Example of an infinite group where every element except identity has order 2

Question

Find an infinite group, in which every element g not equal identity (e) has order 2

Does this question mean this:

the group that fail condition (2) which is no inverse and also that group must have the size 2

My answer:

Z*

Answer 1

No, that’s clearly not what it means: a group of size $2$ is not an infinite group. You’re to find an infinite group $G$ in which every element except the identity has order $2$, meaning that if $g\in G$, and $g$ is not the identity element $1_G$ of $G$, then $g^2=1_G$. Of course $1_G^2=1_G$ as well, so your problem is really to find an infinite group $G$ in which every element satisfies the equation $x^2=1_G$, where $1_G$ is the identity element in $G$.

HINT: First find a finite group $H$ with this property, and then look at the product of infinitely many copies of $H$.

Alternative HINT: Consider the operation of symmetric difference on the set of subsets of some infinite set.

Answer 2

Consider the commutative group of all infinite-dimensional vectors $(x_n)_{n=1}^\infty$ where all entries $x_n$ in a vector are either $0$ or $1$ and you define $1+1 = 0$ and the group addition is entry-wise for the vectors who are elements in your group (and the vector of all zeros is the identity in your group).

Answer 3

Let $X$ be an infinite set and consider the power set $\mathscr P(X)$ with a binary operation symmetric difference defined by $$A+B=(A-B)\cup(B-A)$$ Then, $(\mathscr P(X),+)$ is a group of infinite order (meaning that the set $\mathscr P(X)$ is infinite) while every nontrivial element of it has order 2 (meaning that if $A\in\mathscr P(X)$ and $A\neq\emptyset$, then $A^2=\emptyset$).

1. If $A,B\in\mathscr P(X)$, then $A+B\in\mathscr P(X)$.
2. If $A,B,C\in\mathscr P(X)$, then $(A+B)+C=A+(B+C)$.
3. There exists $E\in\mathscr P(X)$ such that for every $A\in\mathscr P(X)$, $A+E=E+A=E$.
4. For each $A\in\mathscr P(X)$, there exists $(-A)\in\mathscr P(X)$ such that $A+(-A)=(-A)+A=E$.

The proof of 1 is obvious and that of 2 is cumbersome but elementary (if you want, see Associativity of the symmetric difference or An alternative proof from proofwiki. Or, you can simply use the Venn diagram.)

In 3, $E=\emptyset$. That is, the empty set serves as the identity of this group. This is obvious from $A+\emptyset=(A\cup\emptyset)-(A\cap\emptyset)=A-\emptyset=A$ and the like.

In 4, $(-A)=A$. That is, $A$ itself is the inverse element of $A$, which is also obvious from $A+A=(A\cup A)-(A\cap A)=A-A=\emptyset$.

By the last argument above, the group $(\mathscr P(X),+)$ is the desired group each nontrivial element of which is of order $2$. Indeed, $2A=A+A=E$ for every $A\in\mathscr P(X)$ and $2$ is the smallest number $n$ such that $nA=E$ (or $A^n=E$, if you like the multiplicative expression). This is what it means by saying that $A$ is of order 2.

Answer 4

A simple case wherein there is a group of infinite order, but each element has order $1,2$ is:

The set of all nonnegative integers expressed in base 2, with binary operation summing the numbers under bitwise XOR.

The identity is $0$ and the inverse of each positive integer is itself save for $0$. Further XOR is associative.

Answer 5

If $A=\{\sqrt{r}:r\in\mathbb{Q}^+\}$ then $A$ is a group under the usual multiplication operation $\times$ and all non-identity elements of the quotient group $A/\mathbb{Q}^+$ have order $2$ .