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.
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.