[proofplan] We first verify directly that the centre $Z(G)$ is a [normal subgroup](/page/Normal%20Subgroup) of $G$. Since $G$ is simple, this forces $Z(G)$ to be either $\{e\}$ or $G$. The second possibility would mean that every element of $G$ commutes with every other element, making $G$ abelian, contrary to the hypothesis. Hence only the identity-subgroup possibility remains. [/proofplan]

[step:Show that the centre is a normal subgroup of $G$]Let $e\in G$ denote the identity element. Let $Z(G)$ denote the centre of $G$, namely $Z(G)=\{z\in G:zg=gz\text{ for every }g\in G\}$. First, $e\in Z(G)$ because, for every $g\in G$, the identity law gives \begin{align*} eg=g e=g. \end{align*} If $z_1,z_2\in Z(G)$ and $g\in G$, then associativity and the defining commutation property of $z_1$ and $z_2$ give \begin{align*} (z_1z_2)g=z_1(z_2g)=z_1(gz_2)=(z_1g)z_2=(gz_1)z_2=g(z_1z_2). \end{align*} Thus $z_1z_2\in Z(G)$. If $z\in Z(G)$ and $g\in G$, then $zg=gz$. Multiplying on the left by $z^{-1}$ and on the right by $z^{-1}$ gives $gz^{-1}=z^{-1}g$. Hence $z^{-1}\in Z(G)$, so $Z(G)\le G$. Now let $a\in G$ and $z\in Z(G)$. For every $g\in G$, since $a^{-1}ga\in G$ and $z$ commutes with every element of $G$, we have $z(a^{-1}ga)=(a^{-1}ga)z$. Multiplying on the left by $a$ and on the right by $a^{-1}$ gives $(aza^{-1})g=g(aza^{-1})$. Therefore $aza^{-1}\in Z(G)$. Since this holds for every $a\in G$ and every $z\in Z(G)$, the subgroup $Z(G)$ is normal in $G$.[/step]

[guided]We prove normality rather than cite it, so that the only structural input later is the definition of simplicity. The centre is $Z(G)=\{z\in G:zg=gz\text{ for every }g\in G\}$. First we check that this set is a subgroup. The identity element $e$ belongs to $Z(G)$ because, for every $g\in G$, the identity law gives \begin{align*} eg=ge=g. \end{align*} If $z_1,z_2\in Z(G)$, then each of $z_1$ and $z_2$ commutes with every $g\in G$. Therefore, for an arbitrary $g\in G$, associativity and the defining commutation property give \begin{align*} (z_1z_2)g=z_1(z_2g)=z_1(gz_2)=(z_1g)z_2=(gz_1)z_2=g(z_1z_2). \end{align*} Thus $z_1z_2$ also commutes with every element of $G$, and hence $z_1z_2\in Z(G)$. Next let $z\in Z(G)$. Since $zg=gz$ for every $g\in G$, multiplying this equality on the left by $z^{-1}$ and on the right by $z^{-1}$ gives $gz^{-1}=z^{-1}g$. Thus $z^{-1}$ commutes with every $g\in G$, so $z^{-1}\in Z(G)$. We have proved $Z(G)\le G$. To prove normality, we must show that conjugating an element of $Z(G)$ by an arbitrary element of $G$ stays in $Z(G)$. Let $a\in G$ and $z\in Z(G)$. We need to prove $aza^{-1}\in Z(G)$, meaning that $aza^{-1}$ commutes with every $g\in G$. Fix $g\in G$. Since $a^{-1}ga\in G$ and $z$ commutes with every element of $G$, we have $z(a^{-1}ga)=(a^{-1}ga)z$. Multiplying this equality on the left by $a$ and on the right by $a^{-1}$ gives $(aza^{-1})g=g(aza^{-1})$. Because $g\in G$ was arbitrary, $aza^{-1}$ commutes with every element of $G$. Hence $aza^{-1}\in Z(G)$ for every $a\in G$ and $z\in Z(G)$, so $Z(G)\trianglelefteq G$.[/guided]

[step:Use simplicity to restrict the possible centres] Since $G$ is simple and $Z(G)\trianglelefteq G$, the normal subgroup $Z(G)$ must be either the identity subgroup $\{e\}$ or all of $G$, so $Z(G)=\{e\}$ or $Z(G)=G$.[/step]

[step:Exclude the possibility that the centre is all of $G$] Suppose, for contradiction, that $Z(G)=G$. Then every element of $G$ lies in the centre. Hence for all $x,y\in G$, $xy=yx$. Therefore $G$ is abelian. This contradicts the hypothesis that $G$ is non-abelian. Thus $Z(G)\ne G$. [/step]

[step:Conclude that the centre is the identity subgroup] The only possibilities from simplicity were $Z(G)=\{e\}$ and $Z(G)=G$. The second has been excluded by the non-abelian hypothesis. Therefore $Z(G)=\{e\}$. This proves the theorem. [/step]