Question on Cyclic Subgroups

Question

Suppose we have a finite group, $G$ with some element, $g$. Is it plausible to say that, $\forall g \in G$, there exists a cyclic subgroup, $\langle g\rangle$, of $G$?

I know, by definition, what $\langle g\rangle$ amounts to and that it's order is identical to that of the element $g$. Is this necessarily a group for any given group element? And, for example, is the group is trivially the identity, would we simply assert that $\langle g\rangle$ is $\langle e\rangle$?

It seems to me that the properties are satisfied. It's surely closed, as $g \in G$ and $G$ is closed per the definition of a group. We can make a similar argument for associativity. The identity is the element $g^0 = 1$, which is surely in this group. The only property I struggle with is inverses. Surely, if $<g>$ has order $m$, then $g^m = 1$. I believe there's a plausible proof outline I 've seen for this wherein we 'multiply' (or apply the group operation) on the left by the inverse of $g$, which exists by the definition of the subgroup (and sine $g^m = g^{m-1} g = 1$, so $g^{m-1}$ is by definition the inverse of $g$. Clearly, $m \geq 0$ as it represents a cardinality. If the subgroup is trivially the identity, then $m = 0$, so $g^{m-1} = g^{1-1} = g^0 = 1$. If the subgroup is not trivial, $m - 1 > 1$. From here, I'm a bit lost on whee to continue.

So, first, is this notion correct? Second, how do we verify that an inverse exists for each element in the subgroup?

Thanks.

Answer 1

When you use the notation $\langle S \rangle$, you're referring to the smallest subgroup $K$ of $G$ that is generated by $S$. So, what you're looking for is in fact right there in the definition.

$\langle S \rangle$ means that you first consider all subgroups of $G$ containing $S$ and then you take the intersection over all of them. You can see that the intersection of subgroups is always a new subgroup and of course, since all of them contain $S$, their intersection contains $S$ as well. Therefore, $\langle S \rangle$ is indeed a subgroup of $G$ and this is why it's called the 'subgroup' generated by $G$.

$$\langle S \rangle = \bigcap\{K: \text{$K$ is a subgroup of $G$ and $S \subseteq K$}\}$$

You can even represent it nicely like this:

$$\langle S \rangle = \{ g_1^{a_1}\star g_2^{a_2}\star \cdots \star g_k^{a_k} : g_i \in S, a_i \in \mathbb{Z} \text{ and } k\geq 0\}$$

Convince yourself that when $G$ is not finite, $k$ can vary depending on the element you choose. If we can always choose a finite number of elements, i.e. $k$ in my notation has a maximum, we say that $G$ is finitely generated. Particularly, finite groups are finitely-generated and $k \leq |G|$.

Also, remember that an element $h \in \langle S \rangle$ can have more than one such representation. So, we do not have uniqueness of representation in general.