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In an arbitrary group with identity element $e$, the equation $x^2 = e$ always has one solution, namely, $x = e$. Is there always a second (different) solution? Prove or give a counterexample.

The problem I have with this question is understanding what my professor is trying to say simply because he did not describe what the binary operation of this group is. $e$ is different for any binary operation, example for multiplication it is 1 and addition is 0. Either way, 1 or 0 works in this case, namely the case where $x = e$ , I can't think of any identities in any group that this would not work.

So now, the real question is:

Is there always a second (different) solution? Either i'm misunderstanding his question, but his arbitrary group only had one solution, so I could just be sneaky and use that as my counterexample?

If I narrow it down, I can say, the group of integers under multiplication where the identity is 1, for example: if $x=-1$ or $x=1$ then the only possible case is when $x=e$ or $x=1$ is when $x=1$ since $-1 \neq e$

So my question is can someone dissect what my professor is trying to say in this problem because it is rather confusing. Thanks.

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    $\begingroup$ The problem is asking if there's always another solution no matter what the group is. Also, when the statement says the equation has one solution in every group, it means at least one solution. Because of course it's not stating there's precisely one solution no matter what the group is! $\endgroup$ – whacka Feb 23 '15 at 5:01
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In a finite group $G$ there is a solution to $x^2=e$ other than $x=e$ if and only if the order of the group is even (You can prove one implication with Lagrange's theorem and the other with Cauchy's theorem).

As for the counterexample take any finite group of odd order (Like $\mathbb Z_3$).

For an infinite example take $\mathbb Z$ with addition.

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  • $\begingroup$ I think I understand a bit more now but I'm confused on the $\mathbb{Z_3}$. So for the group of $\mathbb{Z}$ under addition, if we have x+x=0 then the only solution is x=0 therefore, there is no second solution but how can this be shown for odd orders or why does it only work for odd orders? $\endgroup$ – Justin Feb 24 '15 at 0:51
  • $\begingroup$ It follows from Lagrange's theorem. The order of every element $g$ in a group divides the order of the group $G$. Therefore $g$ cannot have order $2$ if $G$ is odd, and so $g^2=e$ if and only if $g=e$ $\endgroup$ – Jorge Fernández Hidalgo Feb 24 '15 at 0:52
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    $\begingroup$ I shall look into Lagrange's theorem then. thanks a lot! $\endgroup$ – Justin Feb 24 '15 at 0:55
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The binary operation of a group is part of the definition of that group. So when considering a group there is a given binary operation and your professor is talking about that operation.

Given that operation is there always an element $x \neq e$ such that applying the binary operation to the pair $(x, x)$ yields the element $e$?

No. Consider the group $\mathbb Z/3\mathbb Z$ under addition.

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The reason your professor did not define the operation is because it is arbitrary. If the claim is true that there is always a non-trivial element of a group $G$ such that $x^2 = e$, then that claim must hold no matter what operation makes your set a group. What your professor is asking you to do is try to come up with a specific group where this claim fails. If you cannot figure out a counter-example, then the claim may warrant a formal proof. However, there are plenty of readily available counter examples. As I said in my comment before deleting to opt for an answer, the group $\mathbb{Z}_3$ defined with the set $\{0,1,2\}$ under addition suffices. This one came to mind first because $3$ is odd and prime. More generally, $\mathbb{Z}_p$ will work for any prime greater than $2$. But you just need one counter example to disprove a claim.

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