# If $a^{24}=e$ in a group $G$, what are the possible orders of $a$? [duplicate]

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.

Thanks!

• If $a^{24}=e$, then the order of $a$ divides $24$ – J. W. Tanner Jan 19 at 2:50
• 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)\ \$ – 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.

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