Proof of 1st Sylow Theorem by induction on $|G|$. Theorem: Let $G$ be a finite group and $p$ be a prime dividing the order of $G$. Then $G$ contains a Sylow $p$-subgroup.
I'm getting stumped at one part of the proof.

Proof:   Let $G$ be a finite group with $|G| = n$ and let $p$ be a prime dividing $|G|$. To prove this theorem we will induct on the order of $G$, so let $P(n)$ be the statement that $G$, a group of order $|G| = n$, has a Sylow p-subgroup. First note that the base case is $n = p$ since $p$ is assumed to divide $n$, in this case the entire group is a Sylow $p$-subgroup so the base case is satisfied. Now assume that the statement holds for all finite groups with order less than $|G| = n$, where we write $n = p^rm$ with $r \geq 1, p \nmid m$. To finish the induction, we need to consider the centre of $G$. This leads to two cases, first assume that $p \mid |Z(G)|$. By Cauchy's theorem this means that there exists some $g \in Z(G)$ with $|g| = p$. Note that if $r = 1$, then $\langle g \rangle $ is our Sylow $p$-subgroup. If $r >1$, note that by  it follows that $\langle g \rangle \unlhd G$ and so the quotient group is well defined, and in particular
$$
  |G/\langle g \rangle| = \frac{p^rm}{p} = p^{r-1}m.
  $$
But then by our inductive hypothesis $G/\langle g \rangle $ is a finite group of order less than $n$ and so we can find some $H \in G/\langle g \rangle$ such that $H$ is a Sylow p-subgroup; i.e $|H| = p^{r-1}$. Then if you let $K$ be the preimage of $H$ in $G$ it follows by Lagrange's theorem that
$$
|K| = |\langle g\rangle ||H'| = pp^{r-1} = p^r,
$$
and so $K$ is a Sylow $p$-subgroup.

I have no idea why the cardinality of the preimage is what it is. I'm sure it's simple but for the life of me I can't see it and it's driving me crazy. I know that $K \leq G$ but why can we conclude this way? Thanks in advance.
 A: If $G$ is a group and $N$ is a normal subgroup then there is a correspondence between:

*

*Subgroups $K \subseteq G$ such that $N \subseteq K \subseteq G$ and

*Subgroups $H$ in $G/N$.

The correspondence is defined by:

*

*to 2.: Given $K$ define $H = K/N$, which makes sense since $N\subseteq K$.

*to 1.: Given $H$ define $K$ to be its preimage under the natural projection $G \rightarrow G/N$.

Under this correspondence, you see $|H| = [K:N]$, or $|K| = |H|\cdot|N|$.
That is a rather general theorem, which typically appears in texts prior to the Sylow theory. You may find it in your reference. In your case, $N$ is the cyclic group $\langle g \rangle$ generated by $g \in G$.
For completeness, let me just sketch the part of the calcluation you seem to be asking about: if $H$ is a subgroup of $G/N$, then its preimage $K$ has the property that $K/N = H$. (So $|K|=|H||N|$.) For notation, let $\pi: G \rightarrow G/N$ be the projection map $\pi(g) = gN$. So, $K = \{g \in G \mid \pi(g) \in H\} \supseteq N$.

*

*$K/N \subseteq H$. Indeed, if $g \in K$ then $\pi(g) \in H$. Therefore, the coset $gN$ belongs to $H$. So, this shows $K/N \subseteq H$.

*$H \subseteq K/N$. If $gN \in H$ then $g \in K$, by definition of $K$. Therefore, $H \subseteq K/N$.

Ask if there are doubts still.
