# Proof of condition for distinct left cosets.

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

## 2 Answers

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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