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From Prof. Charles Pinter's "A Book of Abstract Algebra"'s Chapter 4 exercises:

For each of the following rules, either prove that it is true in every group $G$, or give a counter-example.

$$ \text{if } x^{2}=a^{2}, \text{then } x=a$$

I believe that this is true by:

$$xx = aa$$

by cancellation,

$$x=a$$

Is that right? Also, when the word "prove" is used, does that mean to use theorems to prove?

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    $\begingroup$ You can't use cancellation like that. You need to cancel the same thing on either side. In any case, this is not true. (Hint: $(-1)^2 = 1$) $\endgroup$ – Prahlad Vaidyanathan May 13 '15 at 15:15
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    $\begingroup$ Disproof can be by counterexample, for example: $x=-1 , a=1$ $\endgroup$ – Joffan May 13 '15 at 15:25

10 Answers 10

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What is it that you are canceling there? It looks to me like you used that $x=a$ to show that $x=a$, which is not a good plan.

Think: is this true even in the real numbers?

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In standard real numbers:

$$x^2 = a^2 \implies x^2-a^2=0$$ We can then factor this polynomial as $$(x-a)(x+a)=0$$ Thus $x=a$ or $x=-a$.

Thus in the group of real numbers $\mathbb{R}\setminus\{0\}$ with multiplication we see that this claim is not true.

In particular, $x^2=4=2^2$ has two solutions: $x=\pm 2$.

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Look for a group with two distinct elements $a, b$ of order $2$.

Then $a^2 = b^2 = e$, but $a \neq b$.

For example, in the symmetric group $S_4$, we have $a = (1, 2), \;\;b = (1,3)(2,4)$. Each, when composed with itself yields $a^2 = b^2 = e$.

More immediately, in the Klein-4 group (order $4$) $\{e, a, b, c\}$, we have $a^2 = b^2 = e$, but $a\neq b$

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Consider the group of the symmetries in the plane. There are infinitely many of these symmetries.

Take any of them $s$. We have that $s^2=i^2$ ($i$ is the identity). "Therefore", $s=i$. That is, every symmetry is the identity.

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Noting all the other proofs that it is not a general result, your argument holds for odd cyclic groups (eg. $C_3, C_5$), where every square is distinct.

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Just think $(R,*)$ as a counter-example. $(R,*)$ is a Group. then $2^2={(-2)}^2$, however $2 \neq -2$.

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If you press twice the swith of the lamp in your bedroom, you get the same as if you do the same in the kitchen: nothing. (Ignore the cost of electricity).

But that does not mean that the lamp of your room and the lamp of your kitchen are the same lamp!

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Cancellation is not possible, you must take the root of both sides. Square roots have the property $q = \pm\sqrt{q^2}$, therefore $a = \pm x$, which is not equivalent to $a = x$. The original statement has thus been disproved.

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The free product $\Bbb Z_2\ast\Bbb Z_2$ of $\Bbb Z_2$ with itself has as a presentation $$\langle a, x\mid a^2=x^2=e\rangle,$$ from which it is clear that, if $x=a$, then the presentation becomes $$\langle a\mid a^2\rangle,$$ a presentation for $\Bbb Z_2$, implying $\Bbb Z_2\ast\Bbb Z_2\cong \Bbb Z_2,$ a contradiction.

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You cannot cancel in the way you do, as you can only cancel the same element. However, what you can do is transform the equation to get more of an insight into what is needed for this to happen.

Note $a^2 = x^2$ is equivalent to $a^2 x^{-2} = e_G$ and further to $(ax^{-1})^2= e_G$.

Now $a=x$ is equivalent to $ax^{-1}=e_G$. So the question reduces to answering if there can be a group with an element $b \neq e_G$ such that $b^2= e_G$.

The answer to which is: yes. I assume you can find an explicit example.

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