I need to show that $ \mathbb{Z}_7 \cong S $, where $S = \{0,2,4,6,8,10,12 \}$ using the First Isomorphism Theorem. So I've gathered that I must find some ring $R$ and a ring homomorphism $\phi$ such that $R/ \ker \phi = \mathbb{Z}_7$, however I don't really see how this is possible. Finding $\phi$ is easy enough I thought, I just used $\phi(a) = 2a\bmod14 $ and this appears to work. Since $\ker\phi = (7)$, I'm struggling to find an $R$ such that $R/ (7) = \mathbb{Z}_7$, as it doesn't seem clear to me how $\mathbb{Z}_7$ could be a quotient group.
Note that $S$ is a commutative ring with addition and multiplication modulo 14, with multiplicative identity 8.
Best Answer
Since you are looking at this as an isomorphism of rings (given the tags), and in $\mathbb{Z}_7$ there is only one nonzero element whose square equals itself (namely $1$), you want to identify some element in $2\mathbb{Z}/14\mathbb{Z}$ whose square equals itself. That will indicate where to try to map $1\in\mathbb{Z}$ to get the desired morphism.
A little bit of trial and error will show that $8^2\equiv 8\pmod{14}$, and no other element of $S$ has a square equal to itself. Which you already noted.
So we want to try to define the morphism $\phi$ by mapping $1\in\mathbb{Z}$ to $8\in\mathbb{Z}_{14}$. This completely determines the value of the morphism on the underlying abelian groups, since $\mathbb{Z}$ is cyclic. The kernel is also easy to verify as being $7\mathbb{Z}$. So this gives an isomorphism of abelian groups between $\mathbb{Z}/\ker(\phi)=\mathbb{Z}/7\mathbb{Z}$ and $S=\langle 8\rangle\subseteq \mathbb{Z}_{14}$.
So the only thing that remains is to verify that this map is not just an additive map, but in fact a multiplicative map (that is, it is a ring homomorphism). So you need to verify that if $a,b \in\mathbb{Z}$, then $\phi(ab)=\phi(a)\phi(b)$. Since $a\mapsto 8a\bmod 14$ and $b\mapsto 8b\bmod 14$, you need to verify that $(8a)(8b)\equiv 8ab\pmod{14}$.
If that holds, then this will establish what you want, using the First Isomorphism Theorem.