Let $H,K \leq G$ s.t $H,K$ two different Sylow $p$-subgroups of $G$. Prove $HK$ isn't a subgroup of $G$.

Question

First of all I wish to explain why I'm opening a new thread:

1. I want to know if my proof is correct.
2. The only proof I could find relies on the assumption $H,K$ are distinct which isn't always the case:

If H,K are two Sylow p subgroups of a group G, then can we say something about the existence of the subgroup HK?

EDIT

I was mistaken and the other post solves this question without any other assumption. I will still leave it for others to learn.

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Question:

Let $H,K \leq G$ s.t $H,K$ two different Sylow $p$-subgroups of $G$. Prove $HK$ isn't a subgroup of $G$.

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So this is my proof:

Assume with contradiction $HK \leq G$. Denote:

$HK \leq G \iff HK=KH \iff (\forall h \in H, \forall k \in K):$

$hk=kh \iff hkh^{-1} = k \iff hkh^{-1} \in K \iff K \lhd H$

We know $|G| = p^rm$ for some $r \in \mathbb{N}$ and $|H|=|K| =p^r$.

This implies for reasons of size and containment that $H=K$ which derives contradiction to the datum.

EDIT

I'm giving this another try:

Since $|H|=|K| = p^r$ and we know $H \neq K$ then $H \cap K \leq HK$ satisfies $|H \cap K| \leq p^{r-1}$.

This implies we have $p$ unique elements in $H \setminus K$ and $p$ unique elements in $K \setminus H$.

Let $k \in K \setminus H$ and $h \in H \setminus K$.

$kh \in KH \iff (kh)^{-1} \in KH \iff h^{-1}k^{-1} \in KH$

$\iff h^{-1} \in K$ and $k^{-1} \in H \iff k \in H$ and $h \in K$ which is a contradiction.

Answer 1

The proof in the linked question does not use the hypothesis $H \cap K = 1$. It just uses the fact that the cardinality of the set $HK$ is $|H||K| / |H\cap K|$. Since $H \cap K$ is a proper subgroup of $K$, $|K|/|H\cap K|$ is an integer divisible by $p$, so $|HK|$ is divisible by a higher power of $p$ than $|G|$, so $|HK|$ does not divide $|G|$. In particular $HK$ cannot be a subgroup.

Proof that $|HK| = |H||K|/|H\cap K|$: Consider the map $f:H\times K \to HK$ defined by $(h,k) \mapsto hk$. It is surjective by definition. Consider a point $h_0k_0 \in HK$. The number of preimages is the number of solutions to $hk = h_0k_0$, or $h_0^{-1} h = k_0 k^{-1}$, and the number of solutions is clearly $|H \cap K|$. Therefore the map $f$ is precisely $|H\cap K|$-to-$1$.

Alternative: Consider the action of $H$ on $\Omega = G/K$ by left-translation. The stabilizer of $K \in G/K$ is $H \cap K$. The orbit of $K$ consists of the cosets $hK$ with $h \in H$, so the orbit has size $|HK|/|K|$. Therefore by the orbit--stabilizer theorem $|H : H \cap K| =|HK|/|K|$.

Answer 2

Let $|G|=p^a n,$ where $p$ does not divide $n$

Then the orders of the Sylow $p$ -subgroups $H,K$ are $p^a.$

Since the intersection $H \cap K$ is a subgroup of $H$ , the order of $H \cap K$ is $p^b$ for some integer $ b \leq a$ by Lagrange’s theorem. As $H$ and $K$ are distinct subgroups, we must have $b<a.$

Then the number of elements of the product $HK$ is $$|HK|=\dfrac{|H||K|}{|H∩K|}=p^{2a−b}.$$

Since $b<a$ , we have $2a−b>a$ .

It follows that the product $HK$ cannot be a subgroup of $G$ since otherwise the order $|HK|=p^{2a−b}$ divides $|G|$ by Lagrange’s theorem but $p^a$ is the highest power of $p$ that divides $|G|.$