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We were told to give a proof for the following statement:

Let G be a group, where $ r \in G $, and suppose $ r \in G \setminus \langle s \rangle $. If $ \langle r \rangle \cap \langle s \rangle = \{e\}$ and $ |r|=k $, then $ \langle s \rangle $, $ r\langle s \rangle $, $r^2\langle s \rangle$,...,$r^{k-1}\langle s \rangle$ are distinct left cosets.

I was wondering if my proof below is a proper proof, and if there is perhaps a more simple, intuitive proof of this fact. Thanks.

$ Proof $. Let $ i=1,2,...k $, and suppose towards a contradiction that $ r^{i-1}\langle s \rangle = r^i\langle s \rangle$ for some $ i $. This implies that $ r^{i-1}s^t = r^is^u $, for some integers $ u $ and $ t $. This in turn implies that $ r^{-1} = s^{u-t} $. This is a contradiction as $ \{r,r^2,...,r^{k-1} \} \notin \langle s \rangle$, from $ \langle r \rangle \cap \langle s \rangle = \{e\}$. Therefore $ r^{i-1}\langle s \rangle \neq r^i\langle s \rangle$, or rather $ \langle s \rangle $, $ r\langle s \rangle $, $r^2\langle s \rangle$,...,$r^{k-1}\langle s \rangle$ are distinct left cosets.

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One more remark - your book has probably proved a theorem in generality about $gH = kH \Leftrightarrow k^{-1}g \in H$. You are basically reproving that theorem in this special case; if this is a homework problem, it might be cleaner just to quote it. – user29743 Nov 2 '12 at 16:31
up vote 2 down vote accepted

There is a minor problem. You have shown that $r^{i-1}\langle s \rangle \neq r^i \langle s \rangle$, but this does not imply that $r^{i}\langle s \rangle \neq r^j \langle s \rangle$ for arbitrary $i, j$. So you should start from this assumption, and then deduce that $r^{i - j} \in \langle s \rangle$, which forces $i = j$.

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Sorry, I'm a little confused. So you are saying as opposed to making the assumption that $ r^{i-1}\langle s \rangle = r^i\langle s \rangle$, I should make the assumption that $ r^j\langle s \rangle = r^i\langle s \rangle$? – mkeachie Nov 2 '12 at 16:45
That's right. Your proof only shows that two "neighboring" cosets can't be equal, and you need to show two arbitrary ones can't be. – user29743 Nov 2 '12 at 17:28
So would this answer be correct? – mkeachie Nov 2 '12 at 20:02
yes, I like the new answer. – user29743 Nov 2 '12 at 20:33
thank you very much. – mkeachie Nov 2 '12 at 20:44

Suppose that that $ r^i \langle s \rangle = r^j \langle s \rangle $, for where $ i $, $ j =0,1,2,...,k-1 $. This implies that $ r^is^t = r^js^u $, for some integers $u $ and $ t $, which in turn implies $ r^{i-j} = s^{u-t} $. Thus $ r^{i-j} \in \langle s \rangle$. From $ \langle r \rangle \cap \langle s \rangle = \{e\}$, we wee that $r^{i-j}=e$. So $k$ divides $i-j$. As $i-j \lt k $, $i-j=0$ or rather $i=j$. This proves the claim.

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