I am trying to solve the following problem from Dummit and Foote. The problem is from the section where semidirect product is introduced.
Show that there are exactly $4$ distinct homorphisms from $Z_2$ into $\operatorname{Aut}(Z_8)$. Prove that the resulting semidirect products are the groups: $Z_8 \times Z_2, D_{16}$, the quasidihedral group $QD_{16}$ and the modular group $M$ (c.f. the exercises in Section 2.5).
Notationally, let's let $x$ be a generator for $Z_2$ and $y$ a generator for $Z_8$.
I have shown the first part: namely that there are exactly $4$ homomorphisms $\phi: Z_2 \to Aut(Z_8)$ (each is completely determined by $\phi_x(y)$, which must be in $\{y,y^3,y^5,y^7\}$. I now want to show that the semidirect products actually yield th groups listed above.
I see why the homomorphism $a$ such that $a_x(y)=y$ gives us $Z_8 \rtimes_ a Z_2 \cong Z_8 \times Z_2$. But what about the other homomorphisms?
Now suppose $b$ is the homomorphism such that $b_x(y)=y^3$. How would we go about showing $Z_8 \rtimes_ b Z_2 \cong D_{16}$? (If this is even true).
I have tried setting $r:=(y,1)$ and $s:=(1,x)$. From this I see $r^2=(y,1)(y,1)=(y b_1(y),1)=(y^2,1)$ and by similar logic we actually have $r^i=(y^i,1)$, hence $r$ has order 8. By similar logic, $s$ has order 2. However, $rsr=(y,1)(1,x)(y,1)=(y,1)(b_x(y),x)=(y,1)(y^3,x)=(yb_1(y^3),x)=(y^4,x)$. I was hoping we would have $rsr=s$ as in the dihedral group, but apparently this is not the case.
And even if we chose $r,s$ differently so that $rsr=s$ (which I'm not claiming we can do per se), is there a non-tedious way to show that our resuling semi-direct product is in fact isomorphic to $D_{16}$? What about for $QD_{16}$ and $M$? (which I have never dealt with before)
Let $Z_2=\langle x\rangle$ and $Z_8=\langle y\rangle$. The first homomorphism is $\phi_1(y)=xyx^{-1}=y$. This leads to the abelian group of $Z_8 \rtimes_{\phi_1} Z_2 \cong Z_8 \times Z_2$.
Let $\phi_2(y)=xyx^{-1}=y^{-1}=y^7$. By the presentation of $D_{16}$, this leads to $Z_8 \rtimes_{\phi_2} Z_2 \cong D_{16}$.
Let $\phi_3(y)=xyx^{-1}=y^{3}$. This leads to $Z_8 \rtimes_{\phi_3} Z_2 \cong QD_{16}$ (the quasi-dihedral group of $16$).
Let $\phi_4(y)=xyx^{-1}=y^{5}$. This leads to $Z_8 \rtimes_{\phi_4} Z_2 \cong M_{16}$ (the modular group of $16$).
Since each group has a different number of elements with certain orders ($2,4$ or $8$), each is non-isomorphic from the others.