How can I show that every cyclic group is isomorphic to $Z/nZ$, using the First Isomorphism Theorem? This theorem is pretty hard for me to get a grasp on, considering I just learned it.
Also, what happens when $n=0$?
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Widawensen
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3Let $C$ be a cyclic group, and $g$ a generator of $C$. Consider $f \colon \mathbb{Z} \to C; f(n) = g^n$. – Daniel Fischer Sep 14 '13 at 23:09
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@PVAL If it's an infinite group, then take $n = 0$. – Sep 14 '13 at 23:10
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If you have a finite cyclic group $G$ with generator $a$, let $n$ be the least positive integer for which $na = 0$. Now consider a map $\Bbb Z \to G$ defined by $k \to ka$.
To check that it's onto $G$, note that the generator $a$ lies in the image of the map, since it's the image of $1$.
To check that the kernel is $n\Bbb{Z}$, note that the kernel is a cyclic subgroup of $\Bbb{Z}$ containing $n$. If $0 < k < n$ lies in the kernel, this will contradict the choice of $n$ being the smallest positive integer killing $a$.
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Is the homomorphism f(x) = g^(x mod n), where n is the order of the group, because it can't just be g^x because you don't necessarily show that G is cyclic, you have to be able to get back to g^0? – Sep 15 '13 at 02:00