# Understanding subgroups and how to (always) generate them.

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?

Footnote:

$$_0$$. 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

• There is an error in your computation of $H_6$. Jul 4, 2022 at 12:17
• 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 $g^2=(132)$. Jul 4, 2022 at 12:18
• @quanticbolt can you please direct me with where the error is? I just tried again with $a^n$ and I got the same answer. Jul 4, 2022 at 12:22
• 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. Jul 4, 2022 at 12:31
• For $S_3$ the operation is composition of permutations. With $\phi(n)$ I mean Euler's totient function. So $\phi(3)=2$. Jul 4, 2022 at 12:54

Notice that if $$H$$ is a proper subgroup of an arbitrary group $$G$$, then $$G=\langle G \setminus H \rangle$$.
• This is true for cyclic $G$. Jul 4, 2022 at 19:42
• Oh yeah. Then $|G\setminus H|\ge1/2|G|$. So Lagrange finishes it. Since we can get $e$. Jul 4, 2022 at 21:04