Suppose that $\left | G \right | =24$ and G is cyclic. If $a^{8}\neq e$ and $a^{12}\neq e.$ Show that $\left \langle a \right \rangle=G.$

I think using the Fundamental theorem of cyclic group is essential here but I wasn't able to go far after spending a bit of time on this.


The order of subgroup of the cyclic group G are divisors k of n=24. This means the order of the subgroups are $k=\left \{ 1,2,3,4,6,8,12,24 \right \}$

Also, for each divisor k of n=24, the cyclic group has subgroup of order k; namely, $\left \langle a^{\frac{n}{k}} \right \rangle$.

The subgroups for each divisors k of n=24 are

$\left \langle a^{24} \right \rangle,\left \langle a^{12} \right \rangle,\left \langle a^{8} \right \rangle,\left \langle a^{6} \right \rangle,\left \langle a^{4} \right \rangle,\left \langle a^{3} \right \rangle,\left \langle a^{2} \right \rangle,\left \langle a^{1} \right \rangle$

Thanks in advance.

  • $\begingroup$ The \langle \rangle fairy smiles upon you! $\endgroup$ – Patrick Stevens Jun 2 '16 at 10:02

It's sufficient to prove that the order of $a$ is not any of $1,2,3,4,6,8,12$. By assumption, the order of $a$ is neither $8$ nor $12$ and obviously not $1$.

If $|a|\in \{2,3,6\}$, then it follows that $a^{12}=e$, contradictory to assumption.

If the order of $a$ is $4$, then $a^8=e$, also contrary to our assumption. Hence the order of $a$ is 24.

  • $\begingroup$ Sorry I do not understand where the order of elements came from. I am very weak in cyclic group. A more verbose explanation would be ideal. Should we not be talking about order of subgroup? Why are we talking about order of an element. $\endgroup$ – Mathematicing Jun 2 '16 at 12:16
  • $\begingroup$ The order $|a|$ of an element $a$ coincides with the order of the subgroup generated by $a$. It's defined as the least positive integer for which $a^n=e.$ $\endgroup$ – sqtrat Jun 2 '16 at 13:31
  • $\begingroup$ The fundamental theorem of cyclic group says the cyclic group has subgroup of order n. Using the theorem:$\left | \left \langle a \right \rangle \right |=\left | a \right |$ We have $\left | \left \langle a \right \rangle \right |=\left | a \right |=\left \{ 1,2,3,4,6,8,12,24 \right \}$ $\endgroup$ – Mathematicing Jun 2 '16 at 13:53
  • $\begingroup$ Does this make sense? $\endgroup$ – Mathematicing Jun 2 '16 at 13:53
  • 1
    $\begingroup$ Because then $a=a^1=e$ which implies that $a^8=e^8=e$. $\endgroup$ – sqtrat Jun 2 '16 at 14:09

Hint: We know that $|\langle a \rangle|$ divides $24$, but that it doesn't divide either $8$ or $12$.


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