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

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 $$1.$$

• 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. Commented Aug 19, 2023 at 14:20
• By definition $a$ has order $1$ means $a^1 = a = e$, this does not really require a proof? Commented Aug 19, 2023 at 14:21
• It is always true that $a = a^1$. If $a$ has order 1, then $a^1=e$. By transitivity of equality, it follows that $a=e$. Commented Aug 19, 2023 at 14:29
• After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. Commented Aug 19, 2023 at 14:58

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.

• 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? Commented Aug 19, 2023 at 15:09
• Yes, that is correct.
– Javi
Commented Aug 21, 2023 at 14:23

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 $$a=a^1=e$$.