# How can there be an element in a group such that $a^2 = e$, but $a \neq e$?

I'm rather confused. Two of the questions I'm on assume two elements such that $a^2 = b^2 = e$ (identity), but $a \not= b$. Isn't this impossible? Shouldn't they both be the identity?

If $a^2 = e$, then $a = a^{-1}$. Same for $b$.

Then $aabb=bbaa=e$. Then $bb=(aa)^{-1} = a^{-1}a^{-1} = aa$, so $a = b = e$.

No?

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No. $aa=bb$ does not imply that $a=b$. Consider the case with $a$ and $b$ being reflections in the Euclidean plane across different lines. –  George Lowther Sep 29 '11 at 0:15
A similar question is "Are there non-trivial elements such that a^2=a?" The answer to this is no, as $a=a^2\Rightarrow a^{-1}a=a^{-1}a^2\Rightarrow 1=a$. –  user1729 Sep 29 '11 at 11:00

No. $bb=aa$ is the same thing as $b^2=a^2$, which is the same thing as $e=e$ by the assumption that $a^2=b^2=e$. The statement that $e=e$ certainly is true, but does not imply that $a=b$ :)

For example, in $\mathbb{R}^\times$ under multiplication, $1^2=(-1)^2=1$, but $1\neq -1$.

Another example would be the group $$\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}=\{(0,0),(1,0),(0,1),(1,1)\}$$ under addition: we have that $$(1,0)+(1,0)=(0,0)=(0,1)+(0,1),$$ but $(1,0)\neq(0,1)$.

Another example would be the symmetric group $S_3$, consisting of permutations of three objects, named $1$, $2$, and $3$. Consider the permutation $\sigma=(12)$ that switches $1$ and $2$, and does nothing to $3$, and the permutation $\tau=(13)$ that switches $1$ and $3$, and does nothing to $2$. Then $\sigma^2=\tau^2=$ the identity (the "do-nothing" permutation), because if you switch two objects and switch them again they end up where they started, but $\sigma\neq\tau$.

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Oh, ok. Thanks. –  iDontKnowBetter Sep 29 '11 at 0:27
Glad to help! The argument in your post fails in the sentence "Then $bb=(aa)^{-1} = a^{-1}a^{-1} = aa$, so $a=b=e$." The first bit is correct, but there is no way to conclude that $a=b=e$ from it (it might be good to examine your thought process at this stage, and see how you thought you would be able to derive $a=b=e$). –  Zev Chonoles Sep 29 '11 at 0:33

The Klein four-group (also known as $\mathbb{Z}/ 2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$) provides an example of the situation you describe: the elements $(1,0)$ and $(0,1)$ both square to the identity but are visibly non-equal. Your mistake is in assuming that $b^2 = a^2$ implies that $a = b$. This isn't even true in more familiar circumstances: another example of a group where this fails is the nonzero reals under multiplication, as Zev points out in his answer.

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However, in a finite group of odd order your statement holds true: then $a^2=b^2$ implies $a=b$.

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