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I was given the following question:

Suppose that $G = {a, b, c, d, e}$ has some binary operation in which it is a group, with e being the identity. (The five elements are all distinct.)

We look at the sequence $ \{ a, a^2 , a^3 , a^4 , a^5 , a^6 \} $ For the following sequence below where each element corresponds to $ \{ a, a^2 ,... $ decide whether or not it can occur and give a reason.

$a=a, a^2 =b,a^3 = c,a^4 = b, a^5 =c, a^6 =b$

It actually had several different lines that were fairly easy to disprove. i feel like the order of an element shouldn't be bigger than G but other than im that im not sure how to disprove this but im also not able to think of an example...

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    Hint: what you have written would imply that $a^2=a^4$. What would that imply? – 2017-01-24
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    yeah and $a^3 = a^5 $ – 2017-01-24
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    Right. So what does that tell you about $a^2$? – 2017-01-24
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    sorry, but to be honest your question seems incomprehensible in my head. Can you write it maybe a bit better? – 2017-01-24
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    Uh i'll try to rewrite it – 2017-01-24
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    Your group is $\mathbb{Z}_5$ so I think in any case you can work through in a more intuitive example to sort out your query! – 2017-01-24
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    Perhaps that $e=a^2$ if we multiplied by $a^{-1}$ twice – 2017-01-24
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    Exactly. So we have $e=a^2$, but we already had $b=a^2$ so... – 2017-01-24

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Note that $c=a^5=a^2\cdot a^3=b\cdot c$

Now in a group that implies something about $b$. Which element is the identity?

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    b or $a^2$ must be the identity but e is the only identity – 2017-01-24
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    wait no why does it have to be an identity? – 2017-01-24
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    @Faust7 Depending on where you are in your group theory development, the fact that the identity is unique might be assumed. In fact identities in a group are unique, but the group axioms are sometimes written in a form in which uniqueness has to be proved. Here you have $b\cdot c=c = e\cdot c$. The existence of a (right) inverse for $c$ will give you $b=e$. It depends a bit what you are allowed to assume. – 2017-01-24
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    NB If your axioms give you only left inverses, I am sure you can work out the modifications necessary to use this fact. – 2017-01-24
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    well if b is an identity even if it was somehow unique from e i got e a =c or a=c so its pretty broken no matter what since i know c cannot equal a by the definition of the group having 5 unique elements – 2017-01-24