1. ## Generated subgroup

Hi,

So... I've decided to study some algebra, to see if I can handle it... lol

I've come accross this (very basic) question, which may rather be set theory, but the original question is algebra :
Let G be a group. Let H be a subgroup containing a subset P.
Prove that $\langle P\rangle_H=\langle P\rangle_G$

So basically, I have to prove that $\bigcap_{I\in\mathcal{I}} I=\bigcap_{H\in\mathcal{H}} H$

Where $\mathcal{I}=\{\text{subgroups of H containing P}\}$ and $\mathcal{H}=\{\text{subgroups of G containing P}\}$

It is clear that $\mathcal{I}\subseteq \mathcal{H}$

So it is very logical to have the equality $\bigcap_{I\in\mathcal{I}} I=\bigcap_{H\in\mathcal{H}} H$

But I didn't succeed in finding a formal proof

Thanks,

2. Originally Posted by Moo
Hi,

So... I've decided to study some algebra, to see if I can handle it... lol

I've come accross this (very basic) question, which may rather be set theory, but the original question is algebra :
Let G be a group. Let H be a subgroup containing a subset P.
Prove that $\langle P\rangle_H=\langle P\rangle_G$

So basically, I have to prove that $\bigcap_{I\in\mathcal{I}} I=\bigcap_{H\in\mathcal{H}} H$

Where $\mathcal{I}=\{\text{subgroups of H containing P}\}$ and $\mathcal{H}=\{\text{subgroups of G containing P}\}$

It is clear that $\mathcal{I}\subseteq \mathcal{H}$

So it is very logical to have the equality $\bigcap_{I\in\mathcal{I}} I=\bigcap_{H\in\mathcal{H}} H$

But I didn't succeed in finding a formal proof

Thanks,
I'll try not to take the set theory approach...

Consider the following.

If $H\leq G$ and $P\subseteq H$, then as you've observed, $P\subseteq H\subseteq G$.

If I understand correctly, $\left_H=\{p\in P^n=e_H\}" alt="\left<P\right>_H=\{p\in P^n=e_H\}" />. Since $H\leq G, e_H=e_G$. So $\left_H=\left_G$

This seems a bit hard to find a really formal proof since its a pretty straightforward concept...I'm not sure if this is what you're looking for.. xD

3. It's actually not much work to get the result, and perhaps easier to just use minimality as opposed to the intersections. In other words, I just use the fact that the subgroup generated by a set is a subgroup of any group containing the set.

Since $\langle P \rangle_H$ is the smallest subgroup of $H$ containing $P$, it is a subgroup of $G$ containing $P$, and thus by minimality of $\langle P \rangle_G$ in G we must have $\langle P \rangle_G \leq \langle P \rangle_H$. Similarly, since $P\subset H$, by minimality in $G$ we have $\langle P \rangle_G\leq H$, so by minimality in $H$ we have $\langle P \rangle_H\leq \langle P \rangle_G$ thus equality!

4. Originally Posted by Chris L T521
I'll try not to take the set theory approach...

Consider the following.

If $H\leq G$ and $P\subseteq H$, then as you've observed, $P\subseteq H\subseteq G$.

If I understand correctly, $\left_H=\{p\in P^n=e_H\}" alt="\left<P\right>_H=\{p\in P^n=e_H\}" />. Since $H\leq G, e_H=e_G$. So $\left_H=\left_G$

This seems a bit hard to find a really formal proof since its a pretty straightforward concept...I'm not sure if this is what you're looking for.. xD
Hmm no it's not, sorry

I'm technically now allowed to use generators (p), because it was further in the book

Originally Posted by siclar
It's actually not much work to get the result, and perhaps easier to just use minimality as opposed to the intersections. In other words, I just use the fact that the subgroup generated by a set is a subgroup of any group containing the set.

Since $\langle P \rangle_H$ is the smallest subgroup of $H$ containing $P$, it is a subgroup of $G$ containing $P$, and thus by minimality of $\langle P \rangle_G$ in G we must have $\langle P \rangle_G \leq \langle P \rangle_H$. Similarly, since $P\subset H$, by minimality in $G$ we have $\langle P \rangle_G\leq H$, so by minimality in $H$ we have $\langle P \rangle_H\leq \langle P \rangle_G$ thus equality!
Okay, got it ! Thanks

Now, I'd like to know if there's a formal proof for the set theory problem generated by this question

5. I'm not sure what you mean by the set theory part. Is it the implication that the intersection of a subcollection of a collection of sets is equal to the intersection of the collection? That is not necessarily true. As a silly example take the sets {1,2} and {1}. Then the intersection over {1,2} is not equal to the intersection over {1,2} and {1}, you see what I mean?

If this is not the question though, please clarify

6. Originally Posted by siclar
I'm not sure what you mean by the set theory part. Is it the implication that the intersection of a subcollection of a collection of sets is equal to the intersection of the collection? That is not necessarily true. As a silly example take the sets {1,2} and {1}. Then the intersection over {1,2} is not equal to the intersection over {1,2} and {1}, you see what I mean?

If this is not the question though, please clarify
Okay, I see your point !
My intuition was completely misleading me... no wonder I didn't succeed in proving a false thing
I hope I understood my own problem, actually... XD

Thanks !

Oh, another question... How can one prove that $\langle \emptyset\rangle_G=e_G$ ? I'm having trouble understanding this...
Because I don't know how to interprete that a subgroup contains $\emptyset$

7. From the minimality condition, what is the smallest subgroup containing the empty set? Well, as all sets contain the empty set, this is simply the smallest subgroup which is always just the identity, the trivial subgroup!

Oh as to "containing the empty set", this is vacuously always true. The subset statement is "If x is in A, then x is in B" for A being a subset of B. Perhaps its easier to understand from the contrapositive: "If x is not in B, then x is not in A". If A is the empty set, then the "then" part is always true hence the empty set always is a subset. I hope this clarifies things