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I think similar questions may have been asked, so apologies if this is the case. I haven't been able to find one that helps me yet though.

Anyway, for an assignment question I've figured out a solution but it relies on my assumption that an isomorphism between cyclic groups maps generators to generators. I've seen this stated before but I'm not quite sure how to prove it.

Could anyone help me? I attempted using the idea that the identity gets mapped to the identity and trying to generalise to generators but got stuck.

Thanks in advance!

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    $\begingroup$ This is certainly true for any isomorphism, but for general homomorphisms it need not be the case (consider the 'trivial homomorphism' $\phi :G\rightarrow H$ where the kernel of $\phi$ is all of $G$). $\endgroup$ – Justin Benfield Aug 25 '16 at 1:16
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If $g$ generates $G$, then the elements of $G$ are precisely $g,g^2,\ldots,g^n$ where $n$ is the order of $g$.

If $\phi : G \to G'$ is an isomorphism, then consider where an arbitrary element of $G$ is sent. An arbitrary element of $G$ is of the form $g^k$, so $\phi(g^k)=\phi(g)^k$. Since $\phi$ is a bijection, the elements of $G'$ are precisely $\phi(g),\phi(g)^2,\ldots,\phi(g)^n$.

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  • $\begingroup$ Ahh, I should've seen that thank-you very much! $\endgroup$ – CAPM Aug 25 '16 at 1:19
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No. Consider the trivial homomorphism $\phi: Z_n\to Z_m$ that takes $g\in Z_n$ to $e\in Z_m$. $e$ isn't a generator of $Z_m$, but $\phi(g_1g_2)=e$ and $\phi(g_1)\phi(g_2)=e^2=e$, so $\phi$ is a homomorphism.

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