Centre of non-Abelian Group G, Order 6, is Trivial Claim: Let $G$ be a non-Abelian group, $\lvert{G}\rvert=6$. Then $Z(G)=1$.
Attempt so far:
We know by Lagrange's Theorem that, since $Z(G)\leq G$, we have that $n:= \lvert G\rvert\in\{1,2,3,6\}$, and that since the centre is Abelian that $Z(G)\neq G$, i.e. $n\neq 6$. We then need to rule out the cases where $n=2,3$?
I know that
$$
Z(G)=\bigcap_{g\in G} C_G(g)=\bigcap_{g\in G}\{h\in G:hg=gh\},
$$
but I'm not entirely sure how to use this. I guess it is sufficient to find $g\in G$ such that $C_G(g)=1$, however I'm not sure how to go about doing so.
Thanks in advance. 
 A: Another approach uses the following well-known fact which is easy to prove:

If $G/Z(G)$ is cyclic then $G$ is abelian

Indeed, if $G$ has order $6$ and $Z(G)$ is not trivial, then $G/Z(G)$ has order $1$, $2$, or $3$. In all cases, $G/Z(G)$ is cyclic and so $G$ is abelian.
A: Apparently we get to use Lagrange's Theorem but no other theorems. So here is an argument that doesn't use anything else. 
By pairing every element off with its inverse, we see that the identity and therefore (6 is even) at least one other element is self-inverse. So let $t\not=1$ be self inverse, that is of order 2.
By Lagrange every element $g\in G$ has order dividing $6$: that is 1,2,3, or 6.
If any element $g$ is of order 6, then $G=\{1,g,g^2,g^3,g^4,g^5\}$ is abelian - which it is not. 
If every non-identity element is order 2, then for any $x,y\in G$ we have $(xy)^2=1$, $x^2=1$, and $y^2=1$
so that
$$
yx=xx(yx)yy=x((xy)^2)y=xy.$$ 
But $G$ is not abelian so this does not happen.
So we have that $G$ must have an element $d$ of order $3$ and an element $t$ of order $2$. 
If $C(d)>\{1,d,d^2\}$ then by Lagrange $C(d)=G$. In that case $d,t$ commute and so $dt$ has order 6, which as we've seen is not so.
If $C(t)>\{1,t\}$ then by Lagrange  $C(t)=G$. In that case $d,t$ commute and so $dt$ has order 6, which as we've seen is not so.
Hence $Z(G)\leqslant C(d)\cap C(t)=\{1,d,d^2\}\cap\{1,t\}=\{1\}$ as required.
A: Let $G$ be a group of order $6$, and suppose it is not abelian. By Cauchy's theorem, there is an element $s$ of order $2$ and an element $r$ of order $3$. If these commute, then $rs$ has order $6$ and $G$ is abelian, so they cannot commute. Moreover, since $r$ has order three, it has index two in $G$ and thus generates a normal subgroup. It follows that $srs^{-1}$ has order $3$ and belongs to $\langle r\rangle$, so by our restrictions the only option is that $srs=r^{-1}$. This shows that $G$ is isomorphic to $S_3=D_3=\langle r,s\mid r^3,s^2,(sr)^2\rangle$, and every symmetric group of order $>2$ is centerless: if $\sigma$ is a non-identity permutation then there are $i,j$ such that $\sigma(i) = j$. Pick $k$ different from these, and note that if $\tau=(ik)$, then $\sigma\tau(j) = i$ while $\tau\sigma(j)=k$. Alternatively, $D_n$ is centerless for $n$ odd and has center generated by $r^{n/2}$ for $n$ even. 
