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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 $\left$ of order $7$, we know that $\left\leq G$, therefore we know

$$|G|=28\nmid 24=4!=[G:\left]!$$

then $\left$ must contain a non-trivial normal subgroup of $G$.

Therefore, $\left$ 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.

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    How does $4!$ come in? – 2017-02-04
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    well, that's some of the things i don't know, that's why i think, he's using a theorem or something, that says something about the factorial, i just know that the order of $[G:\left]$ is $4$. – 2017-02-04
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    The usual way to prove $\langle x\rangle$ is via Sylow theorems. – 2017-02-04
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    https://groupprops.subwiki.org/wiki/Order_of_simple_non-abelian_group_divides_factorial_of_index_of_proper_subgroup – 2017-02-04
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    this problem comes before the Sylow Theorems, in Topics in Algebra, so, i think it would be nice to have an alternative proof, without the Sylow Thoerems. – 2017-02-04
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    @GastónBurrull Thanks! that's exactly what i was looking for. – 2017-02-04

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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.

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    +1, and a comment: I could remember the result that "if $p$ is the smallest prime dividing $|G|$ and $G$ has a subgroup of index $p$, then $G$ can't be simple" but I could never remember its proof. By now, I can remember at least that it uses basically these ingredients :) – 2017-02-05
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By Cauchy's theorem there is a subgroup $H$ of order $7$.

If $K$ is another subgroup of order $7$, then $H\cap K=1$ (why?).

But then $|HK|=|H|.|K|/|H\cap K|=7.7$ contradiction.

So subgroup of order $7$ is unique hence normal.