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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?

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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.

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  • $\begingroup$ Thank you, the final paragraph gives great perspective! Sometimes, we may be fortunate enough to identify the direct factors as living in some larger group. Tough question: is there an example where we can ensure the direct factors do not live in some larger group and the operation is necessarily "unnatural"? $\endgroup$ – mathemusician Sep 17 '16 at 2:40
  • $\begingroup$ @mathemusician That question seems too subjective. $\endgroup$ – arctic tern Sep 17 '16 at 2:43
  • $\begingroup$ Fair enough - thanks! $\endgroup$ – mathemusician Sep 17 '16 at 2:48

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