If the order of a group G is divisible by $p^n$ for some prime p and natural number n, then prove G has a subgroup of order $p^n$.
My try has been by induction on the exponent of divisor $p^n$. I use Cauchy's Theorem for Abelian case which basically is a corollary of Theorem of Finite Abelian Groups. Note that I want to prove this without Sylow's theorems.
$n = 1$ it is clearly true, as Cauchy's Theorem for Abelian Groups says that if order of G is divisible by p, then it contains an element a of order p which clearly generates a cyclic group $<a>$ of order p. Inductive assumption will then of course be that if order of G is divisible by $p^{n-1}$, then G contains a subgroup of this order.
Assume order of G is divisible by $p^n$. Then it certainly also is divisible by $p^{n-1}$, so it contains a nontrivial subgroup of order $p^{n-1}$. Call this subgroup for S. Then, since G is abelian, quotient group $G/S$ certainly exists, and we have relation of orders of groups:
$|G| = |G/S||S|$
Since order of G is divisible by $p^n$, then it follows from the relation that p divides $|G/S|$ and by Cauchy's Theorem, G/S contains an element of order p. Could I somehow 'construct' a group of order $p^n$ by direct product of the S and the group generated by element of order P in G/S?
I'm pretty much stuck. Is there a way to go on, or have I hit a wall?
If somebody knows, could you point me to a place on the internet where this or similar proof has been written out?
Best Answer
You are almost there. So you have a surjective homomorphism of abelian groups $G\longrightarrow G/S$, which simply sends $g$ to the coset $gS$. The group $G/S$ has a subgroup or order $p$. What can you say about the pre-image of this subgroup under the above homomorphism?