# $S$ is a subgroup of index $f$. Show that $fxS=S$ where $fx$ is $x^{f}$ and $fxS$ the left coset.

I'm having a bit of trouble understanding the second half of the proof presented here:

In particular, why is it that $f x S=S$ ? I do not understand why the index of $S$ justifies this.

Thank you!

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If I'm reading the linked proof correctly, I think there is a typo in that the expression $fxS$ should be $fx + S$. Similarly, $f(a+b\omega)S$ should be $f(a + b\omega) + S$, and so on, in that paragraph. Making this change, and carrying on as shown in the blog, the argument there seems to make a bit more sense.
Since the index of $S$ is equal to $f$, therefore, $fx + S = f(x + S) = S$, for any $x$. (Think of these as elements of the additive quotient group $\mathbb{Z}[\omega]/S$, which has order $f$.) –  James Jan 15 '13 at 19:16
The expression $fx + S$ means $x + x + \cdots + x + S$, where the $x$ appears $f$ times (remember $f$ is a positive integer). This can be rewritten as $(x+S) + (x+S) + \cdots + (x+S)$, with $f$ summands, so this is $f(x+S)$. But $f$ times any coset of $S$ is equal to $S$, since $S$ is the zero of the additive quotient group $\mathbb{Z}[\omega]/S$, and the order of any element of this quotient group divides the order ($f$) of the group. Does this help? –  James Jan 16 '13 at 1:08