Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am looking for an example of a group in which the equation $x^2=e$ has more than two solutions, where $e$ is the identity element.

Groups with two solutions are easy to find:

  • nonzero reals under multiplication
  • cyclic group $\mathbb Z/2\mathbb Z$ under addition
  • more generally, cyclic group of even order

But none of these have more than two solutions.

share|cite|improve this question
@abiessu i cannot think of a point to start. – Aman Mittal Sep 27 '13 at 13:39
@AmanMittal: You could start with trying groups of order${}\leq2$ (not much chance there) then those of order $3,4,5,\ldots$; you will find examples before you get to the cases where there are lots of groups of a given order. – Marc van Leeuwen Sep 27 '13 at 14:08
up vote 3 down vote accepted

What about the group $\;C_2\times C_2\;,\;\;C_2=$ the cyclic group of order two ?

share|cite|improve this answer
oh yes, any transposition raised to 2 will be an identity . – Aman Mittal Sep 27 '13 at 13:40
Or, in fact, any product of disjoint transpositions, @AmanMittal – DonAntonio Sep 27 '13 at 13:41
$D_4$ = Dihedral Group too. – blondy Sep 27 '13 at 13:41
@blondy how is $D_4$ defined ? – Aman Mittal Sep 27 '13 at 13:42
@Aman : $$D_4=\langle\;s,t\;;\;s^2=t^4=1\;,\;sts=t^3\;\rangle$$ – DonAntonio Sep 27 '13 at 13:43

Permutation group $S_n$ has a lot of solutions for such equation. For example, any transposition will work.

share|cite|improve this answer
Thanks, i have noted this down !!! – Aman Mittal Sep 27 '13 at 14:05

Consider Example:

Suppose $G =D_3$ is the dihedral group of order 6 (gp. of symmetries of an equilateral triangle).Then, as we know, there are exactly three rotations$(R_0,R_{120},R_{240})$ and exactly three reflections ($p_1,p_2,p_3$). Also $R_0=e$ is the identity element and ${R_0}^2=R_0=e,{p_1}^2=e,{p_2}^2=e$ and ${p_3}^2=e$.

share|cite|improve this answer

In the group of 2x2 invertible matrices, consider the diagonal matrices $\mbox{diag}(1,-1)$, and $\mbox{diag}(-1,-1)$, $\begin{pmatrix} 0 & -1\\ -1 & 0\end{pmatrix}$. Find plenty more by using the inverse of $\begin{pmatrix} a & b\\ c & d\end{pmatrix}$.

share|cite|improve this answer
but that will be just one solution no ? – Aman Mittal Sep 27 '13 at 14:00
that is just 2, we need more than 2 solutions for $x$ – Aman Mittal Sep 27 '13 at 14:04
@AmanMittal: edited – Alex R. Sep 27 '13 at 14:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.