up vote 1 down vote favorite

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.

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  • @abiessu i cannot think of a point to start. – Aman Mittal Sep 27 '13 at 13:39
  • 1
    @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
  • See also: math.stackexchange.com/questions/1443487/… – Martin Sleziak Dec 1 '15 at 15:17
up vote 3 down vote accepted

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

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  • 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
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    @Aman : $$D_4=\langle\;s,t\;;\;s^2=t^4=1\;,\;sts=t^3\;\rangle$$ – DonAntonio Sep 27 '13 at 13:43
up vote 8 down vote

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

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  • Thanks, i have noted this down !!! – Aman Mittal Sep 27 '13 at 14:05
up vote 2 down vote

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 ${p_3}^2=e$.

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up vote 0 down vote

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 $\begin{pmatrix} a & b\\ c & d\end{pmatrix}$.

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  • 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

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