Vtgyjfy

Im trying to prove theory:
Let x be subset of G. There is a group generated by x which is the intersection of all subgroups of G containing x.

Its not easy for me to do this at all.
I tried many different routes but they seem to go nowhere.
Thanks in advance for anyone who is able to help me proving that theory.

The first thing that's helpful to notice is that, if $A$ and $B$ are subgroups of some group $G$, then so is $Acap B$.

(you can skip down to the next paragraph if you're already familiar with the proof) This is because, firstly, if $ain Acap B$, then $ain A$ and $ain B$. But $A$ and $B$ are themselves groups, so it must be the case that $a^{-1}in A$ and $a^{-1}in B$ both also hold, i.e. $a^{-1}in Acap B$. So $Acap B$ is closed with respect to inverses. And, secondly, if $y$ and $z$ are in $Acap B$, then $y$ and $z$ are in $A$ and $y$ and $z$ are in $B$. Again, since $A$ and $B$ are subgroups, $yzin A$ and $yzin B$ both must hold so $yzin Acap B$ must also hold, i.e. $Acap B$ is closed with respect to products. Finally, since both $A$ and $B$ are subgroups, $e$ (the identity of $G$) is $in A$ and $in B$, so $ein Acap B$, making $Acap B$ non-empty. Combining these three facts, we see that, if $A$ and $B$ are subgroups of $G$, $Acap B$ is also a subgroup of $G$.

We can extend this line of reasoning to any arbitrary number of subgroups so that, if $A_1,A_2,...,$ and $A_n$ is some list of $n$ subgroups of $G$, then $A_1cap A_2cap...cap A_n$ is also a subgroup of $G$.

So the intersection of subgroups of the $G$ in your question that all contain the set $x$ (from your question) will produce a subgroup containing $x$. Since a subgroup is nothing but a group that is a subset of another group (G, in this case), this is a group containing $x$.

Denote this group by $I$ and let $I=bigcap_{s=1}^r i_s$, where ${i_1,i_2,...,i_r}$ is the set of all subgroups containing $x$.

Now, why is this group that contains $x$ generated by $x$?

Let $x={x_1,x_2,...,x_m}$ and denote the group generated by $x$ by $X$.

On one hand, given that all groups are closed with respect to inverses and multiplication, any group, $i_s$ containing $x$ would have to contain $X$. This is because, since $xsubset i_s$, (by definition) $x_1,x_2,...,x_min i_s$ so every possible product with elements of $x$ and their inverses (like $x_2^5x_7^{-3}x_3$) must be in $i_s$, i.e. $X$ must be in $i_s$. So $Xsubset i_s$ for each $k$ meaning $Xsubset bigcap_{s=1}^r i_s=I$.

On the other hand, notice that that $X$ is, well, a subgroup of $G$ (because each element in $X$ is of the form $x_1^{k_1}x_2^{k_2}...x_m^{k_m}$, the product of any two such elements or the inverse of any such element will also be of this form). And, this ($X$) is a subgroup that contains $x$. So $X$ is in fact equal to one of the $i_s$ (as $X$ is a subgroup of $G$ containing $x$). Since, in general, $Asupset Acap B;;$, $Xsupsetbigcap_{s=1}^r i_s=I$.

In the end, we have $Xsubset I$ and $Xsupset I$ which give us $X=I$.