An example of a group G that satisfies $x^2 =e$ , for all $x \in G$

Question

I thought for a long time about which group will have this property but I didn't get it.

Is this a valid one

so if G = {a,b} then ab = b and b is the identity for example then aa = b , ab = b and bb = b , ba = b ?

any hints ?

Answer 1

Here I give two results related to the groups with above property.

If a $G$ satisfies $x^2=e$ for all $x\in G,$ then $G$ should be abelian.
proof:
Note that the given condition implies $x=x^{-1},\ \forall x\in G.$ $$xy=(xy)^{-1}=y^{-1}x^{-1}=yx,\ \forall x,y\in G.$$

If a $G_1,\ G_2$ to be two groups satisfies $x^2=e$ for all $x\in G_1,G_2$ then $G_1\times G_2$ also has this property.
Proof:
Suppose $G_1,G_2$ be two groups such that $x^2=e$ for all $x\in G_1,G_2.$
Then we know that $G_1\times G_2$ is also a group. Take any $(x,y)\in G_1\times G_2$ $$(x,y)^2=(x^2,y^2)=(e,e)$$ is the identity element of $G_1\times G_2.$ Hence $G_1\times G_2$ also has this property.

Simple examples of this type of groups are
1) $G=\{1\}$
2) $Z_2=\{0,1\}$
3) $Z_2\times Z_2=\{(0,0), (0,1), (1,0), (1,1)\}$
4) $Z_2\times Z_2\times Z_2.$

Answer 2

Given $m,n\ge 1$, the set of all the $m\times n$ matrices with entries in the finite field $\Bbb F_2$ is, under addition, an abelian group which is also unipotent ($x+x=0$ for all $x$). The order of this group is $2^{m\cdot n}$.