Proving that $\langle{M}\rangle$ is a subgroup of $G$ I found a task suggestion in my lecture book that I unfortunately can't properly solve. It translates into:
$M$ is a non-empty subset of the group $(G,*)$. Prove that $\langle{M}\rangle$ is a subgroup of $G$.
It looks like something extremely trivial but I haven't encountered generators so far and just started learning to write proofs which I am still terrible at.
From what I know the generator's elements look like $m_1*m_2*\cdots*m_n$ (abbreviated) and if $M$ is a subset of $G$, every element of $M$ - including the neutral element $e$ of $G$ - is also in $G$. I can already pick up the scent of a solutions but I'm not sure where to go from here and how to formulate these thoughts in proper "math speak".
I appreciate any help.
EDIT: The definition of $\langle{M}\rangle$ is hard to translate but I will try: 
$\langle{M}\rangle$ is called the subgroup (generated by $M$) of $G$.
Cheers
 A: Use the following definition
$$\langle M\rangle:=\{m_1^{\epsilon_1}\cdot\ldots\cdot m_n^{\epsilon_n}:n\in \mathbb{N},m_j\in M,\epsilon_j=\pm 1\}$$
1) neutral element: you get it for $n=0$
2) closure: product of two elements of the form $m_1^{\epsilon_1}\ldots m_n^{\epsilon_n}$ is again of the same form, hence is in $\langle M\rangle$
3) the inverse of $m_1^{\epsilon_1}\ldots m_n^{\epsilon_n}$ is $m_n^{-\epsilon_n}\ldots m_1^{-\epsilon_1}$, which is clearly of the right form to be an element of $\langle M\rangle$
This proves that $\langle M\rangle$ is a group (a subgroup of $G$). Now consider the intersection of all subgroups $H$ of $G$ containing the subset $M$ of $G$:
$$I:=\displaystyle\bigcap_{H\supseteq M}H$$
Every subgroup $H$ of $G$ containing $M$ needs to contain also inverses of elements of $M$ and their products, hence $I\supseteq \langle M\rangle$. Conversely, we proved that $\langle M\rangle$ is a subgroup of $G$, (obviously containing $M$) hence $\langle M\rangle$ is one of the $H$'s above, so that $I\subseteq \langle M\rangle$. So $I=\langle M\rangle$, thus $\langle M\rangle$ is the smallest subgroup of $G$ containing $M$
A: You need to show three things:


*

*The identity is in $\langle M \rangle$.  Easy enough if $M$ is not empty.

*The inverse of every element of $\langle M \rangle$ is in $\langle M \rangle$.    Write an arbitrary element and then write its inverse.  Check that its inverse can be generated.

*$\langle M \rangle$ is closed under the group operation.  You get this one for free from the definition of generators.

