Changing perspective from external to internal direct product...

Question

When changing perspective from external to internal direct product, what is the operation for the internal direct product? For example, the operations on $S_3 \times \mathbb{Z}_4$ are defined componentwise, yet I cannot think of a natural binary operation between $S_3$ and $\mathbb{Z}_4$. What results from the "multiplication" of a permutation and a number modulo 4? In general? Or is it case by case?

Answer 1

Or is it case by case?

Well, yes, in the sense that the group operation will depend on what group you're working with - even if two groups are isomorphic they and their operations may "look" very different.

As an example, $(\mathbb{R}^{>0},\times)$ (positive reals under multiplication) and $(\mathbb{R},+)$ (all real numbers under addition) are isomorphic, but they look pretty different. The (well, a) isomorphism by the way is given by the exponential function $\exp:(\mathbb{R},+)\to(\mathbb{R}^{>0},\times)$.

I cannot think of a natural binary operation between $S_3$ and $\mathbb{Z}_4$.

If $H$ and $K$ are two groups, then there is a copy of $H$ within the external direct product $H\times K$ given by the subset $H\times\{e_K\}=\{(h,e_K):h\in H\}$. Similarly, $\{e_H\}\times K$ is an isomorphic copy of the group $K$ sitting inside $H\times K$. Explicitly, the isomorphism between $H$ and its copy $H\times\{e_K\}$ is given by $h\mapsto (h,e_K)$, and similarly for $K$.

Since (i) the elements $(h,e_K)$ and $(e_H,k)$ commute for every $h\in H,k\in K$, (ii) their intersection is $\{(e_H,e_K)\}$ which is the trivial subgroup of $H\times K$ and (iii) every $(h,k)\in H\times K$ is uniquely expressible as $(h,e_K)(e_H,k)$, we may conclude that the external direct product $H\times K$ is actually an internal direct product of its subgroups $H\times\{e_K\}$ and $\{e_H\}\times K$.

Let's apply this understanding to $S_3\times\mathbb{Z}_4$. Call the binary operation $\bullet$, so that

$$ (\pi,a)\bullet(\sigma,b)=(\pi\circ\sigma,a+b),$$

where $\pi\circ\sigma$ denotes function composition and $a+b$ denotes addition of integers mod $4$. Then the binary operation between $S_4\times\{0\}$ and $\{\mathrm{id}\}\times\mathbb{Z}_4$ is this "$\bullet$" one. Whether or not it's "natural" I'll leave open for interpretation.

It can look different though. In fact, we can interpret $S_3$ and $\mathbb{Z}_4$ as trivially intersecting subgroups of $S_7$. That is, let $H$ be the set of all permutations of $\{1,\cdots,7\}$ which fix the elements $4,5,6,7$ (i.e. $f(k)=k$ for those values of $k$). Then $H\cong S_3$, so we can think of $H$ as a copy of $S_3$ sitting inside the full $S_7$. And then let $K$ be the subgroup of $S_7$ generated by the cycle $(4567)$ (so if we arrange $4,5,6,7$ in a circle, this permutation rotates them once, and $K$ is comprised of all possible rotations of these numbers along the circle). Since $(4567)$ has order $4$, $K\cong \mathbb{Z}_4$, so that $K$ is a copy of $\mathbb{Z}_4$. Finally, the group $G$ generated by both $H$ and $K$ is an internal direct product of them, and its group operation is just function composition.