Centre of non-Abelian Group G, Order 6, is Trivial

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.

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.