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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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1  
What have you tried so far? What elements are easy to show as solutions? –  abiessu Sep 27 '13 at 13:38
    
@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

4 Answers 4

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

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

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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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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