Let $G$ be a finite 2-group such that $\left|\dfrac{G}{Z(G)}\right|=4$, $Z(G)$ is not cyclic and $Z(G)$ has at least one element of order 4. Then prove that there exists an automorphism $\alpha$ of $G$ such that $\alpha(z)\neq z$ for some $z\in Z(G)$.
Do the proof in My attempt 2 is true?
My attempt1:
Let $\dfrac{G}{Z(G)}=\{Z(G), aZ(G), bZ(G), abZ(G)\}$ and $T$ be the set of minimal generators of $G$. Then $G=\langle a , b, x_{1},..., x_{t}\rangle$, where any $x_{i}\in Z(G)\cap T.$
Now if $|a|, |b|\leq 4$, then we define $\alpha(a)=a, \alpha(b)=b$ and $\alpha(x_{i})=x_{i}^{-1}$. So the statement in this case is true. But if $|a|$ or $|b|\geq 8$, then i do not know how we can define the automorphism $\alpha$.
Please guide me that how continue the proof.
My attempt2: Here we have $\Phi(G)\leq Z(G)$. If $\Phi(G)=Z(G)$, then $G$ is a 2-generated and the problem is easy. If $\Phi(G)<Z(G)$, then there exists $Y=\{y_{1},...y_{t}\}$ such that $\dfrac{Z(G)}{\Phi(G)}=\langle y_{1}\Phi(G),... ,y_{t}\Phi(G)\rangle$. Now we define $\alpha(a)=a, \alpha(b)=b, \alpha(y_{1})=y_{1}h$(for some $h\in\Phi(G) ,|h|=2$) and $\alpha(y_{i})=y_{i}(2\leq i\leq t)$. if this $\alpha$ is automorphism, then the proof is completed.
Is this proof correct? Is $\alpha$ is automorphism?
Thank you
Am I mistaken if I define $\alpha$ as follows?:
As you wrote, we have a partition of $G$ into 4 cosets: The restriction of $\alpha$ to the center $Z(G)$ is the inverse map. The restriction of $\alpha$ to $aZ(G)$ is $\alpha(ag)=ag^{-1},\;g\in Z(G)$. Similarly $\alpha(bg)=bg^{-1}$ and $\alpha ab(g)=abg^{-1}$. Does this $\alpha$ work?