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The following quote is from A Course in Modern Mathematical Physics by Peter Szekeres:

The factor group $\mathbb{Z}/2\mathbb{Z}$ has just two cosets $[0]=0+2\mathbb{Z}$ and $[1]=1+2\mathbb{Z}$, and is isomorphic to the additive group of integers modulo 2 denoted by $\mathbb{Z}_{2}$.

The groups $\mathbb{Z }/2\mathbb{Z}$ and $\mathbb{Z }_{2}$ seem to be the same to me and hence are isomorphic but how can I show this formally?

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    $\begingroup$ As far as I'm concerned, the "group of integers modulo 2" is precisely defined as the factor group $\mathbb Z/2\mathbb Z$. $\endgroup$ – Dustan Levenstein Mar 2 '14 at 17:06
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Hint: Show that the function $f$ defined by $f([0]) = \overline 0$ and $f([1]) = \overline 1$ is an isomorphism from $\Bbb{Z}/2\Bbb{Z}$ to $\Bbb Z_2$.

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  • $\begingroup$ What does the "conjugation hat" means in this case? $\endgroup$ – Mark A. Ruiz Mar 2 '14 at 17:01
  • $\begingroup$ Just that the element is a member of the integers modulo $2$. So in general, $\mathbb{Z}_n = \{\overline 0, \overline 1, \dots , \overline {n-1}\}$. This notation helps distinguish between $[1]$ and $\overline 1$. $\endgroup$ – Omnomnomnom Mar 2 '14 at 17:03
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Hint $\ $ A ring $R$ generated by $1$ with characteristic $m$ is isomorphic to $\,\Bbb Z/m\Bbb Z,\,$ by applying the First Isomorphism Theorem to the natural image of $\,\Bbb Z\,$ in $R.$

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I think the Omnom's post gives us the idea, however; Bill's extended that. This below Cayley table makes the first post idea visualized:

enter image description here

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  • $\begingroup$ Very nice table, B.S.! $\endgroup$ – Namaste Mar 3 '14 at 14:02

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