Finding the order of Z(G) in a non-Abelian group of order 8 Let $G$ be a non-Abelian group of order $8$. Prove that $|Z(G)|$ is less or equal to $2$.
First I must say this is a question about normal subgroups. I haven't yet studied homomorphisms or more complex material.
This is my try to solve the problem. Since $Z(G)$ is a subgroup of $G$, its order should be $8, 4, 2$ or $1$. Because $G$ is non-Abelian, it can't be $8$. Then I assumed the order is $4$ and tried to get to a contradiction. Since $|Z(G)|=4$, there are only two cosets of $Z(G)$ in $G$. One of them is $Z(G)$ itself, and the other one is all the elements that are not in $Z(G)$.
Here I am stuck. Maybe it's not the right way to think about this problem..
Help would be greatly appreciated.
 A: First of all $Z(G)$ is a proper subgroup of $G$ (since $G$ is not abelian), so it's order is $1, 2$ or $4$. If it's $4$, then $G/Z(G)$ is cyclic, so $G$ is abelian, contradiction.
A: In this case, I think it might be easier working in greater generality first.  Try proving that, in general, if $G/Z(G)$ is cyclic, then $G$ is abelian.
To get you started, let $G$ be any group, and suppose that $G/Z(G)$ is a cyclic group.  Then there is a coset $gZ(G)$ for some $g \in G$ such that $\langle g Z(G) \rangle = G/Z(G)$.  That is to say, for any other coset $hZ(G)$, we have $hZ(G) = (gZ(G))^n = g^nZ(G)$ for some $n \in \mathbb{N}$.  
Now choose any $x, y \in G$.  We have $x \in g^nZ(G)$ and $y \in g^mZ(G)$ for some $n, m \in \mathbb{N}$.  Carefully apply the definitions here to conclude that $xy = yx$.  

Once you have this result, you can finish up by showing that $|G| = 8$ and $|Z(G)| > 2 \implies G$ is abelian (very simple argument).
A: Actually non abelian group of order 8 is either D8(dihedral group) or quarternion group( up to isomorphism and you can show it using cauchy's thm). 
Center of D8 equals identity and rotation, center of Q8 is 1 and -1, which proves the statement.
