1. ## Normal subgroups

Let H and K be distinct subgroups of G. Then the set $HK = \{hk : h \in H, k \in K\}$ is a subset of G by the closure of a group. Prove that HK is a subgroup of G if and only if at least one of H or K is a normal subgroup of G. I'm just kind of stuck with this - I know that a subgroup H is a normal subgroup of G iff $ghg^{-1} \in H$ for all g in G and h in H. I also know that a group H is a subgroup iff $xy^{-1} \in H$ for all x,y in H. But I'm having trouble using the definition of HK to show these properties, both in the forward and backward direction of the "if and only if". Can someone just point me in the right direction?

2. "Then the set $HK = \{hk : h \in H, k \in K\}$ is a subgroup of G by the closure of a group."

What do you mean by this? Isn't this the very thing you are trying to prove?

3. Originally Posted by Halmos Rules
"Then the set $HK = \{hk : h \in H, k \in K\}$ is a subgroup of G by the closure of a group."

What do you mean by this? Isn't this the very thing you are trying to prove?
Oops, I meant HK is a subset of G. I fixed it in the original post.

4. Lets see if this helps..... Suppose that H is normal.
If x,y$\epsilon$HK then x=h$\cdot$k and y=h'$\cdot$k', where h,h'$\epsilon$H and k,k'$\epsilon$K.

x$\cdot$y= (h$\cdot$k)$\cdot$(h'$\cdot$k')= h$\cdot$(k$\cdot$h'$\cdot$$k^{-1}$)(k.k')

What can you say about $k.h'.k^{-1}$ ?

5. Ok, here's my proof for the right-to-left direction of the statement ("if at least one of H or K is a normal subgroup of G, then HK is a subgroup of G"):

Assume H is a normal subgroup of G, and let $h_1,h_2 \in H$ and $k_1,k_2 \in K$, so that $h_1k_1,h_2k_2 \in HK$. Then $(h_1k_1)(h_2k_2) = h_1k_1h_2k_1^{-1}k_1k_2 = h_1(k_1h_2k_1^{-1})k_1k_2$. But since H is a normal subgroup, then $k_1h_2k_1^{-1} \in H$, so $(h_1)(k_1h_2k_1^{-1}) \in H$ by closure of a group, and $k_1k_2 \in K$ by closure of a group, so $(h_1k_1h_2k_1^{-1})(k_1k_2) \in HK$, and HK is closed under the group operation. An analogous argument can be made if K is assumed to be a normal subgroup of G. Now let $h \in H$ and $k \in K$ so that $hk \in HK$. Then $(hk)^{-1} = k^{-1}h^{-1}$, and since K is a subgroup of G, then $k^{-1} \in K$, and since H is a subgroup, $h^{-1} \in H$, so $k^{-1}h^{-1} \in HK$ and HK is closed under inverses. Since HK is closed under the group operation and inverses, HK is a subgroup of G.

I think that's pretty sound. I'm not sure how to go in the other direction though: If HK is a subgroup of G, then at least one of H or K is a normal subgroup of G.

6. Originally Posted by dancavallaro
Now let $h \in H$ and $k \in K$ so that $hk \in HK$. Then $(hk)^{-1} = k^{-1}h^{-1}$, and since K is a subgroup of G, then $k^{-1} \in K$, and since H is a subgroup, $h^{-1} \in H$, so $k^{-1}h^{-1} \in HK$ and HK is closed under inverses.

That isn't correct..... $k^{-1}h^{-1}\epsilon KH$, not HK!

Try using the same trick you used for closure under the group operation....

7. Originally Posted by Halmos Rules
That isn't correct..... $k^{-1}h^{-1}\epsilon KH$, not HK!

Try using the same trick you used for closure under the group operation....
Assume H is normal: $(hk)^{-1} = k^{-1}h^{-1} = k^{-1}h^{-1}kk^{-1} = (k^{-1}h^{-1}k)k^{-1}$, and since H is normal, and $h^{-1} \in H$, then $k^{-1}h^{-1}k \in H$, and $k^{-1} \in K$, so $(k^{-1}h^{-1}k)k^{-1} \in HK$. Likewise if K is assumed normal. So HK is closed under inverses.