I am having trouble solving this basic question in Abstract Algebra.

If $a^{24}=e$ for some group $G$, what are the possible orders of $a$?

This is in the chapter on Cyclic Groups, so I assume that $G$ is cyclic. I first started looking for an answer using a theorem from my textbook:

Theorem: Let $G$ be a cyclic group of order $n$ and suppose that $a\in G$ is a generator of the group. If $b= a^{k},$ then the order of $b$ is $n/d$ where $d= \gcd (k,n)$.

I let $b= a ^{24}$ and saw that $\gcd (24,n)=d$.

This got me nowhere, because $n$ is not given. I then realized that 24 can be factored as $3\cdot 2^3$, giving $a^{3\cdot 2^3}$. I figured that the order of $a$ must be 3 or 2. I don't know if this is right, and if it is I have no clue on how to prove that's the case. I also figured that maybe the order of $a$ would be some combination of the two, as in the following:

2, 2*2, 2*2*2, 3, 2*3, 2*2*3, 2*2*2*3.

I do not know how to approach the problem, because no matter what I do I get stuck.

I especially do not know how to solve this in general, for non cyclic groups.

Any help is appreciated, and I would really like a step by step walkthrough on how these types of problems are tackled.


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    $\begingroup$ If $a^{24}=e$, then the order of $a$ divides $24$ $\endgroup$ – J. W. Tanner Jan 19 at 2:50
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    $\begingroup$ The proofs in the dupe are given in $\Bbb Z_n^*$ but they work in any group (here we only need to work in the cyclic (sub)group generated by $a)\ \ $ $\endgroup$ – Bill Dubuque Jan 19 at 5:09

$a^{24}=e$. This means:

$a$ generated a group of order $24$.

Or this could imply that $(a^{2})^{12}=a^{24}=e$ which means that it could be that $a^2=e$ so $a$ generated a group of order $2$ (or a subgroup).

Or this could be $a^4=e$ because $(a^4)^6=e$ so $a$ has order four in this case.

Same goes for all the other divisors of $24$.

Note that your group doesn’t have to be cyclic at all.

  • $\begingroup$ Thank you. I guess I was on the right track but didn't realize it. $\endgroup$ – shm614 Jan 19 at 3:03

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