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Let $G$ be a group and $a ∈ G$. If $a^{12}= e$, what can we say about the order of $a$? What can we say about the order of $G$?

We know that $|a|$ divides $12$, but what can we say about the order of $G$? Would it then be true that $|G|= 12$? Why or why not?

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    $\begingroup$ It may be that $a=1$, and then not much can be said. If you stipulate that $a\neq 1$, then $a$ has order $2,3,4,6$ or $12$, and in each case $|G|$, if finite, is divisible by this number. For example, $\Bbb Q^\times$ contains $-1$ and this has order $2$, but $|\Bbb Q|$ is infinite. $\endgroup$
    – Pedro
    Dec 17, 2014 at 4:01
  • $\begingroup$ Data: there's only 5 group with 12 elements groupprops.subwiki.org/wiki/Groups_of_order_12 $\endgroup$
    – janmarqz
    Dec 17, 2014 at 16:31

2 Answers 2

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We know that the order of any subgroup must divide the order of the whole group by Lagrange's Theorem. Since we don't know the order of the element, but we do know it must divide 12, by the reasoning above, we know the order of the element must be $1,2,3,4,6,12$. So the best we can say about the group is NOTHING at all. If we know that $a\ne e$, then we could say $G=2k$ at least. And if we know the order of $a$ is $12$, then we can say the order of the group is $12k$(if the group is finite).

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    $\begingroup$ The question does not assert that $G$ has an element of order 12, but rather that it has an element of order dividing 12. It could thus have no element with order dividing twelve; or any element of order 3; or any element of order 2. In fact the statement is trivially satisfied by the identity element in every group. $\endgroup$ Dec 17, 2014 at 4:05
  • $\begingroup$ I see that, sorry. $\endgroup$
    – H_B
    Dec 17, 2014 at 4:08
  • $\begingroup$ So could we then say that |G| divides |a|? $\endgroup$
    – Kaitlyn
    Dec 17, 2014 at 4:11
  • $\begingroup$ I think that is possible for only the trivial group or a prime power cyclic group. $\endgroup$
    – H_B
    Dec 17, 2014 at 4:12
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    $\begingroup$ Not at all, @H_B :answering, answering and being wrong, correcting, etc. one can learn a lot. Don't worry about that, and I can tell you this because I've made tons of mistakes here. Just learn and correct and go on. $\endgroup$
    – Timbuc
    Dec 17, 2014 at 4:37
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The order of $a$ is at most $12$, because $a^{12}=e$ and $|a| = \min\{k \ge 1 : a^k=e\}$. In fact, the order of $a$ is a divisor of $12$, because if we use the Euclidean algorithm to divide 12 by $|a|$ we get $$12 = q|a| + r$$ with $0 \le r < |a|$. Exponentiating yields $e = a^{12} = a^{q|\,a|+r} = (a^{|\,a|})^qa^r=a^r$; since $|a|$ is the least positive exponent killing $a$, we must have $r = 0$. Thus, $12 = q|a|$ so $|a|\in\{1,2,3,4,6,12\}$.

The only thing we can say about the order of $G$ is that if $G$ is finite, then only $|a|$, not necessarily $12$, divides $|G|$. For example, $7 + 14\mathbf{Z}$ has order $2$ in the order-$14$ group $\mathbf{Z}/14\mathbf{Z}$. If $G$ is a group with order $1$, $2$, $3$, $4$, $6$, or $12$ (and there are $1 + 1 + 1 + 2 + 2 + 5 = 12$ such groups up to isomorphism) then every element $a$ in each of these groups satisfies the equation $a^{12}=e$. The additive group $\mathbf{Z}/24\mathbf{Z}$ contains an element of order $12$ (e.g. $2 + 24\mathbf{Z}$) but has order $24$. The symmetric group $S_4$ also has order $24$, but no elements of order $12$ -- the biggest possible order of a permutation of four things is $4$; note, however, that the least $n$ such that $a^n=e$ for all $a$ in $S_4$ is, in fact, $12$ -- this is called the exponent of the group. The infinite group $\mathbf{Q}/\mathbf{Z}$, written additively, has many solutions $\frac{a}{b}$ to $12\big(\frac{a}{b}+\mathbb{Z}\big) = 0+\mathbb{Z}$, like $\frac{1}{3}$ or $\frac{7}{12}$; the precise number is the length of the Farey sequence $\mathscr{F}_{12}$.

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