Cyclic Group Problem

Question

If G is cyclic group of 24 order, then how many element of 4 order in G? I can't understand how to find it, step by step.

Answer 1

Lemma: If $o(g)=n$, then $o(g^k)=\dfrac{n}{gcd(n,k)}$.

Assume $G=\langle g \rangle$ and $o(g)=24$, notice that every element of $G$ has the form $g^k$ for some integer $k$, by lemma, $o(g^k)=\dfrac{24}{gcd(24,k)}$. Let $\dfrac{24}{gcd(24,k)}=4$, i.e., $gcd(24,k)=6$, we can easily conclude this only holds for $k=6$ or $k=18$. Hence there is only two elements in $G$ of order $4$, $g^6,g^{18}$.

Answer 2

Hints:

1) A cyclic group of order $\;n\;$ has exactly $\;\varphi(n)\;$ generators.

(2) A cyclic group of order $\;n\;$ has exactly one unique subgroup of order $\;d\;$ for any divisor $\;d\;$ of $\;n\;$

Answer 3

Any element of order 4 will generate a cyclic group of order 4. Inside any finite cyclic group, there is a unique cyclic subgroup of any order dividing the order of the group. Therefore, there is a unique group of order 4 inside this group of order 24 and every element of order 4 is inside this subgroup. In a group of order 4, the number of elements of order 4 is $\varphi(4) = 2$. So there are precisely two elements of order 4 in a cyclic group of order 24. In the usual representation of this as $\mathbf{Z}/24\mathbf{Z}$, these would be the elements $18$ and $6$.

Answer 4

If $G=\langle a\rangle =\{a^0,a^1,a^2,\cdots,a^{23}\}$ then $$|a|=24\\|a^2|=k\to 24\mid2k\to12|k\to k=12.~\text{(because 12 is the least positive integer in this case)}\\|a^3|=k\to 24|3k\to8|k\to k=8.~\text{(because 8 is the least positive integer in this case)}\\|a^4|=k\to 24|4k\to6|k\to k=6.~\text{(because 6 is the least positive integer in this case)}\\|a^5|=k\to 24|5k,~\text{but} \gcd(24,5)=1\to 24|k\to\\ k=24.~\text{(because 24 is the least positive integer in this case)}\\ \vdots\\$$ By this mechanical way you can see why above lemma works.