Groups of order 1016 This is an exercise in Group Theory that has been suggested during a course. Let $G$ be a group of order $1016$ with an Abelian $2$-Sylow. Show that the order of $Z(G)$, the center of $G$, is divisible by $4$ and that there can be no elements in $G$ of order $508$. I was trying to argue using the quotient $G/Z(G)$, looking at the wrong cases, i.e. when $|Z(G)|=1,2,127,254$. By Lagrange, if I can exclude these cases, I am done, but I can't find a good way to do the job. Do you have any guess? Thanks!
 A: We have the prime factorization $1016=2^3\cdot127$. We also know that the number of Sylow $127$-subgroups is equal to $1$. So there exists a normal subgroup $H\unlhd G$ such that $H\cong C_{127}$. 
You can make progress as follows.


*

*Show that $Aut(H)\cong C_{126}$, so $H$ has a unique automorphism of order two, and no automorphisms of order $4$. Why?

*Let $P$ be a Sylow $2$-subgroup. There exists a homomorphism $f:P\to Aut(H)$. How is that defined?

*Show that $\operatorname{ker}(f)$ has order $\ge4$.

*Show that all the elements of $\operatorname{ker}(f)$ commute with all of $P$ and all of $H$, and thus $\operatorname{ker}(f)\le Z(G)$.

A: Let $Q\in Syl_{127}(G)$. By counting argument, $Q\unlhd G$. $G/C_{G}(Q)$ is isomorphic to a subgroup of $Aut(Q)$. $Q\cong \mathbb{Z}/127\mathbb{Z}$. $Aut(Q)\cong (\mathbb{Z}/127\mathbb{Z})^{\times}$. $|Aut(Q)|=126=2\cdot 3^{2}\cdot 7$. $|G/C_{G}(Q)|$ divides $2\cdot 3^{2}\cdot 7$. But it also divides $|G|=2^{3}\cdot 127$. It equals $1$ or $2$. If it is $1$, $G=C_{G}(Q)$, $Q\leq Z(G)$. Let $P\in Syl_{2}(G)$. $G=PQ$. Let $x\in P$. Since $P$ is abelian, $x$ commutes with all elements in $P$ and $Q$. $x\in Z(G)$. $P\leq Z(G)$. We are done. If $|G/C_{G}(Q)|=2$, $|C_{G}(Q)|=2^{2}\cdot 127$. Let $R\in Syl_{2}(C_{G}(Q))$. $|R|=4$. $R$ is a $2$-subgroup of $G$. $R\leq P'$ for some $P'\in Syl_{2}(G)$. $G=P'Q$. $P'$ is abelian. Let $x\in R$. $x$ commutes with all elements in $P'$ and $Q$. $R\leq Z(G)$. $4$ divides $|Z(G)|$.
By the comments above, $G$ may have an element of order $508$. I guess maybe we just need to find a counter example. Then $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/127\mathbb{Z}$ would work.
