semidirect product, should the morphism act on the first or second element?

Question

Let us take two groups, $N$ and $H$, as well as a non-trivial homomorphism $\varphi:H\to\mathrm{Aut}(N)$.

Now let me define two further groups: $$ G_R=(N\times H,\circledR)\,,\\ G_S=(N\times H,\circledS)\,, $$ i.e., the underlying set is always $N\times H$, but we have two different group operations: $$ (n_1,h_1)\mathbin{\circledR}(n_2,h_2)=\bigl(n_1\circ\varphi(h_1)(n_2),h_1\circ h_2\bigr)\,,\\ (n_1,h_1)\mathbin{\circledS}(n_2,h_2)=\bigl(\varphi(h_2^{-1})(n_1)\circ n_2,h_1\circ h_2\bigr)\,. $$ As far as I can tell both $G_R$ and $G_S$ are well defined groups. Now some questions: What is the relation between the two definitions? Are they isomorphic or could they lead to different groups? (If yes, what is the isomorphism?) Is either of the two definitions considered more natural? (Would you define one or the other as $N\rtimes_\varphi H$?) [Or is this like with sesquilinear products where mathematicians/physicists seem to prefer having linearity in either the first or the second component?]

-- Edited to correct my mistake as pointed out by Derek Holt.

Answer 1

• $G_R$ and $G_S$ are isomorphic.
• $i:G_R\to G_S,\; (n,h)\mapsto(\varphi(h^{-1})(n),h)$ defines a possible isomorphism.
• $G_R$ could be considered more natural; its elements $(n,h)$ correspond to the ordering $n\circ h$ when viewed as an internal semidirect product, such that $(n_1\circ h_1)\circ(n_2\circ h_2)=(n_1\circ[h_1\circ n_2\circ h_1^{-1}])\circ(h_1\circ h_2)$ and therefore $\varphi(h_1)(n_1)=h_1\circ n_2\circ h_1^{-1}$.