Sylow's theorem proof Sylow's theorems state that the $p$-Sylow subgroups exist for a group G of order $p^km$, where $p$ is prime and does not divide $m$. My question is how to prove that there is at least 1 subgroup of order $p^n$ exists for every non-negative integer $n \le k$.
 A: Actually, there exists a normal subgroup of order $p^n$ for any group of order $p^k$, where $n\leq k$. We know that the center of a nontrivial $p$-group is nontrivial, then you can use induction to prove the fact.
A: The answer is by induction. The statement for $n=1$ is given by Cauchy's theorem, so let us assume we proved this assumption for $n -1 < k$ and prove it holds for $n$. By induction there exists a subgroup $H$ of order $p^{n-1}$. The group $H$ acts  of the set of right cosets $Q = G/H$ by: $\forall h\in H : (Hh')h = Hh'h$. The size $q$ of $Q$ is $q = \frac{p^km}{p^{n-1}}=p^{k-n+1}m$, a multiple of $p$. The sizes of the orbits are (by the orbit stabilizer theorem) divisors of $|H|$, so multiples of $p$, but at least one of them has size $1$ (the orbit of $H$). This implies that all other orbits can't have sizes that are a multiple of $p$ otherwise $q$ can't impossibly be a multiple of $p$ unless $k-n+1=0$ but we assumed that $n -1 < k$. Let $x \in G$ be an element whose orbit has size $s$ (which is a power of $p$) that is not a multiple of $p$ then $s$ must be one. Such an element satisfies $Hxh \in H$ or equivalently $x \in \cal{N}(H)$, the normalizer $N$ of $H$ in $G$, from which we conclude that the number of orbits of size $1$ is a multiple of $p$, if $q$ has to be a multiple of $p$. It is not hard to show that the number of orbits of size one is exactly $|N/H|$. Since this number is a multiple of $p$ by Cauchy there is a subgroup $H'$ of $N$ of order $p|H|$ which answers the question.
EXAMPLE:
$G = S_6$, the symmetric group. Then $G$  has order $p^km$ with $p=2, k=4$ and $m = 45$. 
Step 1: Let the induction start with the group $H = <(1,2)>$  of order $2$. There are $720/2 = 360$ right cosets of $H$. The orbits of $Q$ splits into $24$ of size $1$ and $168$ of size $2$. The normalizer $N$ of $H$ in $G$  is the group $< (1,2), (5,6), (4,6,5), (3,6)(4,5), (3,5)(4,6) >$ of order $48$. As expected there are $|N/H| = 24$ orbits of size $1$. One can chose $H' = <(1,2), (3,6)(4,5)>$ as the next step in the induction.
Step 2: Let $H =<(1,2), (3,6)(4,5)>$ of order $4$. There are $720/4 = 180$ right cosets of $H$. The orbits of $Q$ splits into $4$ of order $1$, $8$ of order $2$ and $40$ of order $4$. The normalizer $N$ of $H$ in $G$  is the group $< (3,4,6,5), (3,6), (4,5), (1,2) >)$ of order 16. As expected there are $|N/H| = 4$ orbits of size $1$. One can chose $H' = < (1,2), (3,6)(4,5), (3,4,6,5) >$ as the next step in the induction.
Step 3: Let $H = < (1,2), (3,6)(4,5), (3,4,6,5) >$ of order $8$. There are $720/8 = 90$ right cosets of $H$. The orbits of $Q$ splits into $2$ of size $1$, $2$ of size $4$ and $10$ of size $8$. The normalizer $N$ of $H$ in $G$  is the group $< (4,5), (3,4,6,5), (3,6)(4,5), (1,2) >$ of order $16$. As expected there are $|N/H| = 2$ orbits of size $1$. One can chose $H' = < (1,2), (3,6)(4,5), (3,4,6,5), (4,5) >$ as the next step in the induction.
Step 4: Let $H' = < (1,2), (3,6)(4,5), (3,4,6,5), (4,5) >$ of order $16$. There are $720/16 = 45$ right cosets of $H$. The orbits of $Q$ splits into $1$ of size $1$, $2$ of size $2$  , $2$ of size $4$,  $2$ of size $8$ and $1$ of size $16$. But now we have the remarkable fact that $q = 45$ is not a multiple of $2$ anymore so the restriction on the number of obits of size $1$ being a multiple of $2$ is not valid anymore so the induction stops here. 
