Classifing groups of order 56: problems with the semidirect product

Solution 1:

I suppose your confusion is due to the double representation of the semidirect product: internal vs external.

The semidirect product's theorem states that if you have a group $G$ having two subgroups $H,K < G$ such that $H$ is normal in $G$, $H \cap K = \{1_G\}$ and $G=HK$ then there's an isomorphism $G \cong H \rtimes_\psi K$, for a certain $\psi \colon K \to \text{Aut}(H)$.

By the theorem we can represent every element of $G$ as a pair $(h,k) \in H \times K$ (which is the support of the group $H \rtimes K$).

Consider the two subgroups $\bar H = \{(h,1_K) | h \in H\} \leq H \rtimes K$ and $\bar K = \{(1_H,k)|k \in K\} \leq H \rtimes K$, these subgroups correspond, via the isomorphism, to the subgroups $H$ and $K$ of $G$.

In $H \rtimes K$ we have that for every $h \in H$ and $k \in K$ $$(1_H,k) * (h,1_K) *(1_H,k)^{-1} = (\psi_k(h),1_K)$$ if we identify every $h \in H$ with its corresponding element $(h,1_K)$ and every $k \in K$ with $(1_H,k)$ then this equality become (internally in $G$) $$k*h*k^{-1}=\psi_k(h)$$

The $\psi$ which determine the operation in the semidirect product is exactly the homomorphism sending every $k \in K$ in the (restriction to $H$ of the) automorphism $\psi_k \colon H \to H$ which send $h \in H$ in $khk^{-1}$ (this is clearly well defined because $H$ is normal in $G$.

Hope this help.