HK < G implies HK = KH

• Apr 4th 2011, 05:22 PM
topsquark
HK < G implies HK = KH
Quote:

Let H, K be subgroups of a group G. Then HK is a subgroup of G iff HK = KH.
I can show that if H, K < G and HK = KH then HK < G. That's easy. I am having trouble with the other direction: If H, K < G and HK < G then HK = KH.

Here's what I have. (The language may not be quite as precise as it needs to be.)

We know that $\displaystyle hk \in HK$, where h is in H and k is in K, by definition. Since HK is a group we also know that $\displaystyle (hk)^{-1} \in HK$. But $\displaystyle (hk)^{-1} = k^{-1}h^{-1} \in KH$. Since h and k are arbitrary in H and K respectively we can state that HK = KH.

My difficulty with this line of reasoning is that I suspect myself of assuming that KH is also a subgroup of G. Does my argument have any flaws? Thanks.

-Dan
• Apr 4th 2011, 05:59 PM
Drexel28
Quote:

Originally Posted by topsquark
I can show that if H, K < G and HK = KH then HK < G. That's easy. I am having trouble with the other direction: If H, K < G and HK < G then HK = KH.

Here's what I have. (The language may not be quite as precise as it needs to be.)

We know that $\displaystyle hk \in HK$, where h is in H and k is in K, by definition. Since HK is a group we also know that $\displaystyle (hk)^{-1} \in HK$. But $\displaystyle (hk)^{-1} = k^{-1}h^{-1} \in KH$. Since h and k are arbitrary in H and K respectively we can state that HK = KH.

My difficulty with this line of reasoning is that I suspect myself of assuming that KH is also a subgroup of G. Does my argument have any flaws? Thanks.

-Dan

Right, $\displaystyle HK=\left(HK\right)^{-1}=K^{-1}H^{-1}=KH$.
• Apr 4th 2011, 07:19 PM
topsquark
Quote:

Originally Posted by Drexel28
Right, $\displaystyle HK=\left(HK\right)^{-1}=K^{-1}H^{-1}=KH$.

W o w. (Surprised)

Ummm...To the best of my knowledge I've never come across that one before. Under what conditions can we treat H and K that way? I'm presuming because H and K are groups, but do they have to be subgroups of the same group G or anything?

-Dan

After thinking about it for a couple of minutes it looks to me like you are simply using a short-hand way of what I did in my post. Is this correct? ie. $\displaystyle \{ hk|h \in H \text{ and } k \in K \}$ is HK and $\displaystyle \{ (hk)^{-1}|h \in H \text{ and } k \in K \}$ is $\displaystyle (HK)^{-1}$.
• Apr 4th 2011, 08:49 PM
Drexel28
Quote:

Originally Posted by topsquark
After thinking about it for a couple of minutes it looks to me like you are simply using a short-hand way of what I did in my post. Is this correct? ie. $\displaystyle \{ hk|h \in H \text{ and } k \in K \}$ is HK and $\displaystyle \{ (hk)^{-1}|h \in H \text{ and } k \in K \}$ is $\displaystyle (HK)^{-1}$.

Yes, that is correct.
• Apr 5th 2011, 12:38 AM
Deveno
more naively, we want to show HK = KH iff HK is a group. well if HK = KH, then (hk)(h'k') = h(kh)k' = h(h"k")k' = (hh")(k"k'), so we have closure.

and (hk)^-1 = k^-1h^-1, which is in KH = HK, so we have inverses.

it gets a bit more involved the other way: if HK is a group, then (hk)(hk) = h'k', so kh = (h^-1h')(k'k^-1), which is in HK, so KH is contained in HK.

but since every hk in HK has an inverse in HK, say h'k', then hk = (h'k')^-1 = k'^-1h'^-1, which is in KH, so HK = KH.

the argument HK = (HK)^-1 = K^-1H^-1 = KH is very succint, but relies on knowing a priori that the statement (HK)^-1 = K^-1H^-1 is indeed an equality of sets (in other words, it follows from the fact that every (hk)^-1 is of the form k^-1h^-1 and vice versa, not from "group multiplication on subgroups". there are many situations in groups where one can treat subgroups as if they were elements, but one must be careful about this sort of thing. for example, one cannot conclude from HK = KH that HK is an abelian group, although one can conclude that HK is always a group if the parent group G is abelian).