I'm trying to solve this group theory problem from my abstract algebra course. It goes like this:
Prove that any group of order 945 has at least one subgroup of order 9.
First, I noticed that $945=3^3\cdot5\cdot7$, and obviously $9=3^2$. I guess the prove will use Sylow theory, finding Sylow $3,5,7$-subgroups, but I'm not sure where to start. If i'm not wrong, there are either $1$ or $7$ Sylow $3$-subgroups in $G$, but I don't know if this is even useful in this problem. Any help will be appreciated, thanks in advance.
It is sufficient to prove that if $G$ is a group of order $p^{n}$, $p$ a prime and $n\geqslant0$, then $G$ has a subgroup of order $p^{m}$ for all $0\leqslant m\leqslant n$. In fact, we prove by induction in $|G|$ a stronger version of this: If $G$ is a $p$--group of order $p^{n}$, then $G$ has a chain of normal subgroups $1=P_{0}\unlhd P_{1}\unlhd P_{2}\unlhd\cdots\unlhd P_{n}=G$ such that $|P_{i}|=p^{i}$ for all $i=0,\ldots,n$. If $|G|\leqslant p$ we have nothing to prove. If $|G|=p^{n}\geqslant p^{2}$, we assume that the result holds for $p$--groups whose order are less than $|G|=p^{n}$. Since all non-trivial finite $p$--groups have non trivial center, we consider $g\in Z(G)$ whose order is equal $p$. Then $\langle g\rangle\unlhd G$ and we can consider the group $G/\langle g\rangle$, whose order is $p^{n-1}$. By induction we can found a normal chain $$1=\langle g\rangle/\langle g\rangle=P_0/\langle g\rangle\unlhd P_{1}/\langle g\rangle \unlhd\cdots \unlhd P_{n-1}/\langle g\rangle=G/\langle g\rangle$$ such that $|P_{i}/\langle g\rangle|=p^{i}$ for all $i=0,\ldots,n-1$. It follows that for all $i=0,\ldots,n-1$ we have $|P_i|=p^{i+1}$. Writing $H_i=P_{i-1}$ for all $i=1,\ldots, n$ and putting $H_0=1$, we get the chain $1=H_0\unlhd H_{1}=P_{0}=\langle g\rangle\unlhd H_{2}=P_{1}\unlhd\cdots\unlhd H_{n}=P_{n-1}=G$, such that $|H_{0}|=|\{1\}|=1$ and for all $i=1,\ldots,n$ we have $|H_{i}|=|P_{i-1}|=p^{(i-1)+1}=p^{i}$. The claim is now proved.
Finally, since $27=3^{3}$ divides $945$, a group of order 945 has at least one subgroup of order 27 (Sylow theorem), say $H$. By the previous paragraph, $H$ has a subgroup of order 9, say $K$. Then, $K$ is a subgroup of the whole group of order 9.