In a bid to understand group theory, I would like to ask the following questions:

What ways can subgroups of a group $G$ be generated (that is guaranteed to always work)? Is it (only) by the elements of $G$ (rule of $a^n$) or by the rule of subgroups generated by subsets? By the way, it does not seem like $a^n$ always work; please see the next section.

This Wikipedia section here: https://en.wikipedia.org/wiki/Subgroup#Example:_Subgroups_of_Z8 mentions that the non-trivial subgroups of $G = \{0,4,2,6,1,5,3,7\}$ are $\{0,4\}$ and $\{0,4,2,6\}$. (Note that the operation here is addition modulo 8.)

Personally, I tried using the elements of $G$ to generate subgroups. I obtained the following (alleged$_0$) subgroups by raising each element to the power $k$; where $k$ $\in \mathbb{N}_1$ :

  • $H_0 = \{0\}$
  • $H_4 = \{4,0\}$
  • $H_2 = \{2,4,0\}$
  • $H_6 = \{6,4,0\}$
  • $H_1 = \{1, ..., 1\}$ # I kept on getting 1 without hitting a $0$ (the identity element)
  • $H_5 = \{5,1,...,5,1\}$ # I kept on getting 5,1 without hitting a $0$ (the identity element)
  • $H_3 = \{3,1,...,3,1\}$ # kept on getting 3,1 without hitting a $0$
  • $H_7 = \{7,1,...,7,1\}$ # kept on getting 7,1 without hitting a $0$ .....

In my understanding, here, every $H_a$ do not qualify as nontrivial subgroups of $G$ except $H_4$; where $a$ is each element in $G$.

If using elements of $G$ (can allegedly) generate a subgroup, why am I not able to generate the subgroup $\{0,4,2,6\}$ like in the Wikipedia example?


$_0$. Please bear with me. I say alleged because this is how I initially understood things, which seem to be false understanding. I am willing to unlearn and relearn.

$_1$. Natural numbers starting from 1

  • 1
    $\begingroup$ There is an error in your computation of $H_6$. $\endgroup$ Commented Jul 4, 2022 at 12:17
  • $\begingroup$ Title question: You can always generate a subgroup by all of its elements. For example, take a subgroup $H=\{1,(123),(132)\}$ of $S_3$. It is generated by these three elements. But since $H$ is cyclic, you can do better, with only one generator. Here there are $\phi(d)$ choices, if $d=|H|$. To be explicit, we have $2$ choices here for a generator, either $g=(123)$ or $g^2=(132)$. $\endgroup$ Commented Jul 4, 2022 at 12:18
  • $\begingroup$ @quanticbolt can you please direct me with where the error is? I just tried again with $a^n$ and I got the same answer. $\endgroup$
    – Joker
    Commented Jul 4, 2022 at 12:22
  • 3
    $\begingroup$ The group composition is addition for subgroups of $\Bbb Z/8$. So we have $na$ instead of $a^n$. So $H_6=\{6,12,18,24\}$ where $12=4$, $18=2$, $24=0$. So $H_6=\{6,4,2,0\}$ as claimed in the wikipedia link. $\endgroup$ Commented Jul 4, 2022 at 12:31
  • 1
    $\begingroup$ For $S_3$ the operation is composition of permutations. With $\phi(n)$ I mean Euler's totient function. So $\phi(3)=2$. $\endgroup$ Commented Jul 4, 2022 at 12:54

1 Answer 1


Notice that if $H$ is a proper subgroup of an arbitrary group $G$, then $G=\langle G \setminus H \rangle$.

  • $\begingroup$ This is true for cyclic $G$. $\endgroup$ Commented Jul 4, 2022 at 19:42
  • $\begingroup$ And for all groups. Not only cyclic ones. $\endgroup$ Commented Jul 4, 2022 at 20:05
  • $\begingroup$ Is this obvious? $\endgroup$ Commented Jul 4, 2022 at 20:26
  • $\begingroup$ It is a consequence of the fact that no group can be the union of two of its proper subgroups. You can find many proofs here on MathStackexange. $\endgroup$ Commented Jul 4, 2022 at 20:50
  • $\begingroup$ Oh yeah. Then $|G\setminus H|\ge1/2|G|$. So Lagrange finishes it. Since we can get $e$. $\endgroup$ Commented Jul 4, 2022 at 21:04

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