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Question- Let $G$ be a group and $S < G$ (proper subgroup) then prove that $G$\ $S$ generates $G$.

TRY- I only have to prove that $S$ $\subset$ <$G$ \ $S$>. So to the contrary let $s \in S$ but not in <$G$ \ $S$> be an element , so .........

where do I go from here? I was also considering action of G or S on set of cosets of S in G but that was not fruitful either.

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    $\begingroup$ @Kaj Hansen, if you take in your example $(1,1,0,0,....)$ and $(0,1,0,....)$ then they are not in $S$ but their product is. $\endgroup$ – Clément Guérin Mar 31 '15 at 7:37
  • $\begingroup$ Ah! You are quite right @ClémentGuérin. I'll delete my comment. $\endgroup$ – Kaj Hansen Mar 31 '15 at 7:38
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By assumption, there is some $g ∈ G\setminus S$. Then $g^{-1} ∈ G\setminus S$ as well.

And for any $s ∈ S$, you also have $sg ∈ G\setminus S$, so just write $s = sg·g^{-1} ∈ (G \setminus S) · (G\setminus S)$.

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I think I would begin with $G$ finite-case. Set $H:=<G\setminus S>$. Well its cardinal is at least $\#((G\setminus S)\cup \{1_G\})=\#G-\#S+1$, because of Lagrange's theorem and because $S$ is a proper subgroup, you have $\#S\leq \#G/2$. Finally :

$$\#H\geq \frac{\#G}{2}+1>\frac{\#G}{2}$$

But $H$ must also verifies Lagrange's theorem so $H$ cannot but be $G$ (because of its cardinal).

This proves it in the finite case. Well maybe it is a little too sophisticate because the following proof works very well, the cardinal of $G$ being finite or infinite... Set $H:=<G\setminus S>$.

Take $s\in S$ and $h\notin S$ then $hs\notin S$ (otherwise $h$ would be in $S$) so you have :

$$h\in H \text{ and } hs\in H\text{ thus } s=h^{-1}(hs)\in H $$

You have that $S\subseteq H=<G\setminus S>$.

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    $\begingroup$ @Derek Holt. According to (what I've understood) about the philosophy of this site and also with what I've seen with my students, I prefer to give my whole reasoning rather (as it came to me at first) than a too short good answer... I think this may be "pedagogical", that's the point about it. But k.stm's answer is undoubtly much better answer than mine (I upvoted it). $\endgroup$ – Clément Guérin Mar 31 '15 at 8:06

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