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> [!definition] Definition. ([[Frobenius product of subgroups]])
> For $H,K$ are [[subgroup]]s of a [[group]] $G$, define their **Frobenius product** $HK$ to be $HK=\{ hk:h \in H, k \in K \}.$
> **Remark.** $HK \leq G$ **if and only if** $HK=KH$.
> [!basicexample] Example. ([[Symmetric group]] on $3$ letters)
> Take $H:=\{ e,\tau,\tau^{2} \} \leq S_{3}$ and $K:=\{ e,\sigma \}\leq S_{3}$. $HK$ looks like $\begin{array}{c|cc}
HK & e & \sigma \\
\hline
e & e & \sigma \\
\tau & \tau & \tau \sigma \\
\tau^2 & \tau^2 & \tau^2 \sigma \\
\end{array} = \begin{array}{c|cc}
HK & e & \sigma \\
\hline
e & e & \sigma \\
\tau & \tau & \sigma \tau^{2} \\
\tau^2 & \tau^2 & \sigma \tau \\
\end{array}.$
$KH$ looks like $\begin{array}{c|cc}
KH & e & \tau & \tau^2 \\
\hline
e & e & \tau & \tau^2 \\
\sigma & \sigma & \sigma \tau & \sigma \tau^2 \\
\end{array}.$
observe that $HK=KH=G$. This isn't surprising because we know that $\tau$ and $\sigma$ [[generating set of a group|generate]] $G$, and since $H$ contains all powers of $\tau$ and $K$ contains all powers of $\sigma$ their product will contain all combinations of powers of $\tau$ and $\sigma$ (which is the def. of [[group]] generated by $\tau$ and $\sigma$).
\
In contrast, if we let $L=\{ e, \sigma \tau \}$, then $KL$ is $\begin{array}{c|cc}
KL & e & \sigma\tau \\
\hline
e & e & \sigma\tau \\
\sigma & \sigma & \sigma^2\tau \\
\end{array}=\begin{array}{c|cc}
KL & e & \sigma\tau \\
\hline
e & e & \sigma\tau \\
\sigma & \sigma & \tau \\
\end{array}$
while $LK$ is $\begin{array}{c|cc}
LK & e & \sigma \\
\hline
e & e & \sigma \\
\sigma\tau & \sigma\tau & \sigma\tau\sigma \\
\end{array}=\begin{array}{c|cc}
LK & e & \sigma \\
\hline
e & e & \sigma \\
\sigma\tau & \sigma\tau & \tau^{2} \\
\end{array}.$
We see that $KL \neq LK$, and neither of them are [[subgroup]]s.
> [!justification]
> We must demonstrate the **remark**ed equivalence.
> \
> $\to.$ Assume $HK=KH$. We want to show $HK \leq G$. Since $e \in H$ and $e \in K$, $e=e^{2} \in HK$. Next let $h_{1}k_{1}, h_{2}k_{2} \in HK$. Since $HK=KH$, we have that $h_{1}k_{1}=k_{3}h_{3}$ for some $k_{3}h_{3} \in KH$. We have $\begin{align}
(h_{1}k_{1})(h_{2}k_{2})= & k_{3}\overbrace{h_{3}h_{2}}^{\in H}k_{2} \\
= & \overbrace{k_{3}h^{*}}^{\in KH}k_{2} \text{ for some } h^{*} \in H \\
= & (k^{*}h^{*})k_{2} \text{ for some } k^{*} \in K \\
= & (hk')k_{2} \text{ for some } h \in H, k' \in K \text{ (since } HK=KH \text{)} \\
= & h(k'k_{2}) \text{ for some } h \in H, k' \in K \text{(associativity)} \\
= & hk^{} \text{ for some } h \in H, k^{} \in K \text{($K$ closed under $\cdot$)}
\end{align},$
hence $HK$ is closed under $G