Example where $x^2 = e$ has more than two solutions in a group Show by means of an example that it is possible for the quadratic equation $x^2 = e$ to have more than two solutions in some group $G$ with identity $e$.
 A: Hint: Try $S_3$. Here $(1\:2), (1\:3), (2\:3)$ are solutions of $x^2=e$.
A: In $S_n$, any  transposition $\tau$ satisfies $\tau^2=e$. There are $\dbinom n2$ transpositions in $S_n$.
A: Try $(\mathbb Z/2\mathbb Z)^n$ for $n\ge 2$.
A: There's an example in $(\mathbb{Z} / n \mathbb{Z})^*$ with $n \leq 10$.  (This shows that the phenomenon can occur in abelian groups as well.)
A: It is possible for every element of a group to satisfy $x^2=e$. Take subsets of a set of size $n$ - there are $2^n$ of these - and have the group operation as disjoint union ($a*b$ consists of the elements of $a$ or $b$ which are not elements of both). The identity is the empty set. This construction works for infinite sets too.
Alternatively the Klein 4-group, or the odd integers modulo $8$ under multiplication.
A: Hint: consider $(1\,2)(3\,4)$, $(1\,3)(2\, 4)$ and $(1\,4)(2\,3)$ in $S^4$.
A: Consider integers mod 12 with multiplication. You have 1, 5, 7 and 11 that square to 1.
A: Another example (and the smallest example possible) is the Klein four-group (that is, $\Bbb Z_2 \times \Bbb Z_2$).  In particular, we have
$$
G = \{e,a,b,ab\}
$$
where every $g \in G$ satisfies $g^2 = e$.
