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Let $S = \{a_{1}, a_{2}, ... a_{n} \}$ be a nonempty subset of $G$, which is a finite group. Also let $S$ be closed under multiplication. If $a_{i} \in S$, then we consider the distinct elements $a_{i}a_{1},...,a_{i}a_{n} \in S$. Because each product is distinct, $a_{i}a_{1} \neq a_{i}a_{k}$. But since I do not know $S$ is a subgroup yet, I cannot assume $S$ contains an inverse element for every $a_{i} \in S$. My question is can I assume $a_{1} \neq a_{k}$ even though I do not know that $a_{i}^{-1} \in S$?

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Yes, since the inverses still exist in G. –  Joe Aug 17 '11 at 16:42
    
The answer to the question you should have asked has been given by @Matt. You want to show that for any $i$, $j$, $k$, if $a_ia_j=a_ia_k$, then $a_j=a_k$. In other words, you want to prove that for fixed $i$, the objects $a_ib$, where $b$ ranges over $S$, are distinct. –  André Nicolas Aug 17 '11 at 16:51

1 Answer 1

Drawing on the comments:

You are assuming that $a_i a_1\neq a_i a_k$. This is a statement about two elements of the group $G$ (whether they are equal does not depend on the choice of subset $S$ at this point). Since $G$ is a group, $a_i$ has an inverse in $G$. Multiplying by this gives that $a_1\neq a_k$ as elements of $G$, as required. However, this still doesn't tell you things about $S$. You don't yet know that $a_i^{-1}\in S$.

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