Multiplication of subgroups

Question

Let $H$ and $K$ be subgroups of a finite group $G$. Define $HK = \{hk\mid h \in H, k \in K\}$ and $KH = \{kh\mid k \in K, h \in H\}$.

a) Show that in general $HK \ne KH$. (For example, consider $G = S_4$ and let $H = (123)$ and $K = (14)$).

b) Prove that if $HK = KH$ then $HK$ is a subgroup of $G$. (Note: to say that $HK = KH$ does not mean that individual $h$ and $k$ commute. Rather it means that $hk = kh$ for some pair $k$, $h$.)

a) I know I have to show that $HK<KH$ and vice versa. But we dont know if $G$ is abelian, which would make this easy. Any ideas on how to go forward?

b) Proof: Assume $HK=KH$. So we have to show $h_1k_1h_2k_2\in HK$. So since we assumed that $HK=KH$, $k_1h_2=h_2k_1$ so $h_1k_1h_2k_2=h_1h_2k_1k_2$. Since $h_1h_2 \in H$ and $k_1k_2 \in K$, thus $h_1h_2k_1k_2 \in HK$. How do I show inverses for this one?

Answer 1

It is impossible to do a) as it is not true in general that $HK=KH$. Indeed, check it for the example you gave and you will see this; by check I mean compute all the products; presumably $H$ and $K$ are the groups generated by those elements it cannot be the elements themselves. Either you transcribed that wrong or there is a typo in the source (or some hypothesis is missing, but that is rather unlikely).

For b) Your argument is somewhat false. Read the remark again. You do not know $k_1h_2=h_2k_1$. But you know $k_1h_2= h' k'$ for some $h'$ in $H$ and $k'$ in $K$ and this is enough. For the inverse, since the groups is finite, once you know it is closed under the operation you already know it is a subgroup.

Answer 2

Here's how the problem is almost surely given:

Let $H$ and $K$ be subgroups of a finite group $G$. Define $HK = \{hk\mid h \in H, k \in K\}$ and $KH = \{kh\mid k \in K, h \in H\}$.

a) Show that in general $HK \ne KH$. (For example, consider $G = S_4$ and let $H = \langle(123)\rangle$ and $K = \langle(14)\rangle$).

b) Prove that if $HK = KH$ then $HK$ is a subgroup of $G$. (Note: to say that $HK = KH$ does not mean that individual $h$ and $k$ commute. Rather it means that $hk = k_1h_1$ for some pair $k_1$, $h_1$.)

The suggestion for (a) is already written out: consider $$ H=\{\mathit{id},(123),(132)\},\qquad K=\{\mathit{id},(14)\} $$ Then $(123)(14)=(1423)\in HK$, but $$ (14)(123)=(1234),\quad (14)(132)=(1324) $$ and so $(123)(14)\notin KH$ (complete the checks, but those actually suffice).

For (b), saying that $HK=KH$ means that

for any $h\in H$ and $k\in K$ there exist $h'\in H$ and $k'\in K$ such that $$kh=h'k'$$ and conversely exchanging the roles of $H$ and $K$.

Now $1\in HK$, obviously. So you just need to show that, for $h_1,h_2\in H$ and $k_1,k_2\in K$, $$ (h_1k_1)(h_2k_2)\in HK $$ (because $G$ is finite); since there are $h'\in H$ and $k'\in K$ such that $k_1h_2=h'k'$, you indeed have $$ (h_1k_1)(h_2k_2)=h_1(k_1h_2)k_2=h_1(h'k')k_2=(h_1h')(k'k_2)\in HK $$

The finiteness assumption is actually unnecessary, because $$ (hk)^{-1}=k^{-1}h^{-1}=h'k'\in HK $$ for suitable $h'\in H$ and $k'\in K$, from $KH=HK$.