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An element in a group has order $1$ iff it is identity.

I want to know if this proof is correct or not,

Let $e$ be an identity of a group $(G,*)$ with $e^1=e$. Now assume that that there exists $a\in (G, *)$ such that $a\neq e$ and $a^1=a\neq e$. Now our assumption implies that $a$ is another identity element of the group $G$ because $a^1=a$ and also due to the definition of the order of the element. But we know that the identity of any group is unique which implies that $a=e$. Hence our assumption is wrong and there doesn't exists such element other than identity which have order $1.$

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  • $\begingroup$ Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read the title. Besides, maybe that you could add the solution-verification tag to your question. $\endgroup$ Commented Aug 19, 2023 at 14:20
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    $\begingroup$ By definition $a$ has order $1$ means $a^1 = a = e$, this does not really require a proof? $\endgroup$ Commented Aug 19, 2023 at 14:21
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    $\begingroup$ It is always true that $a = a^1$. If $a$ has order 1, then $a^1=e$. By transitivity of equality, it follows that $a=e$. $\endgroup$ Commented Aug 19, 2023 at 14:29
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    – Shaun
    Commented Aug 19, 2023 at 14:58

2 Answers 2

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Your proof is not correct.

First of all, if you want to show that having order one implies being the identity, you should start with an element $a$ such that $a^1=e$ (it has order 1), because you want to prove that such $a$ is the identity. Instead, you are taking $a$ such that $a^1\neq e$, which is not what you need.

Then when you say

Our assumption implies that $a$ is another identity element of the group $G$ because $a^1=a$ and also due to the definition of the order of the element.

If by "our assumption" you mean the fact that $a^1\neq e$ (which is what you wrote), then this implication is false. If by "our assumption" you mean the fact that $a^1=e$, then you just need the definition of power, since $e=a^1=a$.

It seems that you are trying a proof by contradiction, which is unnecessarily cumbersome, but if you want to do it you should start by asuming that $a$ is such that $a^1=e$ and $a\neq e$. Then the contradiction is really tautological, because from $e=a^1=a$ you have $a=e$, which contradicts $a\neq e$. And I think it is really obvious at this point that you don't need a contradiction because you get $a=e$ before the contradiction, which is what you want to show.

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  • $\begingroup$ So in conclusion $\forall$ a$\in$G :a$\neq$e: $a^1$=a$\neq$e implies $a^1$$\neq$e i.e there doesn't exist any element other than identity which have order 1, is this correct? $\endgroup$ Commented Aug 19, 2023 at 15:09
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    $\begingroup$ Yes, that is correct. $\endgroup$
    – Javi
    Commented Aug 21, 2023 at 14:23
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Suppose $a=e$. Then $a^1=a=e.$ Since $1$ is the smallest positive integer, this must mean the order of $a=e$ is $1$.

Now suppose the order of $a$ is $1$. Then $a^1=e$. Then $a=a^1=e$.

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