In classifying groups of order 20 using semi direct products, one eventually finds oneself looking for nontrivial homomorphisms $\phi : V_4 \to Z_4$. If $V_4 = \langle a, b \rangle = \{1, a, b, ab \}$ , and $Z_4 = \langle x \rangle = \{1, x, x^2, x^3 \}$, there are three such homomorphisms; $\phi_1$ defined by $\phi_1(a) = x^2, \phi_1(b) = 1$, $\phi_2$ defined by $\phi_2(a) = 1$ and $\phi_2(b) = x^2$, and $\phi_3$ defined by $\phi_3(a) = x^2$ and $\phi_3(b) = x^2$. This collection gives rise to three semi direct products of $Z_5$ with $V_4$. I know that these three semi direct products are all isomorphic to one another using the theorem that says if one can find an automorphism $\tau : V_4 \to V_4$, then $Z_5 \rtimes_\phi V_4 \cong Z_5 \rtimes_{\phi \circ \tau} V_4$, However, this theorem is not stated in my book, and is not even one of the exercises, so my question is as follows:
Is there another way to see the isomorphism between these groups?
Thanks
Any homomorphism from $V_4$ to $Z_4$ has a kernel, which in the semidirect product must be a central element $z$ of order $2$. From looking at the quotient $G/Z_5$ we find three subgroups of order $10$, each necessarily normal in $G$ of order $20$, and they intersect in $Z_5$. Thus one of them, say $H$, cannot contain $z$. Hence $G\cong H\times \langle z\rangle$. The group $H$ is a semidirect product of $Z_5$ by $Z_2$.