What does it mean for a group $G=PQ$ for normal subgroups $P$ and $Q$? What does it mean for a group $G=PQ$ for normal subgroups $P$ and $Q$? I just can't seem to find a formal definition of this, but it pops up in problems a lot. Does it mean each $g \in G$ is of the form $g = pq$ for $p \in P$ and $q \in Q?$ Does this also mean that $g=qp?$
 A: The notation $G=PQ$ is a compact way of writing that every element $g$ in $G$ can be written as $pq$ for some $p\in P$ and $q\in Q$. In general, if $A$ and $B$ are subsets of a group, $AB$ usually the set of products of elements:
$$AB=\{ab: a\in A,b\in B\}.$$
However, if $g=pq$, then $g$ may not also be equal to $qp$. Not necessarily. If $P=Q=G$ and $G$ is non-abelian, then it's possible to pick $p\in P$ and $q\in Q$ with $pq\neq qp$.
However, if every element of $G$ can be represented uniquely as a product $pq$ (this is equivalent to $P\cap Q$ only consisting of the identity), then your statement is true. To see this, consider $x=q^{-1}pqp^{-1}$. Then
$$x\in q^{-1}(pQp^{-1})=q^{-1}Q=Q,$$
and
$$x\in (q^{-1}Pq)p^{-1}=Pp^{-1}=P,$$
so $x\in P\cap Q$, and thus $x$ is the identity. So, $pq=qp$. This implies that $G$ is the direct product of the groups $P$ and $Q$.

A few interesting notes:

*

*For any subgroups $H$ and $K$ of $G$, $G=HK$ if and only if $G=KH$. This may be shown by looking at inverses.


*Above, we shows that $G=PQ$ and $|P\cap Q|=1$ was enough to show that $g=pq$ implies $g=qp$, if $P$ and $Q$ are both normal in $G$. In general, this isn't enough if even one of $P$ and $Q$ is allowed not to be normal. A classic example is $G=S_3$, where $N$ (normal in $G$) is the subgroup of order $3$ and $H$ (not normal) is any subgroup of order $2$. The general name for such constructions is the semidirect product.
