Example where $x^2 = e$ has more than two solutions in a group

Question

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 $e$.

Answer 1

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.

Answer 2

Hint: Try $S_3$. Here $(1\:2), (1\:3), (2\:3)$ are solutions of $x^2=e$.

Answer 3

Hint: consider $(1\,2)(3\,4)$, $(1\,3)(2\, 4)$ and $(1\,4)(2\,3)$ in $S^4$.

Answer 4

In $S_n$, any transposition $\tau$ satisfies $\tau^2=e$. There are $\dbinom n2$ transpositions in $S_n$.

Answer 5

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.)

Answer 6

Consider integers mod 12 with multiplication. You have 1, 5, 7 and 11 that square to 1.

Answer 7

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 $g^2 = e$.

Answer 8

Try $(\mathbb Z/2\mathbb Z)^n$ for $n\ge 2$.