# Proving that identity element is the only element of a group

How to prove that the identity $e$ of a group $G$ is the only element of $G$ if $x^2=x$ for all $x\in G$.

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Multiply both sides by the inverse of $x$. – Gerry Myerson Sep 15 '13 at 11:17

since you are in a group, every element in the group has an inverse. so, you can multiply both sides of $x^2=x$ by $x^{-1}$ to get $x=e$.

However, if you also want to know why the identity element of a group is unique, then it's proved by using the definition of an identity element in a group. By definition, the identity element of a group $G$ satisfies $x*e=e*x=x$ for any $x \in G$. So, if you have two identity elements $e_1$ and $e_2$, by using this property, you'll see that $e_1 = e_1*e_2=e_2$ which proves that talking about 'the' identity element of a group is meaningful because it's unique.

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What do you mean by simplifying? I don't get your question and what that set has to do with this. – some1.new4u Sep 15 '13 at 11:37
I was given the question like this itself. I was told to simplify this part.G is a group – JOEF Sep 15 '13 at 11:41
but the set $G$ you have defined doesn't look like a group, unless you have to guess what group it is first, which is another question. Overall, there are two groups of order 4. One is $\mathbb{Z}_4$ and the other one is Klein's four group. Do you know them? You need to open a new question for that one, because it's a different question. – some1.new4u Sep 15 '13 at 11:46
The other one is the group of integers modulo 4. You might've seen it in number theory. I think you need to open a new question for that one if you need more explanations because it doesn't fit in here. – some1.new4u Sep 15 '13 at 17:38

Denoting by $\,a'\;$ the inverse of an element $\;a\in G\;$ :

$$e=xx'=(xx)x'=x(xx')=xe=x$$

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there are only the identity of a group such that it's order is $1$. Other elements' order is no less than 2, that is to say $$a^t=1(a\in G, t\geq2)$$ that means all the elements except the identity won't satisfy the equation

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