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Denote $R=\langle6\rangle$ and $S=\langle30\rangle$.
Consider the subgroups $R$ and $S$ 0f the set of integers $\mathbb{Z}$.
Show that $R/S \cong \mathbb{Z}_5$.
$(R/S$ is the set of left cosets of $S$ over $R$, $\mathbb{Z}_5$ denotes congruence class $\pmod 5.)$

Define a map $f:R/S \to \mathbb{Z}_5$ by $f(6m+30\mathbb{Z})=(m+5\mathbb{Z})$.

I have shown the following:

  1. $f$ is well-defined.
  2. $f$ is a homomorphism.
  3. $f$ is surjective or onto.

Now, to show that $f$ is injective, I assume that $$f(6m+30\mathbb{Z})\equiv f(6n+30\mathbb{Z}) \pmod 5$$ My question is: will it imply that $m \equiv n \pmod 5$?
If yes, please show me how.

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Unless I am misunderstanding, you are assuming that $$(*)\,\,\,\,\,f(6m+30\mathbb{Z}) = f(6n+30\mathbb{Z})$$ in $\mathbb{Z}/5\mathbb{Z}$. By definition, $f(6m + 30\mathbb{Z}) = m + 5\mathbb{Z}$ and $f(6n+30\mathbb{Z}) = n + 5\mathbb{Z}$, so assumption ($*$) says exactly that $m + 5\mathbb{Z} = n + 5\mathbb{Z}$, or that $m\equiv n\pmod{5}$.

But that doesn't show that $f$ is injective. To show that $f$ is injective, you must show that $6m + 30\mathbb{Z} = 6n + 30\mathbb{Z}$ in $R/S$. Indeed, since $m\equiv n\pmod{5}$, there is some integer $k$ such that $n = m + 5k$. Thus $6n = 6m + 30k$, so that $6m+30\mathbb{Z} = 6n + 30\mathbb{Z}$.

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  • $\begingroup$ You have explained it very well sir @froggie . $\endgroup$ – Philip Benj Dec 17 '12 at 15:16

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