How to efficiently find all elements in $\langle a\rangle$ for all $a$ in $\Bbb Z / 48 \Bbb Z$

Question

We were asked in my class to find all elements of $\langle a\rangle$ for all $a$ in $\Bbb Z / 48 \Bbb Z$.

The answer can be found here, and I understand how the cyclic groups are formed. Rather than tediously writing out all 48 cyclic groups, I am wondering what the pattern is, if there is one. I see that the group $\Bbb Z/48\Bbb Z$ is generated by $\langle1\rangle,\langle5\rangle,\langle7\rangle,\langle11\rangle,\langle13\rangle,\langle17\rangle,\langle19\rangle,\langle23\rangle,\langle25\rangle$, etc.

I know that $\langle a\rangle=\langle a^{-1}\rangle$, so $\langle 1\rangle = \langle 47\rangle, \langle 2\rangle = \langle 46\rangle...$

I feel like a pattern is glaring right at me, but I would love to hear from others! Thanks.

Answer 1

It all depends on whether the number is coprime to $48$. Any number coprime to $48$ will generate the whole group. Any number not coprime to $48$ will generate only multiples of the $\gcd$ of the number and $48$.

Answer 2

One can use the free software GAP to compute the elements of $\langle a\rangle$ for each $a\in\Bbb Z/48\Bbb Z$.

F:=FreeGroup(1);
rels:=[(F.1)^48];
G:=F/rels;
C:=[];
for a in G do
  H:=Subgroup(G, [a]);
  AddSet(C, H);
od;
Print(C);

Answer 3

The number of subgroups of $\mathbb{Z}/48\mathbb{Z}$ is the same as the number of divisors of $48=2^4 \cdot 3$, which is $(4+1)(1+1)=(5)(2)=10$.

The entire group is generated by $a$ for $a \in \{1,5,7,11,13,17,19,23,25,29,31,35,37,41,43,47\}$.

The subgroup $\{0,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46\}$ is generated by $a$ for $a \in \{2, 10,14,22,26,34,38,46\}$.

The subgroup $\{0,3,6,9,12,15,18,21,24,27,30,33,36,39,42,45\}$ is generated by $a$ for $a \in \{3,9,15,21,27,33,39,45\}$.

The subgroup $\{0,4,8,12,16,20,24,28,32,36,40,44\}$ is generated by $a$ for $a \in \{4,20,28,44\}$.

The subgroup $\{0,6,12,18,24,30,36,42\}$ is generated by $a$ for $a \in \{6,18,30,42\}$.

The subgroup $\{0,8,16,24,32,40\}$ is generated by $a$ for $a \in \{8,40\}$.

The subgroup $\{0,12,24,36\}$ is generated by $a$ for $a \in \{12,36\}$.

The subgroup $\{0,16,32\}$ is generated by $a$ for $a \in \{16,32\}$.

The subgroup $\{0,24\}$ is generated only by $24$.

Finally, the trivial subgroup $\{0\}$ is generated only by $0$.