Group of order 90 contains a subgroup of order 10 I’m trying to prove that if $G$ is a group of order $90$, then it contains a subgroup of order $10$. This is what I have so far:
“The prime factorization of $90$ is  $3^2 \cdot 2 \cdot 5$. By the First Sylow Theorem, Sylow $p$-subgroups exist, meaning that there is at least one Sylow $2$-subgroup, say $P$ and there is at least one Sylow $5$-subgroup, say $Q$. Additionally, Sylow $p$-subgroups intersect trivially, meaning that $|P \cap Q|=1$.
Therefore, we have $|PQ|=\frac{|P||Q|}{|P \cap Q|}=\frac{2 \cdot 5}{1}=10$.
And so, $G$ contains a subgroup $PQ$ of order $10$, as was to be shown.”
Is this correct?
 A: While I admire the other poster's persistence, here is a faster way.
Suppose that $n_5=6$. Then $G$ acts on the six Sylow $5$-subgroups. As $|N_G(P)|=15$, and this is the important bit, no element of order $2$ can normalize any Sylow $5$-subgroup. Thus the permutation action of an element of order $2$ must be (up to labelling) $(1,2)(3,4)(5,6)$, an odd permutation. Let $H$ be the set of elements that induce an even permutation on the Sylow $5$-subgroups. Then $|G:H|=2$, and thus all elements of order $5$ lie in $H$. By Sylow's theorem, $n_5=1$ for $H$ (no other option), and therefore $n_5=1$ for $G$, a contradiction.
A: Following David A. Craven's remark on my more arduous proof:
Here you can find a proof that a group of order 4n+2 has a (normal) subgroup of index 2, using little more than Cayley's theorem to imbed into a symmetric group of degree 4n+2.  Now $90=4\cdot 22+2$ is of this form, so a group $G$ of order 90 always has a subgroup of index 2.  So we have a normal subgroup $H$ of order 45, and every Sylow 5-subgroup of $G$ is a Sylow 5-subgroup of $H$ and vice versa.  But an easy application of the Sylow Theorems shows that a group of order 45 always has a normal Sylow 5-subgroup, so $G$ must have a unique and therefore normal Sylow 5-subgroup.
Your argument then carries through, as your product does in fact define a subgroup.
