If $|G|=n$ and $p^\alpha | n$ then $ \exists H\leq G$ with $|H|=p^\alpha$.
I got stuck in one part of the proof, can anyone clarify?
(induction on $n$) Suppose for inductive hypothesis that ^ holds for groups of order less than $n$
Take an $a \in Z(G)$ with $a^p=e$(one exist by cauchy's theorem),
Then considere $G/ \langle a \rangle$ , which contains
a subgroup of order $p^{\alpha-1}$ since $p^{\alpha-1}| \ \ |G/ \langle a \rangle|$
Where I got stuck:
The inverse image in $G$ of this subgroup has order $p^{\alpha}$ a since each coset of $\langle a \rangle$ contains $p$ elements.
I'm not sure how the author arrived at this. I know this much: Let $\pi$ be the surjective natural projection homomorphism $\pi : G\rightarrow G/\langle a \rangle$. And let $K\leq G/\langle a \rangle$ with $|K|=p^{\alpha-1}$Then $\pi ^{-1}(K)=\{x\in G | \pi(x)\in K\}$ is a subgroup of $G$. I don't see why $|\pi ^{-1}(K)|= p^{\alpha}$
I can tell that $p^{\alpha}=p^{\alpha-1}*p = |K||\langle a \rangle|$ but how ?
Thanks :D
If $\pi:G\to G/N$ is the projection and $H\le G/N$ a subgroup then the preimage $\pi^{-1}(H)=\bigcup H$ is the union of the cosets of $N$ contained in $H$. Thus, $|\pi^{-1}(H)|=|H|\cdot|N|$.
More generally let $\pi:X\to Y$ be any map whatsoever. If $A\subseteq Y$ we have
$$\pi^{-1}(A)=\bigcup_{a\in A}\pi^{-1}(a), \qquad |\pi^{-1}(A)|=\sum_{a\in A}|\pi^{-1}(a)|$$
That is, the preimage of a set is the union of the corresponding fibers. (A fiber is a preimage of a single element.) With groups, the preimage of $hN\in H$ is precisely the coset $hN$ itself, and since the cosets are all the same size ($|hN|=|N|$ for any $h$) we have the explicit calculation
$$|\pi^{-1}(H)|=\sum_{hN\in H}|hN|=|N|\sum_{hN\in H}1=|N|\cdot|H|. $$