| = n!$, then $\phi$ cannot be injective (the image of $G$ would give a subgroup of $S_n$ of size $|G|$). Hence the homomorphism must have a nontrivial kernel. This nontrivial kernel is the nontrivial normal subgroup you are looking for.