# Isomorphism betweeen $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}_{2}$

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?

• As far as I'm concerned, the "group of integers modulo 2" is precisely defined as the factor group $\mathbb Z/2\mathbb Z$. – Dustan Levenstein Mar 2 '14 at 17:06

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 $\Bbb Z_2$.
• 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 $\overline 1$. – Omnomnomnom Mar 2 '14 at 17:03
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 $R.$