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With just being done covering groups, there was an exercise in my class notes that we didn't get the chance to cover:

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My instructor gave us sample sequences to test to see if they can occur or not, so I looked at one:

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So I was trying to wrap my head around the sequence of a increasing to the sixth power and viewed them as orders, from order 1 to 6. Is there something I'm overlooking? I am kind of lost as to how someone would go about getting a sequence based on the orders of a and a binary operation.

Thanks for reading and helping!

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    Like in your question [before](http://math.stackexchange.com/questions/2122878/understanding-group-orders), with $a^2=e$ we now have $a^5=e$ for any element of a group with $5$ elements, with neutral element $e$. So $a^6=a$. – 2017-01-31

2 Answers 2

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I'm assuming you mean that in chronological order, these sequences are the same. That is to say:

$a=a$,

$a^2=b$,

$a^3=c$,

$a^4=e$,

$a^5=a$,

$a^6=d$.

From this, we get a series of equalities: $d=a^6=a^4a^2=ea^2=a^2=b$. Given that, what can you conclude about the sequence a, b, c, e, a, d?

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    That is an interesting approach.Looking at Will's reason that this sequence doesn't work since it would imply that $a^4 = e$, so to conclude on the sequence, since d = b, this invalidates the sequence? – 2017-02-01
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    You said all the elements $\{a, b, c, d, e\}$ are distinct. That is to say, $b \neq d$. So the proposed sequence is impossible because it gives rise to a contradiction, namely that $b=d$, and $b \neq d$. – 2017-02-01
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    Ah okay, thanks for clarifying! If you don't mind, I'd like to check two more sequences if you have the time :) for $a,b,c,e,a,e$, I found that e = b using your method. As e is the identity, would it still make it distinct? For $a,b,c,b,c,b$, I determined this one was distinct as I couldn't find a way to find one element equal to another – 2017-02-01
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    Since all $\{a,b,c,d,e\}$ are distinct, every element is not equal. So in particular, $b \neq e$. So if that first sequence implies $b=e$, then that sequence is impossible. For the second sequence, you get: $a^2 = a^4$, so $a^{-1}a^2= a^{-1}a^4$, which implies $a=a^3=c$. Thus $a=c$, which is also contradicts the elements being distinct. – 2017-02-01
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    Thanks for the feedback! Really appreciate it! – 2017-02-01
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I'm not quite sure what you are looking for in an answer, but I can give you some guidance and information that may be useful.

One thing to consider is that the order of $a$ must divide the order of the group by Lagrange's theorem. In your case, the order of the group is $5$, so the order of $a$ must be $1$ or $5$. The order of $a$ is not $1$ because $a\not= e$, so the order of $a$ is $5$. Therefore the example sequence your instructor gave you is impossible since it would imply that $a^4 = e$. I could say some more, but you can probably figure the rest out on your own. Hope this helps.

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    Thanks for the tips Will. As I'm still pretty new to group orders, why is it impossible for $a^4 = e$? As I can see from TASPlasma, d ended up being equal to b. Does it involve any possible sequence breaking? – 2017-02-01
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    Maybe you haven't learned Langrange's theorem yet? Langrange's theorem says that the size of a subgroup of a group divides the size of the group. The order of an element in a group is the same as the size of the subgroup generated by that element. The group you have has size 5, so the only possible orders for elements are 1 and 5. If an element has order 1, then it is the identity. – 2017-02-03