Order of product of elements of two subgroups

Let $\displaystyle G$ be a finite group. Let $\displaystyle H \leq G$, $\displaystyle K \leq G$, and $\displaystyle HK \leq G$ where $\displaystyle \leq$ denotes a subgroup and $\displaystyle HK := \left\{hk | h \in H, k \in K\right\}$. Let $\displaystyle o(x)$ denote the (group or element) order of $\displaystyle x$.

(a) Show that for every $\displaystyle h \in H, k \in K, o(hk) | o(H)o(K)$, where $\displaystyle o(x)$ is the order of $\displaystyle x$.

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

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.

Re: Order of product of elements of two subgroups

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**thehobbit** Let $\displaystyle G$ be a finite group. Let $\displaystyle H \leq G$, $\displaystyle K \leq G$, and $\displaystyle HK \leq G$ where $\displaystyle \leq$ denotes a subgroup and $\displaystyle HK := \left\{hk | h \in H, k \in K\right\}$. Let $\displaystyle o(x)$ denote the (group or element) order of $\displaystyle x$.

(a) Show that for every $\displaystyle h \in H, k \in K, o(hk) | o(H)o(K)$, where $\displaystyle o(x)$ is the order of $\displaystyle x$.

Here's my hint, all you have to do is show that $\displaystyle (hk)^{|H||K|}=e$, can you do this?

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

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 $\displaystyle h\in H$ then $\displaystyle h\in KN$ and so $\displaystyle h=kn$ for [tex]k\in K[/tx] and $\displaystyle n\in N$ and so $\displaystyle hk^{-1}=n$ and so you know that $\displaystyle o(n)=o(hk^{-1})\mid\cdot$...

Re: Order of product of elements of two subgroups

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

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

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

Re: Order of product of elements of two subgroups

Quote:

Originally Posted by

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

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

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

Right!

Re: Order of product of elements of two subgroups