October 2nd
Today I learned the definition of the semidirect product. For groups $H$ and $K,$ we assume that we have a homomorphism $K\to\op{Aut}(H)$ by $k\mapsto\varphi_k.$ (This mapping can be thought of as a way for $K$ to act on $H$ for a coarse multiplication.) Then $H\rtimes_\varphi K$ is defined on $H\times K$ so that multiplication is done by\[(h_1,k_1)(h_2,k_2)=(h_1\varphi_{k_1}(h_2),k_1k_2).\]Very quickly, in the trivial case where $\varphi_\bullet=\op{id},$ we see that $H\rtimes_\varphi K\cong H\times K,$ so we justify somewhat the name "product.'' Showing that this group behaves is somewhat annoying, but it does. We get $(1,1)$ as our identity, but inverses are $(h,k)^{-1}=(\varphi^{-1}_k(h),k^{-1})$ to make the multiplication work. We intentionally avoid talking about associativity.
To further justify the name "product,'' we see that the set of elements $(h,e)$ are isomorphic to $H$ because\[(h_1,e)(h_2,e)=(h_1\varphi_e(h_2),ee)=(h_1h_2,e)\]because we must have $\varphi_e=\op{id}.$ Similarly, we can embed $K$ into $H\rtimes_\varphi K$ by elements $k\mapsto(e,k).$