# Thread: Order of product of elements of two subgroups

1. ## Order of product of elements of two subgroups

Let $G$ be a finite group. Let $H \leq G$, $K \leq G$, and $HK \leq G$ where $\leq$ denotes a subgroup and $HK := \left\{hk | h \in H, k \in K\right\}$. Let $o(x)$ denote the (group or element) order of $x$.
(a) Show that for every $h \in H, k \in K, o(hk) | o(H)o(K)$, where $o(x)$ is the order of $x$.
(b) Let $N \leq G$ such that $o(N)$ is coprime to $o(H)o(K)$. Prove that if $HN = KN$, then $H = K$.

I'm not quite sure how to start on this problem. I don't want a full solution, just some help in figuring it out.

2. ## Re: Order of product of elements of two subgroups

Originally Posted by thehobbit
Let $G$ be a finite group. Let $H \leq G$, $K \leq G$, and $HK \leq G$ where $\leq$ denotes a subgroup and $HK := \left\{hk | h \in H, k \in K\right\}$. Let $o(x)$ denote the (group or element) order of $x$.
(a) Show that for every $h \in H, k \in K, o(hk) | o(H)o(K)$, where $o(x)$ is the order of $x$.
Here's my hint, all you have to do is show that $(hk)^{|H||K|}=e$, can you do this?

(b) Let $N \leq G$ such that $o(N)$ is coprime to $o(H)o(K)$. Prove that if $HN = KN$, then $H = K$.

I'm not quite sure how to start on this problem. I don't want a full solution, just some help in figuring it out.
Let $h\in H$ then $h\in KN$ and so $h=kn$ for [tex]k\in K[/tx] and $n\in N$ and so $hk^{-1}=n$ and so you know that $o(n)=o(hk^{-1})\mid\cdot$...

3. ## Re: Order of product of elements of two subgroups

For (a), since $HK \leq G$ and $hk \in HK$, we have $|HK| = \frac{|H| \cdot |K|}{|H \cap K|}$, thus $(hk)^{|H| \cdot |K|} = (hk)^{|HK| \cdot |H \cap K|} = e^{|H \cap K|} = e$

For (b), picking up where you left off, $o(n) = o(k^{-1}h)\:\big|\:|H| \cdot |K|$. Since $o(n)$ is coprime to $|H| \cdot |K|$, we must have $n=1$, hence $N$ is the trivial subgroup, yielding the desired result.

4. ## Re: Order of product of elements of two subgroups

Originally Posted by thehobbit
For (a), since $HK \leq G$ and $hk \in HK$, we have $|HK| = \frac{|H| \cdot |K|}{|H \cap K|}$, thus $(hk)^{|H| \cdot |K|} = (hk)^{|HK| \cdot |H \cap K|} = e^{|H \cap K|} = e$
For (b), picking up where you left off, $o(n) = o(k^{-1}h)\:\big|\:|H| \cdot |K|$. Since $o(n)$ is coprime to $|H| \cdot |K|$, we must have $n=1$, hence $N$ is the trivial subgroup, yielding the desired result.