Group of order $4$ such that $x^2=1$ for all $x$?

Question

Give an example of an abelian group of order 4 in which every nonidentity element $x$ satisfies $x*x=e$.

Theorem Let $G$ (with operation $*$) y $H$ (with operation $\diamond$) be groups. Define an operation $\bullet$ on $G \times H$ by $(g,h) \bullet (g',h') = (g*g',h \diamond h')$. Then $G \times H$ is a group. If $G$ and $H$ are abelian, then so is $G \times H$. If $G$ and $H$ are finite, then so is $G \times H$ and $|G \times H|=|G||H|$.

I do not know how to solve using the previous theorem. I showed it in a different way, and it is in the following way: As it is an Abelian group then we will have it.

• $b ∗ a = a ∗ b$
• $b ∗ [a ∗ e] = a ∗ b$
• $b ∗ [a ∗ (b ∗ b)] = a ∗ b$
• $b ∗ [(a ∗ b) ∗ b] = a ∗ b$
• $[b ∗ (a ∗ b)] ∗ b = a ∗ b$
• $b ∗ (a ∗ b) = a$
• $[e ∗ b] ∗ (a ∗ b) = a$
• $[(a ∗ a) ∗ b)] ∗ (a ∗ b) = a$
• $[a ∗ (a ∗ b)] ∗ (a ∗ b) = a$
• $a ∗ [(a ∗ b) ∗ (a ∗ b)] = a ∗ e$
• $(a ∗ b) ∗ (a ∗ b) = e$
• $x*x=e$

But my big doubt is: how to use the previous theorem?. and give an example.

Answer 1

You can apply the theorem with $G=H=\mathbb{Z}/2\mathbb{Z}$. Then $G\times H$ has the property that you are interested in.