Order of product of elements of two subgroups

**thehobbit** · September 11th 2011, 04:11 PM

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.

**Drexel28** · September 11th 2011, 06:11 PM

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 $o(n)=o(hk^{-1})\mid\cdot$ ...

**thehobbit** · September 13th 2011, 03:50 PM

I thought about this one while shopping and now the answer seems quite simple.

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.

**Drexel28** · September 13th 2011, 03:56 PM

Originally Posted by thehobbit

I thought about this one while shopping and now the answer seems quite simple.

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.

Right!

**thehobbit** · September 13th 2011, 04:31 PM

Thank you for your help!

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