I'm attempting Problem 60 on this problem set.
- Let $G$ be a finite group with a normal subgroup $N\cong S_3$. Show that there is a subgroup $H$ of $G$ such that $G = N \times H$.
I'm guessing the author meant to say "$G\cong N\times H$" instead of what he wrote? What do you think?
Edit: I just saw he repeats this notation in other exercises, like
- A group $N$ is said to be complete if the center of $N$ is trivial and every automorphism of $N$ is inner. Show that if $G$ is a group, $N\lhd G$, and $N$ is complete, then $G = N \times C_G(N)$.
So maybe it's a lazy notation for $\cong$? Or does it mean something else?
Best Answer
As others have said in the comments, if $N$ and $H$ are two subgroups of a group $G$, then to say $G=N\times H$ means, by definition, that $N$ and $H$ are both normal, $N\cap H=\{1\}$, and $NH=G$.
The distinction between such an internal direct product and just an abstract isomorphism can be important. Consider the following example. Let $G=\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, and let $H=\langle (2,0)\rangle$. This is a subgroup of order $2$, and it is true that $G$ is abstractly isomorphic to a direct product of a group of order $2$ and some other group. But there is no $N\leq G$ such that $G=N\times H$. We say that $H$ has no complement in $G$, which is a property that one is often interested in.