Let $ε$ be the identity of $G$. Prove for $g\in G$ s.t. $g^5=ε$, either $g=ε$, or $ε, g, g^2, g^3, g^4$ are distinct.

Question

Let $\varepsilon$ be the identity of a group $G$. Prove that if $g\in G$ and $g^5=\varepsilon$, then either $g=\varepsilon$ or $\varepsilon, g, g^2, g^3, g^4$ are distinct.

I don't necessarily want an entire proof, I just don't know how to get started with this proof and the direction I should take to complete the proof.

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Jan 27 '21 at 21:58
If $g$ is not the identity, and two of the given elements are the same, show that $g^r$ is the identity for some $1\le r\le 4$ using the group rules. — Mark Bennet, Jan 27 '21 at 21:58
You don't need Lagrange's Theorem, it's completely elementary. — Derek Holt, Jan 27 '21 at 22:14
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score 1 · Answer 1 · answered Jan 27 '21 at 23:18

Suppose $\epsilon, g, g^2, g^3, g^4$ are not unique and prove that this implies $g=\epsilon$.

Let's look at an example of how this argument would go, say $g^2=g^4$. Denote this common element by $x$. Then, $$\epsilon=xx^{-1}=(g^4)(g^2)^{-1}=g^2.$$ Now you know that $g^2=g^4=\epsilon$, so $\epsilon=g^5=g^4g=\epsilon g=g$.

Adapt this argument to the case where $g^k=g^\ell$ for $0\leq k<\ell\leq 4$.

Let $ε$ be the identity of $G$. Prove for $g\in G$ s.t. $g^5=ε$, either $g=ε$, or $ε, g, g^2, g^3, g^4$ are distinct.

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