Direct Product and Semi-direct Product between $S_{3}$ and $Z_{2}$, and Related Problem I am trying to figure out the following question:
0) Is the direct product $S_{3}\times Z_{2}$ isomorphic or non-isomorphic to the semi-direct product $S_{3}\rtimes_{\phi}Z_{2}$ where
$\phi:Z_{2}\rightarrow \operatorname{Aut}(S_{3}),0\mapsto Id,1\mapsto conjugation ~by~(12)$.
The only technique I can think of, beside using brutal force to write down multiplication table, is to calculate the center of these two groups:
First, it is obvious that the center of the direct product $S_{3}\times Z_{2}$ is ${(1)}\times Z_{2}$ which is of cardinality $2$.
Second, consider the center $Z$ of the semi-direct product $S_{3}\rtimes_{\phi}Z_{2}$ where $\phi$ is given as above.
The first claim I made is that: if $(a,0)\in Z$, then $a=(1)$. This can be proved by starting of assuming $(a,0)(b,0)=(ab,0)=(ba,0)=(b,0)(a,0)$ for all $b\in S_{3}$. Since the center of $S_{3}$ is trivial, we conclude that $a=(1)$.
The second claim I made is that: if $(a,1)\in Z$, then $a=(12)$. This can be proved by considering the equation $(a,1)(b,0)=(a(12)b(12),1)=(b,0)(a,1)=(b(12)a(12),1)$. Set $b=(1)$ we have $a=(12)a(12)$. It follows that $a=(1)$ or $(12)$. But if $a=(1)$, setting $b=(23)$ yields $(12)(23)(12)=(23)$ which is a contradiction. We conclude $a=(12)$ by showing that $((12),1)$ does commute with other elements in $S_{3}\rtimes_{\phi} Z_{2}$.
Now the centers of both group $S_{3}\times Z_{2}$ and $S_{3}\rtimes_{\phi}Z_{2}$ have same cardinality. I cannot decide whether they are isomorphic or not.
Any hints, ideas, thoughts are welcome.
Based on this question, I am also wondering:
1) In general how can see a semi-direct product is isomorphic to a direct product? The only result I am quite familiar with is the Proposition 11 of Section 5.5 in Dummit & Foote which says, for a semi-direct product, the following holds:
$$Id:H\rtimes_{\phi}K\rightarrow H\times K~is~isomorphism\Leftrightarrow \phi:K\rightarrow \operatorname{Aut}(H)~is~trival\Leftrightarrow K\unlhd H\rtimes_{\phi}K$$
2) Do we have any results related the question:
$$(H_{1}\rtimes_{\phi_{1}}H_{2})\rtimes_{\phi_{2}}H_{3}\cong^{?}H_{1}\rtimes_{\phi_{3}}(H_{2}\rtimes_{\phi_{4}}H_{3}).$$

$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$Let us consider the second group $H = S_{3}\rtimes_{\phi}Z_{2}$, and let us fix some notation: 


*

*$S_{3} = \Span{a, b}$, where $a = (123)$ and $b = (12)$, and 

*let $g$, an element of order $2$, be a generator of the $Z_{2}$. 


By construction you have in $H$ the following conjugacy relations
$$\tag{rels}
a^{g} = a^{b} = a^{-1}, \qquad b^{g} = b.
$$
Now consider $x = g b \notin S_{3}$. Since by (rels) $g b = b g$, we have $x^{2} = (g b)^{2} = g^{2} b^{2} = e$, the identity of the group. Also
$$
a^{x} = a^{g b} = (a^{g})^{b} = (a^{-1})^{b} = (a^{-1})^{-1} = a,
\qquad
b^{x} = b^{g b} = b,
$$
that is, $x$ commutes with $a$ and $b$, and thus with all elements of $S_{3}$.
Therefore $H$ is the internal direct product $S_{3} \times \Span{x}$, which shows it is isomorphic to the first group.
