Left vs right semi direct products I just want to make sure that I am not doing anything silly here, but if we let $G$ be a group with $H,K$ subgroups, $H\lhd G$, and $\phi:K\rightarrow Aut(H)$, then is $$H\rtimes_\phi K \approx K \ _\phi\ltimes H$$ where the multiplication in the first is given by $$(h_1,k_1)(h_2,k_2) = (h_1\phi_{k_1}(h_2),k_1k_2). $$  Basically does it make complete sense just to switch the "slots" in the order pair?  This idea has come up in a project that I have been looking at for some time and using this notion would help me simplify some calculations greatly.
Thanks in advance.
 A: The free product $A*B$ of two groups is formed by considering all "words" formed using "letters" from $A$ and $B$, subject only to the condition that multiplying two elements of $A$ gives the same result as it does in $A$ itself, and similarly for $B$, but otherwise multiplying one element from $A$ with another from $B$ does not simply. In general, then, elements of $A*B$ look like $a_1b_1a_2b_2\cdots$.
If $\phi:K\to\mathrm{Aut}(H)$, then we may impose the relations $khk^{-1}=\phi_k(h)$ within $H*K$. Formally this means we take the quotient of $H*K$ by the normal subgroup generated by the set of all elements of the form $khk^{-1}\phi_k(h)^{-1}$. Call this quotient group $G$. For any product $kh\in H*K$, its image in $G$ may be equated with $(khk^{-1})k=\phi_k(h)k$. Using this sliding rule, every $h_1k_1h_2k_2\cdots$ can be simplified to just $hk$. But no two elements of the form $hk$ can be equal, for $h_1k_1=h_2k_2$ implies $h_2^{-1}h_1=k_2k_1^{-1}$ which is in $H\cap K=\{e\}$ within $H*K$ and hence in $G$.
For this reason, we may identify $G$ with the cartesian product $H\times K$, but it remains to see what the multiplication operation is. In fact, to evaluate $(h_1k_1)(h_2k_2)$, simply use the sliding rule on the middle two terms $k_1h_2=\phi_{k_1}(h_2)k_1$ to obtain $h_1k_1h_2k_2=h_1\phi_{k_1}(h_2)\cdot k_1k_2$. This is where the multiplication rule in the usual formal definition of $H\rtimes_\phi K$ comes from.
But there was no reason to use $H\times K$ instead of $K\times H$. The sliding rule applies just as well the other way, with $hk=k(k^{-1}hk)$. Then $(k_1h_1)(k_2h_2)=k_1k_2\cdot\phi_{k_2^{-1}}(h_1)h_2$. We can use this if we want to define a $K{}_\phi\ltimes H$ semidirect product. Then $H\rtimes_\phi K$ and $K{}_\phi\ltimes H$ should be isomorphic because they are both just $H*K$ modulo $khk^{-1}=\phi_k(h)$. Indeed, within the latter group we know that $kh=\phi_k(h)k$, so $K{}_\phi\ltimes H\xrightarrow{\sim} H\rtimes_\phi K$ should just be $(k,h)\mapsto(\phi_k(h),k)$.
