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I need a proof for the next statement:

let $s \subseteq A$ then $\langle s\rangle$ is the smallest subgroup of A that contains $s$. This is $\langle s\rangle$ is characterized by: if $B \le A$ such that $ s \subseteq B $ then $\langle s\rangle \le B$.

I just did:

By definition $\langle s\rangle = \bigcap_{B'\in C} B'$ where $C = \{ B': s \subseteq B' \le A \}$ clearly $\bigcap_{B' \in C}B' \subseteq B $ now since $B$ is a group and also is $\bigcap_{B'\in C}B'$ and share the same $0$ thus$\langle s\rangle \le B'$.

But my lecturer say that is missing the reciprocal to complete the proof. Say:

if $B_0 \le A$ such that $ \forall B \le A$ $s \subset B \implies s \subset B_0 \implies B_0 \le B $ then $B_0 = \langle s\rangle$

I do not see why is that needed.

Shaun
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José Marín
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  • You might want to change some of those $B$'s to another letter. You use it in two different senses. – ReverseFlowControl Apr 05 '19 at 03:21
  • Saying it is “characterized” means showing that it has that property, and that anything that has that property is “it”. (In other words, the second part of the statement is actually stronger than the first part; the first part just asks you to prove it has the property; the second asks you to prove it has the property and that anything with that property is equal to $\langle s\rangle$. Yet it is phrased as if the two statements were equivalent (because of the”this is”), which is not quite right...) – Arturo Magidin Apr 05 '19 at 03:22
  • @ArturoMagidin So, how do you proof the other part? or at least what is a simpler statemet? Because I find the reciprocal statement hard to follow. – José Marín Apr 05 '19 at 03:33
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    Suppose $C$ is a subgroup of $G$ such that (i) $s\in C$; and (ii) for all subgroups $B$ of $G$, if $s\in B$ then $C\subseteq B$. We want to prove that $B=\langle s\rangle$. From (i) you can deduce that $\langle s\rangle\subseteq B$, because $\langle s\rangle$ has the property that it is contained in any subgroup that contains $s$. From (ii) and the fact that $\langle s\rangle$ is a subgroup that contains $s$ you can conclude that $B\subseteq \langle s\rangle$. Thus, we get that $B=\langle s\rangle$, as desired. – Arturo Magidin Apr 05 '19 at 03:36
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    See also this answer for general comments about this sort of thing. – Arturo Magidin Apr 05 '19 at 03:38
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    By the way: for anyone who does math regularly, $S$ and $s$ will automatically be assumed to be different objects. So when you write “let $S\subseteq A$ then $\langle s\rangle$”, in the back of my mind I understood $s$ to be something different from $S$. You need to be more careful writing. Just like you can’t use $B$ to represent two different things in the same sentence. – Arturo Magidin Apr 05 '19 at 03:43
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    @ArturoMagidin That was very useful. Thank you for all your help. I need you as a professor. – José Marín Apr 05 '19 at 03:58
  • @ArturoMagidin In your proof, I think you mean: we want to prove that $C=⟨s⟩$ right? Because $C$ in that context be the smallest subgroup that contain s. – José Marín Apr 05 '19 at 04:20
  • Yeah, few letters went the wrong way, and I can’t edit a comment after 5 minutes. Sorry about that. – Arturo Magidin Apr 05 '19 at 04:25
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    @ArturoMagidin Put on the answers so I can put the check on it. – José Marín Apr 05 '19 at 04:36
  • To show that $\langle S\rangle $ is the smallest subgroup of $A$ containing $S$, you just have to show that 1) $\langle S\rangle$ indeed contains $S$, and 2) if $B$ is a subgroup of $A$ such that $B$ contains $S$ (i.e. $S\subseteq B$), then $\langle S\rangle\subseteq B$. To prove 2), note that $B\in C$ (where $C$ is defined in your post), and use the general set property that if $B\in C$, then $\bigcap_{B'\in C} B'\subseteq B$. (Make sure you can prove this set property! In words, this property basically says "the intersection of a bunch of sets is a subset of each set in the intersection".) – Minus One-Twelfth Apr 05 '19 at 11:09
  • @MinusOne-Twelfth I just did that. I was having problem with the reciprocal. – José Marín Apr 06 '19 at 23:31

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