Is every group $G$ a subgroup of index $2$ of some $\tilde G$?

Let $(G,f)$ be a group (so $f$ stands for the group operation), $H\lneq G$ and $\complement_GH:=G\setminus H$. Then:

\begin{alignat}{1} &f(H,H)\subseteq H \\ &f(H,\complement_GH)\subseteq\complement_GH \\ &f(\complement_GH,H)\subseteq\complement_GH \\ &[G:H]=2 \Rightarrow f(\complement_GH,\complement_GH)\subseteq H \\ \tag 1 \end{alignat}

Reminiscent of this fact, I wonder whether a supergroup of $G$ can be built up, say $\tilde G$, such that $[\tilde G:G]=2$. Namely:

Is every group $G$ a subgroup of index $2$ of some $\tilde G$?

I think that's true, based on the following.

Let $G$ be a group and $X_G$ a set such that:

1. $X_G\cap G=\emptyset$;
2. there is a bijection $\alpha\colon X_G\to G$.

Let's define $\tilde G:=G\cup X_G$ and $\cdot : \tilde G\times\tilde G \to \tilde G$ by:

\begin{alignat}{1} &a)\space g\cdot h:=gh, \space\forall g,h \in G \\ &b)\space g\cdot x:=\alpha^{-1}(g\alpha(x)), \space\forall g \in G, x\in X_G \\ &c)\space x\cdot g:=\alpha^{-1}(\alpha(x)g), \space\forall g \in G, x\in X_G \\ &d)\space x\cdot y:=\alpha(x)\alpha(y), \space\forall x,y\in X_G \\ \tag 2 \end{alignat}

• Associativity

It holds in $G$ by definition. Besides, $\forall g,h \in G, \forall x \in X_G$:

\begin{alignat}{1} (g\cdot h)\cdot x &= (gh)\cdot x \\ &= \alpha^{-1}(gh\alpha(x)) \\ \end{alignat}

and

\begin{alignat}{1} g\cdot (h\cdot x) &= \alpha^{-1}(g\alpha(h\cdot x)) \\ &= \alpha^{-1}(g\alpha(\alpha^{-1}(h\alpha(x)))) \\ &= \alpha^{-1}(gh\alpha(x)) \\ \end{alignat}

whence:

$$g\cdot (h\cdot x) = (g\cdot h)\cdot x \tag 3$$

Likewise, $\forall g,h \in G, \forall x \in X_G$:

\begin{alignat}{1} (g\cdot x)\cdot h &= \alpha^{-1}(\alpha(g\cdot x)h) \\ &= \alpha^{-1}(\alpha(\alpha^{-1}(g\alpha(x)))h) \\ &= \alpha^{-1}(g\alpha(x)h) \\ \end{alignat}

and

\begin{alignat}{1} g\cdot (x\cdot h) &= \alpha^{-1}(g\alpha(x\cdot h)) \\ &= \alpha^{-1}(g\alpha(\alpha^{-1}(\alpha(x)h))) \\ &= \alpha^{-1}(g\alpha(x)h) \\ \end{alignat}

whence:

$$g\cdot (x\cdot h) = (g\cdot x)\cdot h \tag 4$$

Furthermore, $\forall g \in G, \forall x,y \in X_G$:

\begin{alignat}{1} (g\cdot x)\cdot y &= \alpha(g\cdot x)\alpha(y)\\ &= \alpha(\alpha^{-1}(g\alpha(x)))\alpha(y)\\ &= g\alpha(x)\alpha(y)\\ &= g(\alpha(x)\alpha(y))\\ &= g\cdot(\alpha(x)\alpha(y))\\ &= g\cdot(x\cdot y)\\ \tag 5 \end{alignat}

Finally, $\forall x,y,z \in X_G$:

\begin{alignat}{1} (x\cdot y)\cdot z &= (\alpha(x)\alpha(y))\alpha(z)\\ &= \alpha(x)(\alpha(y)\alpha(z)) \\ &= x\cdot(y\cdot z) \\ \tag 6 \end{alignat}

• Unit

Note that:

• by $(2$-$b)$: $\space$ $e\cdot x=\alpha^{-1}(e\alpha(x))=\alpha^{-1}(\alpha(x))=x, \space\forall x \in X_G$
• by $(2$-$c)$: $\space$ $x\cdot e=\alpha^{-1}(\alpha(x)e)=\alpha^{-1}(\alpha(x))=x, \space\forall x \in X_G$

and $e$ serves as unit throughout $\tilde G$.

• Inverses

Let's define:

$$x^{-1}:= \alpha^{-1}(\alpha(x)^{-1}), \forall x \in X_G \tag 7$$

Therefore, by $(2$-$d)$ and $(7)$:

• $\space$ $x\cdot x^{-1}=\alpha(x)\alpha(x^{-1})=\alpha(x)\alpha(\alpha^{-1}(\alpha(x)^{-1}))=\alpha(x)\alpha(x)^{-1}=e, \space\forall x \in X_G$

and $x^{-1}$ serves as inverse of the element $x \in X_G$.

Seemingly, $\tilde G$, endowed with the operation $(2)$, is indeed a group, and $[\tilde G:G]=2$.

Why not take $\tilde G = G \times \mathbb{Z}/2 \mathbb{Z}$?