For element of order $2$, $\{1,z\}$ is normal subgroup iff $z \in Z(G)$ 
Let $G$ be a group, and let $z\in G$ be an element of order $2$. Show that $\{ 1,z\}$ is a normal subgroup of $G$ if and only if $z \in Z(G)$.

For the $\rightarrow$ way, I thought about -
If $g\{1,z\}=\{1,z\}g$, then we got 2 options-


*

*$gz=zg$, and we are done.

*$g*1=g=zg$, and then $1=z$, but $z$ is of order 2 and therefore $z$ isn't $1$( which is meanningless as $1 \in Z(G)$ anyway...).
I feel like this is wrong, as we don't really use the assumption that $z$ if of order $2$...
For the other way I could use the same claim, just from the other direction...
 A: You use that fact for knowing that it is a subgroup of G.
A: I think your argument is along the right lines, but the logic in case 2 isn't very clear. Here's another way of writing it:
If $z \not\in Z(G)$ there is an element $x$ such that $zx \neq xz$. But then $x^{-1}zx \neq z$ and $x^{-1}zx \neq 1$ (because $x^{-1}zx$ has the same order as $z$, namely 2). So $x^{-1}\{1, z\}x = \{1, x^{-1}zx\} \neq \{1, z\}$. So $\{1, z\}$ is not a normal subgroup.
A: More conceptual and general view.
Proposition Let $G$ be a finite group and $p$ the smallest prime dividing $|G|$. Let $N$ be a subgroup of order $p$. Then $N$ is normal if and only if $N \subseteq Z(G)$.
Proof (Sketch) The "if" part is trivial. Assume $N \unlhd G$. Then $G$ acts on $N$ by conjugation, inducing automorphisms of $N$. This provides a homomorphical embedding $G/C_G(N) \hookrightarrow Aut(N)$, where $C_G(N)=\{g \in G: n^g=n \text { for all } n \in N\}$. Since $|N|=p$, $Aut(N) \cong C_{p-1}$. But $p$ is the smallest prime dividing $|G|$, whence $G/C_G(N)$ must be trivial, so $G=C_G(N)$, equivalent to $N$ being central. $\square$
