Let $A=\langle a \rangle $ and $B = \langle b \rangle$ be cyclic groups of orders 2 and 8 respectively. Let $\phi: A \rightarrow Aut(B)$ be the homomorphism specified by $\phi(a) : b^i \rightarrow b^{5i}$ for all $0\leq i \leq 7$. Let $H$ be the semi-direct product of $A$ and $B$ with respect to this $\phi$. List all the elements in the centre $Z(H)$ of $H$.
So to begin with I have established the homomorphism. In plain notation, let $\{1,a\}$ and $\{1,b,b^2,b^3,b^4,b^5,b^6,b^7\}$.
$ \phi = \left( \begin{array}{ccc} 1 & b & b^2 & b^3 & b^4 & b^5 & b^6 & b^7 \\ 1 & b^5 & b^2 & b^7 & b^4 & b & b^6 & b^3 \\ \end{array} \right)$
Now, $H=\{ (1,1), (1,b), (1,b^2), (1,b^3), (1,b^4), (1,b^5), (1,b^6), (1,b^7), (a,1), (a,b), (a,b^2), (a,b^3), (a,b^4), (a,b^5), (a,b^6), (a,b^7) \}$.
Now, to list the elements in the centre. This is what I am somewhat confused by. Please let me know if this is right and whether I have missed anything from the centre.
In the case of $\phi(1)$,
$(1,k)(1,k')=(1*1,k^{\phi(1)}k')=(1,k k')=(1,k'k)=(1*1,k'^{\phi(1)}k)=(1*1, k'k)=(1,k')(1,k)$,
where $k,k' \in B$.
$(1,k)(a,k')=(1*a,k^{\phi(a)}k')=(a,k k')=(a,k'k)=(a*1,k'^{\phi(1)}k)=(a*1, k'k)=(a,k')(1,k)$,
provided only that $k', k \in \{1, b^2, b^4, b^6 \}$, and lastly
$(a,k)(a,k')=(a*a,k^{\phi(a)}k')=(1,k k')=(1,k'k)=(a*a,k'^{\phi(a)}k)=(a*a, k'k)=(a,k')(a,k)$,
provided only that $k', k \in \{1, b^2, b^4, b^6 \}$.
Therefore, we have that $Z(H)=\{ (1,1), (1,b^2), (1,b^4), (1,b^6), (a,1), (a,b^2), (a,b^4), (a,b^6) \}$