Does a finite group always contains subgroups generated by its elements Does a finite group contains subgroups generated by its elements. 
I suspect that answer is Yes and  that my question could be trivial. But I am studying algebra for the first time so I wanted to be sure. I would really appreciate your input.
While proving that a group of prime order is cyclic (2.2.4 page 60 Herstein) it has been mentioned that if $a\in G$, $a\neq e$ then the power of $a$ forms a subgroup $(a)$ of $G$ different than $(e)$. And then the the proof is completed easily.  
I agree that the power of $a$ forms a cyclic group $(a)$ but what is the guarantee that $G$ will always contain such a subgroup. 
May be I should edit my question little bit now in light of your answer. If $a \in  G$ then what is the guarantee that a cyclic subgroup $(a)$ exist. Could not " $a$ " be an element of $G$ but not generate any group. Again it could be a trivial thing to ask but better safe than sorry.
 A: Closure under the group operation of $G$ ensures that if $a\in G$, then $a^n\in G$ for all positive integers $n$. Since $a^{-1}\in G$, again by closure we have $(a^{-1})^n \in G$ for all positive integers $n$. Since $(a^{-1})^n = (a^n)^{-1}$ for all $n\in \Bbb N$ (which can be shown by induction), we have that $a^n\in G$ whenever $n$ is a negative integer. So $a^n \in G$ for all $n\in \Bbb N$. Since $G$ contains all powers of $a$, $G$ contains $(a)$.
A: Any collection of elements in a group, finite or infinite, always generates some subgroup. This is in particular always true for a single element of the group. Sometimes, however, the generated subgroup may be the entire group. (But any group is a subgroup of itself, so this is not a contradiction).
To see this, we first need a definition of what it means to generate a subgroup. Typically it would be something like this:

Let $G$ be a group, and let $A\subseteq G$ be any set of elements of $G$ (in particular $A$ may consist of a single element, $A=\{a\}$). We say that the subgroup $H$ of $G$ is generated by $A$ iff

*

*$A\subseteq H$, that is, every element of $A$ is in $H$, and

*For every subgroup $K$ such that $A\subseteq K$, it holds that $H\subseteq K$
(in other words, $H$ must be the smallest subgroup that contains $A$).

With this definition it clear that $A$ generates at most one subgroup (because if $H_1$ and $H_2$ both satisfy the condition, then we must have $H_1\subseteq H_2$ and $H_2\subseteq H_1$, which together mean that $H_1=H_2$). But it is not immediately obvious that every $A$ always generates some subgroup.
In order to prove that it does, we can first prove a lemma: Let $S$ be any collection of subgroups of $G$. Then the set
$$ \bigcap S = \{ x\in G \mid (\forall H\in S) x\in H \}$$
is a subgroup of $G$. This is fairly easy to show -- for example, if $a$ and $b$ are both in $\bigcap S$, then it must be because $a$ and $b$ are both in each subgroup that is in $S$. But then, by definition of subgroup, $ab$ must be in each of the subgroups in $S$, too, so $ab\in\bigcap S$.
Now, given, $A$ apply the lemma with $S$ being the collection of all subgroups that has $A$ as a subset. I claim that $\bigcap S$ is generated by $A$. We know that $\bigcap S$ is a subgroup. Additionally for every $a\in A$, we know by construction that $a$ is in each of the subgroups in $S$, which means that $a$ is also in $\bigcap S$. So $A\subseteq \bigcap S$.
The second condition for being a generates subgroup now requires that $\bigcap S$ must be a subset of $K$ for every $K\in S$. However, this is just a consequence of what $\bigcap S$ means -- if $K$ is some member of $S$, then by definition every element of $\bigcap S$ must be in $K$.
This proves that $\bigcap S$ is a subgroup generated by $A$, as promised.
